WEIGHTAGE FOR CLASS ---XI

One Paper       Three Hours       Max Marks. 100

Units MarksI. SETS AND FUNCTIONS

II. ALGEBRA

III. COORDINATE GEOMETRY

IV. CALCULUS

V. MATHEMATICAL REASONING

VI. STATISTICS AND PROBABILITY

29

37

13

06

03

12  100

Fundamental Trigonometric Identities

Before we start to prove trigonometric identities, we see where the basic identities come from.

Recall the definitions of the reciprocal trigonometric functions, csc θ, sec θ and cot θ

from the trigonometric functions chapter:

After we revise the fundamental identities, we learn about: Proving trigonometric identities

Now, consider the following diagram where the point (x, y) defines an angle θ at the origin,

and the distance from the origin to the point is r units:

From the diagram, we can see that the ratios sin θ and cos θ are defined as:

and

Now, we use these results to find an important definition for tan θ:

Now, also so we can conclude that:

Also, for the values in the diagram, we can use Pythagoras' Theorem and obtain:

y2 + x2 = r2

Dividing through by r2 gives us:

so we obtain the important result:

sin2 θ + cos2 θ = 1

We now proceed to derive two other related formulas that can be used when proving trigonometric identities.

It is suggested that you remember how to find the identities, rather than try to memorise each one.

Dividing sin2 θ + cos2 θ = 1 through by cos2 θ gives us:

so

tan2 θ + 1 = sec2 θ

Dividing sin2 θ + cos2 θ = 1 through by sin2 θ gives us:

so

1 + cot2 θ = csc2 θ

Trigonometric Identities Summary

Proving Trigonometric Identities

Suggestions...

1. Learn well the formulas given above (or at least, know how to find them quickly).

The better you know the basic identities, the easier it will be to recognise what is going on in the problems.

2. Work on the most complex side and simplify it so that it has the same form as the simplest side.

Don't assume the identity to prove the identity.

This means don't work on both sides of the equals side and try to meet in the middle.

3. Start on one side and make it look like the other side.

4. Many of these come out quite easily if you express everything on the most complex side in terms of sine and cosine only.

5. In most examples where you see power 2 (that is, 2 ), it will involve using the identity sin2 θ + cos2 θ = 1 (or one of the other 2 formulas that we derived above).

Using these suggestions, you can simplify and prove expressions involving trigonometric identities.

Prove that

sin y + sin y cot2 y = cosec y

Answer

FunctionAbbreviation

Description

Identities (using radians)

Sine sin

Cosine cos

Tangent tan (or tg)

Cotangent

cot (or ctg or ctn)

Secant sec

Cosecantcsc (or cosec)

Note that these values can easily be memorized in the form

but the angles are not equally spaced.

The values for 15°, 54° and 75° are slightly more complicated. [ by using formulas of sin(A-B),sin(A+B)

Similarly for cosine function & tan function.]

Special values in trigonometric functions There are some commonly used special values in

trigonometric functions, as shown in the following table.

Function

sin

0 1

cos 1 0

tan 0 1

cot 1 0

sec 1 2

csc 21

The sine and cosine functions graphed on the Cartesian plane.

For angles greater than 2π or less than −2π, simply continue to rotate

around the circle; sine and cosine are periodic functions with period 2π:

for any angle θ and any integer k.

The smallest positive period of a periodic function is called the

primitive period of the function.

The primitive period of the sine or cosine is a full circle, i.e.

2π radians or 360 degrees.

Figure 1

If Q(x,y) is the point on the circle where the string ends,

we may think of as being an angle by associating to it the central angle with vertex O(0,0) and sides passing through the points P and Q. If instead of wrapping a length s of string around the unit circle, we decide to wrap it around a circle of radius R, the angle (in radians) generated in the process will satisfy the following relation:

 

Observe that the length s of string gives the measure of the angle only when R=1.

As a matter of common practice and convenience, it is useful to measure angles in degrees, which are defined by partitioning one whole revolution into 360 equal parts, each of which is then called one degree. In this way, one whole revolution around the unit circle measures radians and also 360 degrees (or ), that is:

 

Each degree may be further subdivided into 60 parts, called minutes, and in turn each minute may be subdivided into another 60 parts, called seconds:

 

Angle sum identities

Sine

Illustration of the sum formula.

Draw the angles α and β. Place P on the line defined by α + β at unit distance from the origin.

