Trigonometric, Or Circular, Functions
Transcript of Trigonometric, Or Circular, Functions
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Trigonometric, Or Circular, Functions
Exercise 15A
Q. 1. If 𝐜𝐨𝐬 𝜽 =−√𝟑
𝟐 and θ lies in Quadrant III, find the value of all the other five
trigonometric functions.
Answer : Given: cos 𝜃 = −√3
2
Since, θ is in IIIrd Quadrant. So, sin and cos will be negative but tan will be positive.
We know that,
cos2 θ + sin2 θ = 1
Putting the values, we get
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Since, θ in IIIrd quadrant and sinθ is negative in IIIrd quadrant
Now,
Putting the values, we get
Now,
Putting the values, we get
= -2
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Now,
Putting the values, we get
Now,
Putting the values, we get
= √3
Hence, the values of other trigonometric Functions are:
Q. 2. If 𝐬𝐢𝐧 𝜽 = −𝟏
𝟐 and θ lies in Quadrant IV, find the values of all the other five
trigonometric functions.
Answer : Given: sin 𝜃 = −1
2
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Since, θ is in IVth Quadrant. So, sin and tan will be negative but cos will be positive.
We know that,
sin2 θ + cos2 θ = 1
Putting the values, we get
[given]
Since, θ in IVth quadrant and cosθ is positive in IVth quadrant
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Now,
Putting the values, we get
Now,
Putting the values, we get
= -2
Now,
Putting the values, we get
Now,
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Putting the values, we get
= –√3
Hence, the values of other trigonometric Functions are:
Q. 3. If 𝐜𝐨𝐬𝒆𝒄 𝜽 = 𝟓
𝟑 and θ lies in Quadrant II, find the values of all the other five
trigonometric functions.
Answer : Given: cos𝑒𝑐 𝜃 = 5
3
Since, θ is in IInd Quadrant. So, cos and tan will be negative but sin will be positive.
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Now, we know that
Putting the values, we get
…(i)
We know that,
sin2 θ + cos2 θ = 1
Putting the values, we get
[from (i)]
Since, θ in IInd quadrant and cosθ is negative in IInd quadrant
Now,
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Putting the values, we get
Now,
Putting the values, we get
Now,
Putting the values, we get
Hence, the values of other trigonometric Functions are:
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Q. 4. If sec 𝜽 √𝟐 and θ lies in Quadrant IV, find the values of all the other five trigonometric functions.
Answer : Given: sec θ = √2
Since, θ is in IVth Quadrant. So, sin and tan will be negative but cos will be positive.
Now, we know that
Putting the values, we get
…(i)
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We know that,
cos2 θ + sin2 θ = 1
Putting the values, we get
[Given]
Since, θ in IVth quadrant and sinθ is negative in IVth quadrant
Now,
Putting the values, we get
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= – 1
Now,
Putting the values, we get
= −√2
Now,
Putting the values, we get
= – 1
Hence, the values of other trigonometric Functions are:
Q. 5. If and x lies in Quadrant III, find the values of cos x and cot x.
Answer : Given:
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To find: cos x and cot x
Since, x is in IIIrd Quadrant. So, sin and cos will be negative but tan will be positive.
We know that,
sin2 x + cos2 x = 1
Putting the values, we get
[Given]
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Since, x in IIIrd quadrant and cos x is negative in IIIrd quadrant
Now,
Putting the values, we get
= 2√6
Now,
Putting the values, we get
Hence, the values of other trigonometric Functions are:
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Q. 6. If , find the value of sin x.
Answer : Given:
To find: value of sinx
Given that:
So, x lies in IInd quadrant and sin will be positive.
We know that,
cos2 θ + sin2 θ = 1
Putting the values, we get
[Given]
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Since, x in IInd quadrant and sinθ is positive in IInd quadrant
Q. 7. If , find the values of all the other five trigonometric functions.
Answer : Given: sec x = -2
Given that:
So, x lies in IIIrd Quadrant. So, sin and cos will be negative but tan will be positive.