Let PQ be a perpendicular from P to the line defined by the angle α. OQP is a right angle.

Let QA be a perpendicular from Q to the x axis, and PB be a perpendicular from P to the x axis. OAQ is a right angle.

Draw QR parallel to the x-axis. Now angle RPQ = α (because

or PQ/OP = Sin𝛃 or OQ/OP = Cos𝛃 (if OP ≠ 1)

, so

, so

Or = PBOP = PR+RB

OP = PROP . PQ

PQ + AQOP . OQ

OQ

(∵ RB = AQ , PQ & OQ are the hyp. )

= PRPQ . PQ

OP + AQOQ . OQ

OP .

By substituting − β for β and using Symmetry, we also get:

Cosine

Using the figure above,

OR PQ/OP = Sin𝛃 OR OQ/OP = Cos𝛃

, so

, so

Or = OBOP = OA−BA

OP = OAOP

. OQOQ - RQ

OP . PQPQ

(AB=RQ)

= OAOQ .OQ

OP - RQPQ . PQ

OP .

By substituting − β for β and using Symmetry, we also get:

Also, using the complementary angle formulae,

**Another simple "proof" can be given using Euler's formula known from complex analysis: Euler's formula is:

Although it is more precise to say that Euler's formula entails the trigonometric identities, it follows that for angles α and β we have:

ei(α + β) = cos(α + β) + isin(α + β)

Also using the following properties of exponential functions:

ei(α + β) = eiαeiβ = (cosα + isinα)(cosβ + isinβ)

Evaluating the product:

ei(α + β) = (cosαcosβ − sinαsinβ) + i(sinαcosβ + sinβcosα)

This will only be equal to the previous expression we got, if the imaginary and real parts are equal respectively. Hence we get:

cos(α + β) = cosαcosβ − sinαsinβ

sin(α + β) = sinαcosβ + sinβcosα ]

Tangent and cotangent

From the sine and cosine formulae, we get

Dividing both numerator and denominator by cos α cos β, we get

Similarly (using a division by sin α sin β), we get

Double-angle identities

From the angle sum identities, we get

and

The Pythagorean identities give the two alternative forms for the latter of these:

The angle sum identities also give

**It can also be proved using Eulers Formula

Mulitplying the exponent by two yields

But replacing the angle with its doubled version, which achieves the same result in the left side of the equation, yields

It follows that

By multiplying we get

Because the imaginary and real parts have to be the same, we are left with the original identities

Half-angle identities

The two identities giving alternative forms for cos 2θ give these:

One must choose the sign of the square root properly—note that if 2π is added to θ the quantities inside the square roots are unchanged, but the left-hand-sides of the equations change sign. Therefore the correct sign to use depends on the value of θ.

tan function, we have

If we multiply the numerator and denominator inside the square root by (1 + cos θ), and do a little manipulation using the Pythagorean identities, we get

If instead we multiply the numerator and denominator by (1 - cos θ), we get

This also gives

Similar manipulations for the cot function give

Example. verify the identity

Answer. We have

which gives

But

and since

and , we get finally

Remark. In general it is good to check whether the given formula is correct. One way to do that is to substitute some numbers for the variables. For example, if we take a=b = 0, we get

or we may take . In this case we have

Example. Find the exact value of

Answer. We have

Hence, using the additions formulas for the cosine function we get

Since

we get

Example. Find the exact value for

Answer. We have

Since

we get

Finally we have

Remark. Using the addition formulas, we generate the following identities

Double-Angle and Half-Angle formulas are very useful. For example, rational functions of sine and cosine wil be very hard to integrate without these formulas. They are as follow

Example. Check the identities

Answer. We will check the first one. the second one is left to the reader as an exercise. We have

Hence

which implies

Many functions involving powers of sine and cosine are hard to integrate. The use of Double-Angle formulas help reduce the degree of difficulty.

Example. Write as an expression involving the trigonometric functions with their first power.

Answer. We have

Hence

Since , we get

or

Example: Verify the identity

Answer. We have

Using the Double-Angle formulas we get

Putting stuff together we get

From the Double-Angle formulas, one may generate easily the Half-Angle formulas

In particular, we have

Example. Use the Half-Angle formulas to find

Answer. Set . Then

Using the above formulas, we get

Since , then is a positive number. Therefore, we have

Same arguments lead to

Example. Check the identities

Answer. First note that

which falls from the identity . So we need to verify only one identity. For example, let us verify that

using the Half-Angle formulas, we get

which reduces to

Product and Sum Formulas

From the Addition Formulas, we derive the following trigonometric formulas (or identities)

Remark. It is clear that the third formula and the fourth

are identical (use the property to see it).

The above formulas are important whenever need rises to transform the product of sine and cosine into a sum. This is a very useful idea in techniques of integration.

Example. Express the product as a sum of trigonometric functions.

Answer. We have

which gives

Note that the above formulas may be used to transform a sum into a product via the identities

Example. Express as a product.

Answer. We have

Note that we used .

Example. Verify the formula

Answer. We have

and

Hence

which clearly implies

Example. Find the real number x such that

and

Answer. Many ways may be used to tackle this problem. Let us use the above formulas. We have

Hence

Since , the equation gives

and the equation gives . Therefore, the solutions to the equation

are

Example. Verify the identity

Answer. We have

Using the above formulas we get

Hence

which implies

Since , we get

TRIGONOMETRIC EQUATIONS

Example :        Solve for x in the following equation.

There are an infinite number of solutions to this problem. To solve for x, you must first isolate the tangent term.

= tan(±π

6 ) General solution:

X = nπ ± π6 ∀ n Є Z(integers)

Example :        Solve for x in the following equation.

There are an infinite number of solutions to this problem. To solve for x, set the equation equal to zero and factor.

then

when

when , and when

when and

This is impossible because

The exact value solutions are and

Example :        Solve for x in the following equation.

There are an infinite number of solutions to this problem.

Isolate the sine term. To do this, rewrite the left side of the equation in an equivalent factored form.

The product of two factors equals zero if at least one of the factors equals zeros. This means that

if or

We just transformed a difficult problem into two easier problems. To find the solutions to the original equation,

, we find the

solutions to the equations OR

Sinx = sin(-π/6) x = nπ + (-1)n (π/6) ∀ nЄ Z

OR

sinx = sin(π/2) x = nπ + (-1)n (π/2) ∀ nЄ Z.

Example :        Solve for x in the following equation. (general solution)

There are an infinite number of solutions to this problem.

Cos(3x-1) = cos(π/2) 3x-1 = 2nπ ± π/2 3x = 2nπ ± π/2 +1 x = 1/3(2nπ ± π/2 +1) ∀ n Є Z.

1. Solve the trigonometric equation analytically

4 tan x − sec2 x = 0 (for 0 ≤ x < 2π)

Answer

4 tan x − sec2 x = 0

 

In 0 ≤ x < 2π, we need to find values of 2x such that 0 ≤ 2x < 4π.

So

So

or x = 0.2618, 1.309, 3.403, 4.451

2. Solve the trigonometric equation analytically for 0 ≤ x < 2π:

sin 2x cos x − cos 2x sin x = 0

Answer

We recognise the left hand side to be in the form:

sin(a − b) = sin a cos b − cos a sin b,

where a = 2x and b = x.

So

sin 2x cos x − cos 2x sin x

= sin(2x − x)

= sin x

Now, we know the solutions of sin x = 0 to be:

x = 0, π.

3. Solve the given trigonometric equation analytically and by graphical method (for 0 ≤ x < 2π):

sin 4x − cos 2x = 0

Answer

sin 4x − cos 2x = 0

2sin 2x cos 2x − cos 2x = 0

cos2x (2sin 2x - 1) = 0

EITHER

cos 2x = 0

OR

sin 2x = 1/2

Question Solve the equation tan 2θ − cot 2θ = 0 for 0 ≤ θ < 2π.

Answer

tan2 2θ = 1

tan 2θ = ± 1

 

Since 0 ≤ θ < 2π , we need to consider values of 2θ such that 0 ≤ 2θ < 4π. Hence, solving the above equation, we have:

So

Question Solve the equation for 0 ≤ θ < 2π.

Answer

By using the half angle formula for and then squaring both sides, we get:

= 1 + cos x

So we have:

2 cos2 x + 3 cos x + 1 = 0

(2 cos x + 1)(cos x + 1) = 0

Solving, we get

cos x = − 0.5 or cos x = − 1

Now gives .

However, on checking in the original equation, we note that

but

So the only solution for this part is

Also, cos x = − 1 gives x = π.