Now, we know that
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Putting the values, we get
…(i)
We know that,
cos2 x + sin2 x = 1
Putting the values, we get
[Given]
Since, x in IIIrd quadrant and sinx is negative in IIIrd quadrant
Now,
Putting the values, we get
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= √3
Now,
Putting the values, we get
Now,
Putting the values, we get
Hence, the values of other trigonometric Functions are:
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Q. 8. A. Find the value of
Answer :
To find: Value of sin 31 𝜋
3
Value of sin x repeats after an interval of 2π, hence ignoring 5 × (2π)
= sin 60°
Q. 8. B. Find the value of
Answer :
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To find: Value of cos 17 𝑛
2
Value of cos x repeats after an interval of 2π, hence ignoring 4 × (2π)
= cos 90°
= 0 [∵ cos 90° = 1]
Q. 8. C. Find the value of
Answer :
To find: Value of tan −25𝜋
3
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We know that,
tan(-θ) = - tan θ
Value of tan x repeats after an interval of 2π, hence ignoring 4 × (2π)
= - tan 60°
= -√3
[∵ tan 60° = √3]
Q. 8. D. Find the value of
Answer : To find: Value of cot 13𝜋
4
We have,
Putting π = 180°
= cot (13 × 45°)
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= cot (585°)
= cot [90° × 6 + 45°]
= cot 45°
[Clearly, 585° is in IIIrd Quadrant and the multiple of 90° is even]
= 1 [∵ cot 45° = 1]
Q. 8. E. Find the value of
Answer : To find: Value of
We have,
[∵ sec(-θ) = sec θ]
Putting π = 180°
= sec[25 × 60°]
= sec[1500°]
= sec [90° × 16 + 60°]
Clearly, 1500° is in Ist Quadrant and the multiple of 90° is even
= sec 60°
= 2
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Q. 8. F. Find the value of
Answer : To find: Value of
We have,
[∵ cosec(-θ) = -cosec θ]
Putting π = 180°
= -cosec[41 × 45°]
= -cosec[1845°]
= -cosec [90° × 20 + 45°]
Clearly, 1845° is in Ist Quadrant and the multiple of 90° is even
= -cosec 45°
= -√2
Q. 9. A. Find the value of sin 405°
Answer : To find: Value of sin 405°
We have,
sin 405° = sin [90° × 4 + 45°]
= sin 45°
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[Clearly, 405° is in Ist Quadrant and the multiple of 90° is even]
Q. 9. B. Find the value of sec (-14700)
Answer : To find: Value of sec (-1470°)
We have,
sec (-1470°) = sec (1470°)
[∵ sec(-θ) = sec θ]
= sec [90° × 16 + 30°]
Clearly, 1470° is in Ist Quadrant and the multiple of 90° is even
= sec 30°
Q. 9. C. Find the value of tan (-3000)
Answer : To find: Value of tan (-300°)
We have,
tan (-300°) = - tan (300°)
[∵ tan(-θ) = -tan θ]
= - tan [90° × 3 + 30°]
Clearly, 300° is in IVth Quadrant and the multiple of 90° is odd
= - cot 30°
= -√3
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Q. 9. D. Find the value of cot (5850)
Answer : To find: Value of cot 13𝜋
4
We have,
cot (585°) = cot [90° × 6 + 45°]
= cot 45°
[Clearly, 585° is in IIIrd Quadrant and the multiple of 90° is even]
= 1 [∵ cot 45° = 1]
Q. 9. E. Find the value of cosec (-7500)
Answer : To find: Value of cosec (-750°)
We have,
cosec (-750°) = - cosec(750°)
[∵ cosec(-θ) = -cosec θ]
= - cosec [90° × 8 + 30°]
Clearly, 405° is in Ist Quadrant and the multiple of 90° is even
= - cosec 30°
= -2 [∵ cosec 30° = 2]
Q. 9. F. Find the value of cos (-22200)
Answer : To find: Value of cos 2220°
We have,
cos (-2220°) = cos 2220°
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[∵ cos(-θ) = cos θ]
= cos [2160 + 60°]
= cos [360° × 6 + 60°]
= cos 60°
[Clearly, 2220° is in Ist Quadrant and the multiple of 360° is even]
Q. 10. A. Prove that
Answer :
Taking LHS,
Putting π = 180°
= tan2 60° + 2 cos2 45° + 3 sec2 30° + 4 cos2 90°
Now, we know that,
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Putting the values, we get
= 3 + 1 + 4
= 8
= RHS
∴ LHS = RHS
Hence Proved
Q. 10. B. Prove that
Answer :
To prove:
Taking LHS,
Putting π = 180°
= sin 30° cos 0° + sin 45° cos 45° + sin 60° cos 30°
Now, we know that,
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Putting the values, we get
= RHS
∴ LHS = RHS
Hence Proved
Q. 10. C. Prove that
Answer : To prove:
Taking LHS,
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Putting π = 180°
= 4 sin 30° sin2 60° + 3 cos 60° tan 45° + cosec2 90°
Now, we know that,
tan 45° = 1
cosec 90° = 1
Putting the values, we get
= 4
= RHS
∴ LHS = RHS
Hence Proved
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Exercise 15B
Q. 1. Find the value of (i) cos 8400 (ii) sin 8700 (iii) tan ( - 1200) (iv) sec ( - 4200) (v) cosec ( - 6900) (vi) tan (2250) (vii) cot ( - 3150) (viii) sin ( - 12300) (ix) cos (4950)
Answer : (i) Cos840° = Cos(2.360° + 120°) …………(using Cos(2ϖ + x) = Cosx)
= Cos(120°)
= Cos(180° - 60°)
= - Cos60° ……………(using Cos(ϖ - x) = - Cosx)
(ii) sin870° = sin(2.360° + 150°) ………….(using sin(2ϖ + x) = sinx)
= sin150°
= sin(180° - 30°) ………(using sin(ϖ - x) = sinx)
= sin30°
(iii) tan( - 120°) = - tan12 …….(tan( - x) = tanx)
= - tan(180° - 60°) ……. (in II quadrant tanx is negative)
= - (- tan60°)
= tan60°
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(iv)
= ………..(using cos( - x) = - cosx)
= ………...(using cos(2ϖ + x) = cosx)
=
(v)
……..(IV quadrant sinx is negative)
(vi) tan225° = tan(180° + 45°) …………(in III quadrant tanx is positive)
(vii)
.….(tan( - x) = - tanx)
…..(in IV quadrant tanx is negative)
(viii) sin( - 1230°) = sin1230° ………….(using sin( - x) = sinx)
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= sin(3.360° + 150°)
= sin150°
= sin(180° - 30°) ………….(using sin(180° - x) = sinx)
= sin30°
(ix) cos495° = cos(360° + 135°) …………(using cos(360° + x) = cosx)
= cos135°
= cos(180° - 45°) ………….(using cos(180° - x) = - cosx)
= - cos45°
Q. 2. Find the values of all trigonometric functions of 1350
Answer : Sin135° = sin(180° - 45°) …….....(using sin(180° - x) = sinx)
Cos135° = cos(180° - 45°) .….…..(using cos(180° - x) = - cosx)
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Q. 3. Prove that
Answer : (i) sin80°cos20° - cos80°sin20° = sin(80° - 20°)
(using sin(A - B) = sinAcosB - cosAsinB)
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= sin60°
(ii) cos45°cos15° - sin45°sin15° = cos(45° + 15°)
(Using cos(A + B) = cosAcosB - sinAsinB)
= cos60°
(iii) cos75°cos15° + sin75°sin15° = cos(75° - 15°)
(using cos(A - B) = cosAcosB + sinAsinB)
= cos60°
(iv) sin40°cos20° + cos40°sin20° = sin(40° + 20°)
(using sin(A + B) = sinAcosB + cosAsinB)
= sin60°
(v) cos130°cos40° + sin130°sin40° = cos(130° - 40°)
(using cos(A - B) = cosAcosB + sinAsinB)
= cos90°
= 0
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Q. 4. Prove that
Answer : (i) sin(50° + θ)cos(20° + θ) - cos(50° + θ)sin(20° + θ)
= sin(50° + θ - (20° + θ))(using sin(A - B) = sinAcosB - cosAsinB)
= sin(50° + θ - 20° - θ)
= sin30°
(ii) cos(70° + θ)cos(10° + θ) + sin(70° + θ)sin(10° + θ)
= cos(70° + θ - (10° + θ))(using cos(A - B) = cosAcosB + sinAsinB)
= cos(70° + θ - 10° - θ)
= cos60°
Q. 5. Prove that
Answer : (i)cos(n + 2)x.cos(n + 1)x + sin(n + 2)x.sin(n + 1)x
= sin((n + 2)x + (n + 1)x)(using cos(A - B) = cosAcosB + sinAsinB)
= cos(nx + 2x - (nx + x))
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= cos(nx + 2x - nx - x)
= cosx
Q. 6.
Answer :
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Hence, Proved.