So the solutions for the equation are

GENERAL SOLUTIONS OF TRIGONOMETRIC EQUATIONS

1. If sinx = 0 ⇨ x = nπ, ∀ n є Z

2. If cosx = 0 ⇨ x = (2n+1) π/2, ∀ n Є Z

3. If tanx = 0 ⇨ x = nπ, n Є Z

4. If sinx = six ⇨ x = nπ + (-1)n𝛂, ∀ n Є Z

5. If cosx = cos ⇨ x = 2nπ ±𝛂, ∀ n Є Z

6. If tanx = tan ⇨ x = nπ+𝛂, ∀ n Є Z.

EXAMPLE: Solve the equation: sin3θ + cos2θ = 0

SOLUTION: cos2θ = - sin3θ ⇨ cos2θ = cos (π2 +3θ)

⇨ 2θ = 2nπ± (π2

+3θ) , ∀ nЄZ

-θ = nπ+ π2

and 5θ = nπ- π2 , ∀ nєz.

Question –1 If cos(A+B)=4/5 , sin(A-B)=5/13 and A,B lie between 0 and π/4 , prove that

tan 2A = 56/33.

Answer: Since A-B Є (-π/4 , π/4) and A+B Є (0,π/2) both are positive sin(A+B)=3/5 , cos(A-B)=12/13

tan(A+B)=3/4 , tan(A-B) = 5/12 then

tan2A = tan(A+B+A-B) =

34+ 5

12

1−34

×5

12

=56/33

Question – 2 If cos (A-B)+cos(B-C)+cos(C-A) = -3/2 , Prove that

CosA+cosB+cosC = sinA+sinB+sinC = 0

Answer : From given result we get

2cosAcosB+2sinAsinB+2cosBcosC+2sinBsinC+2cosAcosC+2sinAsinC +3 =0

2cosAcosB+2sinAsinB+2cosBcosC+2sinBsinC+2cosAcosC+2sinAsinC + sin2A+cos2A+sin2B+cos2B

+sin2C+cos2C = 0. ( 3=1+1+1 and 1 can be written as sin2x+cos2x)

(sin2A+sin2B+sin2C+ 2sinAsinB+2sinBsinC+2sinAsinC)

+(cos2A+cos2B+cos2C +2cosAcosB+2cosBcosC+2cosAcosC) = 0

(sinA+sinB+sinC)2+ (cosA+cosB+cosC)2 = 0.

Question – 3 Prove that (1+cos π8 ) (1+ cos

3 π8 ) (1+

cos 5 π8 ) (1+ cos

7 π8 ) =

18 .

Answer : cos 7 π8 = cos( π -

π8 ) = - cos

π8 , cos

5 π8

= cos ( π - 3 π8 ) = - cos

3 π8

L.H.S. (1+cos π8 ) (1+ cos

3 π8 ) (1+ cos

5 π8

) (1+ cos 7 π8 )

= (1 – cos2 π8 ) (1 – cos2

3 π8 ) = sin2 π

8

. sin2 3 π8

= 14 (2sin2 π

8 ). (2sin2 3 π8 ) =

14 (1 -

cos π4 ) ( 1 - cos

3 π4 ) =

18 .

Question – 4 Prove that cos 2 π15 cos

4 π15 cos

8 π15 cos

14 π15 =

116

.

Answer : L.H.S. ⇨ - cos 2 π15 cos

4 π15 cos

8 π15 cos

π15

[∵cos 14 π15 = cos(π -

π15 ) ]

= - [ sin 16 A

24 SinA ] , where A= π15

[∵ all angels are in G.P. , short-cut Method]

= - [sin 16

π15

24 sinπ

15

] = - sin(π+ π

15)

16 sinπ

15

=

sinπ

15

16sinπ

15

= 116

. [ ∵ sin (π + π15

) lies in 3rd quadrant]

OR

L.H.S. ⇨ - 1

2sinπ15

(2 sin π15 cos

π15 ) cos

2 π15

cos 4 π15 cos

8 π15

= - 1

2× 2sinπ

15 ( 2sin

2 π15 cos

2 π15 )cos

4 π15

cos 8 π15

= - 1

2× 2× 2sinπ

15 ( 2sin

4 π15 cos

4 π15 )cos

8 π15

= - 1

2× 2× 2× 2 sinπ

15 (2 sin

8 π15 .cos

8 π15 )

, now you can apply above result.

Similarly you can prove cosAcos2Acos4Acos8A = sin16A/16sinA.