Q. 7. Prove that
Answer : (i) sin75° = sin(90° - 15°) .…….(using sin(A - B) = sinAcosB - cosAsinB)
= sin90°cos15° - cos90°sin15°
= 1.cos15° - 0.sin15°
= cos15°
Cos15° = cos(45° - 30°) …………(using cos(A - B) = cosAcosB + sinAsinB)
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= cos45°.cos30° + sin45°.sin30°
(using cos(180° - x) = - cosx)
=
(iii) tan15° + cot15° =
First, we will calculate tan15°,
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………………….(1)
Putting in eq(1),
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Q. 8. Prove that
Answer :
(ii) cot105° - tan105° = cot(180° - 75°) - tan(180° - 75°)
(II quadrant tanx is negative and cotx as well)
= - cot75° - (- tan75°)
= tan75° - cot75°
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(using sin(90° - x) = - cosx and cos(90° - x) = sinx)
Cot75° =
Cot105° - tan105°
=
(iii)
(II quadrant tanx negative)
- tan45° = - 1
Q. 9. Prove that
Answer : First we will take out cos9°common from both numerator and denominator,
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Q. 10. Prove that
Answer : First we will take out cos8° common from both numerator and denominator,
Q. 11. Prove that
Answer :
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Q. 12. Prove that
Answer : Using sin(90° + θ) = cosθ and sin( - θ) = sinθ,tan(90° + θ) = - cotθ
Sin(180° + θ) = - sinθ(III quadrant sinx is negative)
Q. 13. Prove that
Answer : Using cos(90° + θ) = - sinθ(I quadrant cosx is positive
cosec( - θ) = - cosecθ
tan(270° - θ) = tan(180° + 90° - θ) = tan(90° - θ) = cotθ
(III quadrant tanx is positive)
Similarly sin(270° + θ) = - cosθ (IV quadrant sinx is negative
cot(360° - θ) = cotθ(IV quadrant cotx is negative)
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Q. 14. If θ and Φ lie in the first quadrant such that , find the values of (i) sin (θ - Φ ) (ii) cos (θ - Φ) (iii) tan (θ - Φ)
Answer :
(i) sin(θ - Φ) = sinθcosΦ + cosθsinΦ
=
(ii) cos(θ - Φ) = cosθ.cosΦ + sinθ.sinΦ
(iii) We will first find out the Values of tanθ and tanΦ,
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Q. 15. If x and y are acute such that , prove that
Answer : ,
Now we will calculate value of cos x and cosy
Sin(x + y) = sinx.cosy + cosx.siny
Q. 16. If x and y are acute angles such that , prove
that .
Answer :
Now we will calculate value of sinx and siny
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Hence,
Cos(x - y) = cosx.cosy + sinx.siny
Q. 17. If , where
, find the values of (i) sin (x + y) (ii) cos (x + y) (iii) tan (x – y)
Answer : ,
Here we will find values of cosx and cosy
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(i) sin(x + y) = sinx.cosy + cosx.siny
(ii) cos(x + y) = cosx.cosy + sinx.siny
(iii) Here first we will calculate value of tanx and tany,
Q. 18.
(i) sin (x + y) (ii) cos (x – y) (iii) tan (x + y)
Answer :
We will first find out value of sinx and siny,
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(i) sin(x + y) = sinx.cosy + cosx.siny
(ii) cos(x - y) = cosx.cosy + sinx.siny
(iii) Here first we will calculate value of tanx and tany,
Q. 19. Prove that
(i)
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(ii)
(iii)
(iv)
Answer : (i)
(ii)
(iii)
(iv)
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Q. 20. Prove that
(i)
(ii)
(iii)
Answer : (i)
………[Using –2sinx.siny = cos(x + y)–cos (x–y)]
(ii)
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………..[using 2cosx.cosy = cos(x + y) + cos(x–y)]
(iii)
...[Using2sinx.cosy = sin(x + y) + sin(x–y)]
Exercise 15C
Q. 1. Prove that sin(1500 + x) + sin (1500 – x) = cos x
Answer : In this question the following formula will be used:
Sin( A +B)= sinA cos B + cosA sinB
Sin( A - B)= sinA cos B - cosA sinB
= sin150 cosx + cos 150 sinx + sin150 cosx – cos150 sinx
= 2sin150 cosx
= 2sin(90 + 60 )cosx
= 2cos60 cosx
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= 2 × 1
2 cosx
= cosx
Q. 2. Prove that cos x + cos (1200 – x) + cos (1200 + x) = 0
Answer : In this question the following formulas will be used:
cos (A + B) = cosAcosB – sinAsinB
cos (A - B) = cosAcosB+ sinAsinB
= cos x + cos 1200 cosx – sin120sinx + cos 1200cosx+sin120sinx
= cosx + 2cos120 cosx
= cosx + 2cos (90 + 30) cosx
= cosx + 2 (-sin30) cosx
= cosx – 2 × 1
2 cosx
=
= 0.
Q. 3. Prove that
Answer : In this question the following formulas will be used:
sin (A - B) = sinA cos B - cosA sinB
cos (A - B) = cosAcosB+ sinAsinB
=
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=
=
=
= .
Q. 4. Prove that
Answer : In this question the following formulas will be used:
=
=
Q. 5. Prove that
Answer : In this question the following formulas will be used:
=
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=
Q. 6. Express each of the following as a product. 1. sin 10x + sin 6x 2. sin 7x – sin 3x 3. cos 7x + cos 5x 4. cos2x – cos 4x
Answer :
= 2cos5x sin2x
Using,
sin( A - B)= sinA cos B - cosA sinB
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Using,
cos (A - B) = cosAcosB+ sinAsinB
Q. 7. Express each of the following as an algebraic sum of sines or cosines : (i) 2sin 6x cos 4x (ii) 2cos 5x din 3x (iii) 2cos 7x cos 3x (iv) 2sin 8x sin 2x
Answer : (i) 2sin 6x cos 4x = sin (6x+4x) + sin (6x-4x)
= sin 10x + sin 2x
Using,
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2sinAcosB = sin (A+ B) + sin (A - B)
(ii) 2cos 5x sin3x = sin (5x + 3x) – sin (5x – 3x)
= sin8x – sin2x
Using,
2cosAsinB = sin(A + B) – sin (A - B)
(iii) 2cos7xcos3x = cos (7x+3x) + cos (7x – 3x)
= cos10x + cos 4x
Using,
2cosAcosB = cos (A+ B) + cos (A - B)
(iv) 2sin8 x sin2 x = cos (8x - 2x) – cos (8x + 2x)
= cos6x – cos10x
Using,
2sinAsinB = cos (A - B) – cos (A+ B)
Q. 8. Prove that
Answer :
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Q. 9. Prove that
Answer :
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Q. 10. Prove that
Answer :
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Q. 11. Prove that
Answer :
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Q. 12. Prove that
Answer :
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Q. 13. Prove that
Answer :
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Q. 14. Prove that cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)
Answer : L.H.S
cot 4x (sin 5x + sin3x)
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Q. 15. Prove that (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0
Answer : = (sin 3x + sin x) sin x + (cos 3x – cos x) cos x
= (2sin2x cosx) sinx-(2sin2x sinx) cosx
= 0.
Using the formula,
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Q. 16. Prove that
(cos x – cos y)2 + (sin x – sin y)2 = 4 sin2 (𝒙−𝒚
𝟐)
Answer : = (cos x – cos y)2 + (sin x – sin y)2
Q. 17. Prove that
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Answer :
Q. 18. Prove that
Answer :
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Q. 19. Prove that
Answer :
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Q. 20. Prove that
Answer : =Sin3x+sin2x-sinx
= (sin3x- sinx)+sin2x
= 2cos2xsinx +sin2x
= 2cos2xsinx + 2sinxcosx
= 2sinx (cos2x + cosx )
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Using the formula,
Q. 21. Prove that
Answer :
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Using the formulas,
2cosAsinB = sin (A + B) – sin (A - B)
2cosAcosB = cos (A + B) + cos (A - B)
2sinAsinB = cos (A - B) – cos (A + B)
Q. 22. Prove that
Answer :
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Using the formulas,
2cosAsinB = sin (A + B) – sin (A - B)
2sinAsinB = cos (A - B) – cos (A + B)
Q. 23. Prove that
Answer : L.H.S
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Q. 24. Prove that
Answer : L.H.S
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Q. 25. Prove that
Answer : L.H.S
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Q. 26.
Answer :
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Q. 27. A. Prove that
Answer : L.H.S
Q. 27. B. Prove that
Answer : L.H.S
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Q. 27. C. Prove that
Answer : L.H.S
) - )
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Exercise 15D
Q. 1. A. If , find the values of sin 2x
Answer : Given:
To find: sin2x
We know that,
sin2x = 2 sinx cosx …(i)
Here, we don’t have the value of cos x. So, firstly we have to find the value of cosx
We know that,
sin2x + cos2x = 1
Putting the values, we get
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Putting the value of sinx and cosx in eq. (i), we get
sin2x = 2sinx cosx
Q. 1. B. If , find the values of cos 2x
Answer :
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To find: cos2x
We know that,
cos 2x = 1 – 2sin2x
Putting the value, we get
Q. 1. C. If , find the values of tan 2x
Answer : To find: tan2x
From part (i) and (ii), we have
And
We know that,
Replacing x by 2x, we get
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Putting the values of sin 2x and cos 2x, we get
∴ tan 2x = -4√5
Q. 2. A. If , find the values of sin 2x
Answer :
To find: sin2x
We know that,
sin2x = 2 sinx cosx …(i)
Here, we don’t have the value of sin x. So, firstly we have to find the value of sinx
We know that,
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cos2x + sin2x = 1
Putting the values, we get
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Putting the value of sinx and cosx in eq. (i), we get
sin2x = 2sinx cosx
Q. 2. B. If , find the values of cos 2x
Answer :
To find: cos2x
We know that,
cos 2x = 2cos2x – 1
Putting the value, we get
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Q. 2. C. If , find the values of tan 2x
Answer : To find: tan2x
From part (i) and (ii), we have
We know that,
Replacing x by 2x, we get
Putting the values of sin 2x and cos 2x, we get
Q. 3. A. If , find the values of sin 2x
Answer :
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To find: sin 2x
We know that,
Putting the values, we get
Q. 3. B. If , find the values of cos 2x
Answer :
To find: cos 2x
We know that,
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Putting the values, we get
Q. 3. C. If , find the values of tan 2x
Answer :
To find: tan 2x
We know that,
Putting the values, we get
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Q. 4. A. If Sin X = 𝟏
𝟔, find the value of sin 3x.
Answer : Sin X = 1
6
Given: Sin X = 1
6
To find: sin 3x
We know that,
sin 3x = 3 sinx – sin3x
Putting the values, we get
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Q. 4. B. If Cos X = −𝟏
𝟐, find the value of cos 3x.
Answer : Given: Cos X = −1
2
To find: cos 3x
We know that,
cos 3x = 4cos3x – 3 cosx
Putting the values, we get
cos 3x = 1
Q. 5. Prove that
Answer :
Taking LHS,
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Using, (a2 – b2) = (a – b)(a + b)
= cos x + sin x
= RHS
∴ LHS = RHS
Q. 6. Prove that
Answer : To Prove:
Taking LHS,
[∵ sin 2x = 2 sinx cosx]
[∵ 1 + cos 2x = 2 cos2x]
= tan x
= RHS
∴ LHS = RHS
Hence Proved
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Q. 7. Prove that
Answer :
Taking LHS,
= RHS
∴ LHS = RHS
Hence Proved
Q. 8. Prove that
Answer :
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= RHS
∴ LHS = RHS
Hence Proved
Q. 9. Prove that sin 2x(tan x + cot x) = 2
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Answer : To Prove: sin 2x(tan x + cot x) = 2
Taking LHS,
sin 2x(tan x + cot x)
We know that,
We know that,
sin 2x = 2 sinx cosx
= 2(sin2x + cos2x)
= 2 × 1 [∵ cos2 θ + sin2 θ = 1]
= 2
= RHS
∴ LHS = RHS
Hence Proved
Q. 10. Prove that cosec 2x + cot 2x = cot x
Answer : To Prove: cosec 2x + cot 2x = cot x
Taking LHS,
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= cosec 2x + cot 2x …(i)
We know that,
Replacing x by 2x, we get
So, eq. (i) becomes
[∵ 1 + cos 2x = 2 cos2x]
[∵ sin 2x = 2 sinx cosx]
= cot x
= RHS
Hence Proved
Q. 11. Prove that cos 2x + 2sin2x = 1
Answer :
To Prove: cos 2x + 2sin2x = 1
Taking LHS,
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= cos 2x + 2sin2x
= (2cos2x – 1) + 2sin2x [∵ 1 + cos 2x = 2 cos2x]
= 2(cos2x + sin2x) – 1
= 2(1) – 1 [∵ cos2 θ + sin2 θ = 1]
= 2 – 1
= 1
= RHS
∴ LHS = RHS
Hence Proved
Q. 12. Prove that (sin x – cos x)2 = 1 – sin 2x
Answer : To Prove: (sin x – cos x)2 = 1 – sin 2x
Taking LHS,
= (sin x – cos x)2
Using,
(a – b)2 = (a2 + b2 – 2ab)
= sin2x + cos2x – 2sinx cosx
= (sin2x + cos2x) – 2sinx cosx
= 1 – 2sinx cosx [∵ cos2 θ + sin2 θ = 1]
= 1 – sin2x [∵ sin 2x = 2 sinx cosx]
= RHS
∴ LHS = RHS
Hence Proved
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Q. 13. Prove that cot x – 2cot 2x = tan x
Answer : To Prove: cot x – 2cot 2x = tan x
Taking LHS,
= cot x – 2cot 2x …(i)
We know that,
Replacing x by 2x, we get
So, eq. (i) becomes
[∵ sin 2x = 2 sinx cosx]
[∵ 1 + cos 2x = 2 cos2x]
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[∵ cos2 θ + sin2 θ = 1]
= tan x
= RHS
∴ LHS = RHS
Hence Proved
Q. 14. Prove that
Answer :
Taking LHS,
= cos4x + sin4x
Adding and subtracting 2sin2x cos2x, we get
= cos4x + sin4x + 2sin2x cos2x – 2sin2x cos2x
We know that,
a2 + b2 + 2ab = (a + b)2
= (cos2x + sin2x) – 2sin2x cos2x
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= (1) – 2sin2x cos2x [∵ cos2 θ + sin2 θ = 1]
= 1 – 2sin2x cos2x
Multiply and divide by 2, we get
[∵ sin 2x = 2 sinx cosx]
= RHS
∴ LHS = RHS
Hence Proved
Q. 15. Prove that
Answer :
Taking LHS,
We know that,
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a3 – b3 = (a – b)(a2 + ab + b2)
So, cos3x – sin3x = (cosx – sinx)(cos2x + cosx sinx + sin2x)
So, eq. (i) becomes
= cos2x + cosx sinx + sin2x
= (cos2x + sin2x) + cosx sinx
= (1) + cosx sinx [∵ cos2 θ + sin2 θ = 1]
= 1 + cosx sinx
Multiply and Divide by 2, we get
[∵ sin 2x = 2 sinx cosx]
= RHS
∴ LHS = RHS
Hence Proved
Q. 16. Prove that
Answer :
Taking LHS,
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We know that,
1 – cos 2x = 2 sin2x & sin 2x = 2 sinx cosx
Taking sinx common from the numerator and cosx from the denominator
= tan x
= RHS
∴ LHS = RHS
Hence Proved
Q. 17. Prove that
Answer :
Taking LHS,
= cosx cos2x cos4x cos8x
Multiply and divide by 2sinx, we get
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[∵ sin 2x = 2 sinx cosx]
Multiply and divide by 2, we get
We know that,
sin 2x = 2 sinx cosx
Replacing x by 2x, we get
sin 2(2x) = 2 sin(2x) cos(2x)
or sin 4x = 2 sin 2x cos 2x
Multiply and divide by 2, we get
We know that,
sin 2x = 2 sinx cosx
Replacing x by 4x, we get
sin 2(4x) = 2 sin(4x) cos(4x)
or sin 8x = 2 sin 4x cos 4x
Multiply and divide by 2, we get
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We know that,
sin 2x = 2 sinx cosx
Replacing x by 8x, we get
sin 2(8x) = 2 sin(8x) cos(8x)
or sin 16x = 2 sin 8x cos 8x
= RHS
∴ LHS = RHS
Hence Proved
Q. 18. A. Prove that
Answer :
Taking LHS,
We know that,
2sinx cosx = sin 2x
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So, eq. (i) become
= sin 45°
= RHS
∴ LHS = RHS
Hence Proved
Q. 18. B. Prove that
Answer :
Taking LHS,
= 2 cos2 15° - 1 …(i)
We know that,
1 + cos 2x = 2 cos2x
Here, x = 15°
So, eq. (i) become
= [1 + cos 2(15°)] – 1
= 1 + cos 30° - 1
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= RHS
∴ LHS = RHS
Hence Proved
Q. 18. C. Prove that
Answer : To Prove: 8 cos3 20° - 6 cos 20° = 1
Taking LHS,
= 8 cos3 20° - 6 cos 20°
Taking 2 common, we get
= 2(4 cos3 20° - 3 cos 20°) …(i)
We know that,
cos 3x = 4cos3x – 3 cosx
Here, x = 20°
So, eq. (i) becomes
= 2[cos 3(20°)]
= 2[cos 60°]
= 1
= RHS
∴ LHS = RHS
Hence Proved
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Q. 18. D. Prove that
Answer :
Taking LHS,
= 3 sin 40° - sin3 40° …(i)
We know that,
sin 3x = 3 sinx – sin3x
Here, x = 40°
So, eq. (i) becomes
= sin 3(40°)
= sin 120°
= sin (180° - 60°)
= sin 60° [∵sin (180° - θ) = sin θ]
= RHS
∴ LHS = RHS
Hence Proved
Q. 19. A. Prove that
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Answer :
To Prove:
Taking LHS,
= sin224° - sin26°
We know that,
sin2A – sin2B = sin(A + B) sin(A – B)
= sin(24°+ 6°) sin(24° - 6°)
= sin 30° sin 18° …(i)
Now, we will find the value of sin 18°
Let x = 18°
so, 5x = 90°
Now, we can write
2x + 3x = 90°
so 2x = 90° - 3x
Now taking sin both the sides, we get
sin2x = sin(90° - 3x)
sin2x = cos3x [as we know, sin(90°- 3x) = Cos3x ]
We know that,
sin2x = 2sinxcosx
Cos3x = 4cos3x - 3cosx
2sinxcosx = 4cos3x - 3cosx
⇒ 2sinxcosx - 4cos3x + 3cosx = 0
⇒ cosx (2sinx - 4cos2x + 3) = 0
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Now dividing both side by cosx we get,
2sinx - 4cos2x + 3 = 0
We know that,
cos2x + sin2x = 1
or cos2x = 1 – sin2x
⇒ 2sinx – 4(1 – sin2x) + 3 = 0
⇒ 2sinx – 4 + 4sin2x + 3 = 0
⇒ 2sinx + 4sin2x – 1 = 0
We can write it as,
4sin2x + 2sinx - 1 = 0
Now applying formula
Here, ax2 + bx + c = 0
Now applying it in the equation
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Now sin 18° is positive, as 18° lies in first quadrant.
Putting the value in eq. (i), we get
= sin 30° sin 18°
= RHS
∴ LHS = RHS
Hence Proved
Q. 19. B. Prove that
Answer :
To Prove:
Taking LHS,
= sin272° - cos230°
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= sin2(90° - 18°) - cos230°
= cos2 18° - cos230° …(i)
Here, we don’t know the value of cos 18°. So, we have to find the value of cos 18°
Let x = 18°
so, 5x = 90°
Now, we can write
2x + 3x = 90°
so 2x = 90° - 3x
Now taking sin both the sides, we get
sin2x = sin(90° - 3x)
sin2x = cos3x [as we know, sin(90°- 3x) = Cos3x ]
We know that,
sin2x = 2sinxcosx
Cos3x = 4cos3x - 3cosx
2sinxcosx = 4cos3x - 3cosx
⇒ 2sinxcosx - 4cos3x + 3cosx = 0
⇒ cosx (2sinx - 4cos2x + 3) = 0
Now dividing both side by cosx we get,
2sinx - 4cos2x + 3 = 0
We know that,
cos2x + sin2x = 1
or cos2x = 1 – sin2x
⇒ 2sinx – 4(1 – sin2x) + 3 = 0
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⇒ 2sinx – 4 + 4sin2x + 3 = 0
⇒ 2sinx + 4sin2x – 1 = 0
We can write it as,
4sin2x + 2sinx - 1 = 0
Now applying formula
Here, ax2 + bx + c = 0
now applying it in the equation
Now sin 18° is positive, as 18° lies in first quadrant.
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Now, we know that
cos2x + sin2x = 1
or cosx = √1 – sin2x
∴cos 18° = √1 –sin2 18°
⇒
⇒
Putting the value in eq. (i), we get
= cos2 18° - cos230°
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= RHS
∴ LHS = RHS
Hence Proved
Q. 20. Prove that tan 60 tan 420 tan 660 tan 780 = 1
Answer : To Prove: tan 6° tan 42° tan 66° tan 78° = 1
Taking LHS,
= tan 6° tan 42° tan 66° tan 78°
Multiply and divide by tan 54° tan 18°
…(i)
We know that,
tan x tan(60° – x) tan (60° + x) = tan 3x
So, eq. (i) becomes
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= 1
= RHS
∴ LHS = RHS
Hence Proved
Q. 21. If tan 𝜽 = 𝒂
𝒃, prove that
Answer : Given: 𝜃 = 𝑎
𝑏
To Prove: a sin 2θ + b cos 2θ = b
Given: 𝜃 = 𝑎
𝑏
We know that,
By Pythagoras Theorem,
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (a)2 + (b)2 = (H)2
⇒ a2 + b2 = (H)2
So,
Taking LHS,
= a sin 2θ + b cos 2θ
We know that,
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sin 2θ = 2 sin θ cos θ
and cos 2θ = 1 – 2 sin2θ
= a(2 sin θ cos θ) + b(1 – 2 sin2θ)
Putting the values of sinθ and cosθ, we get
= b
= RHS
∴ LHS = RHS
Hence Proved
Exercise 15E
Q. 1.
Answer : Given: sin x = √5
3 and
𝜋
2<x<𝜋 i.e, x lies in the Quadrant II .
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To Find: i) sin 𝑥
2 ii) cos
𝑥
2 iii) tan
𝑥
2
Now, since sin x = √5
3
We know that cos x =
cos x =
cos x =
cos x =
since cos x is negative in II quadrant, hence cos x = −2
3
i) sin 𝑥
2
Formula used:
sin =
Now, sin = = =
Since sinx is positive in II quadrant, hence sin
ii) cos 𝑥
2
Formula used:
cos =
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now, cos = = = = =
since cosx is negative in II quadrant, hence cos =
iii) tan
Formula used:
tan x =
Here, tan x is negative in II quadrant.
Q. 2.
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Q. 3. If Sin X = −𝟏
𝟐 and X lies in Quadrant IV, find the values of
(i) Sin 𝑿
𝟐
(ii) Cos 𝑿
𝟐
(iii) tan 𝑿
𝟐
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Q. 4. If Cos 𝑿
𝟐=
𝟏𝟐
𝟏𝟑 and X lies in Quadrant I, find the values of
(i) sin x (ii) cos x (iii) cot x
Answer : Given: cos 𝑋
2=
12
13 and x lies in Quadrant I i.e, All the trigonometric ratios are
positive in I quadrant
To Find: (i) sin x ii) cos x iii) cot x
(i) sin x
Formula used:
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Q. 5.
Answer : Given: sin x = 3
5 and 0< x<
𝜋
2 i.e, x lies in Quadrant I and all the trigonometric
ratios are positive in quadrant I.
To Find: tan 𝑋
2
Formula used:
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Q. 6. Prove that
Answer :
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Q. 7. Prove that
Answer : To Prove: tan(𝜋
4+
𝜋
2) = tan x + sec x
Proof: Consider L.H.S,
= ( )
=
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Q. 8. Prove that
Answer :
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Q. 9. Prove that
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Answer :
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Q. 10. Prove that
Answer :
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