Question – 5 (i) Prove that sin200 sin400 sin600 sin800 = 3

16

Answer : L.H.S. sin200 sin400 sin600 sin800

(√3/2¿

⇨ √ 32× 2 (2sin200 sin400 sin800)

⇨ √ 34 [(cos200 – cos600) sin800 ]

⇨ √ 34 [cos200 . sin800– cos600. sin800 ]

(1/2)

⇨ √ 38 [(sin1000+sin600 – sin800 ] = 3

16 ( ∵ sin1000 lies in 2nd quadrant) [sin(1800 - 800)] = sin800

(ii) Prove that: cosπ7 cos

2 π7 cos

4 π7 = -

18 .

[Hint: let x = π7 , then

12 sinx (2sinx cosx cos2x cos4x)]

Sin2x (iii) Prove that: tan200 tan400 tan800 = tan600.[hint: L.H.S. sin 200sin 400sin 800

cos200 cos 400 cos800 solve as above method.] Question – 6 Solve : 2sinx + √3 cosx = 1+ sinx

Answer : sinx + √3 cosx = 1

√ 32 cosx +

12 sinx =

12 [ dividing by

√a2+b2 ,where a = √3 and b = 1]

cosπ6 . cosx + sin

π6 .sinx = cos

π3

⇨ cos(x - π6 ) = cos

π3 ⇨ x - π

6 = 2nπ ±π3 ∀ n Z (integers)Є

⇨ x = 2nπ ±π3 + π

6 = 2nπ + π2 , 2nπ - π

6 ∀ n Z Є(integers).Question – 7 Solve: 3cos2x - 2 √3 sinx .cosx – 3sin2x = 0. Answer: 3 cos2x - 3 √3 sinx .cosx +√3 sinx .cosx – 3sin2x = 0 ⇨ 3cosx ( cosx - √3 sinx) + √3 sinx (cosx - √3 sinx) = 0 ⇨ (3cosx + √3 sinx) (cosx - √3 sinx) = 0 ⇨ (3cosx + √3 sinx) = 0 or (cosx - √3 sinx) = 0

⇨ tanx = - √3 = tan(-π3 ) or tanx = 1

√ 3 =tan(π6 )

⇨ x = nπ+(- π3 ) ∀ n Z Є(integers) or x = nπ + π

6

Question – 8 If 𝛂, are the acute angles and cos2𝛂 = 3 cos2 β−13−cos2 β , show that tan 𝛂 =

√ 2 tan𝛃. Answer: According to required result , we have to convert given part into tangent function By using cos2𝛂 = 1−tan ² α

1+tan ² α ∴ we will get 1−tan ² α

1+tan ² α = 3( 1−tan ² β

1+tan ² β)−1

3−1−tan ² β1+ tan ² β

= 3−3 tan ² β−1− tan ² β3+3 tan ² β−1+ tan ² β

= 1−2 tan ² β1+2 tan ² β

By C & D 1− tan ² α+1+ tan ² α

(1−tan ² α)−(1+ta n2 α) = 1−2 tan ² β+1+2 tan ² β

(1−2 tan ² β)−(1+2 tan2 β) ⇨ 2−2 tan ² α =

2−4 tan ² β

2

⇨ tan ² α = 2 tan ² β ⇨ tan 𝛂 = √ 2 tan𝛃.

Question – 9 Prove that sinA−sin 5 A+sin 9 A−sin 13 A

cosA−cos 5 A−cos 9 A+cos13 A = cot4A.

Answer : (sin 9 A+sinA )−¿¿ = 2 sin5 Acos4 A−2 sin 9 Acos 4 A

−[−2sin 5 Asin4 A ]+[−2 sin 9 Asin 4 A]

[∵ sinA+sinB = 2sin(A+B)/2. cos(A-B)/2 & cosA-cosB = - 2sin(A+B)/2.sin(A-B)/2 ]

= 2cos 4 A (sin 5 A−sin 9 A)2 sin 4 A(sin 5 A−sin 9 A)

= cot 4A.

Question – 10 Prove that √2+√2+√2+2 cos 8θ = 2COSθ. Answer: √2+√2+√2(1+cos8 θ) = √2+√2+√2×2cos ²4 θ

√2+√2(1+cos 4 θ) = √2+√2× 2cos ²2θ = √2(1+cos2θ) = 2cosθ. [By using 1+cos2 θ= 2cos2θ] Question – 11 Find the general solution of the following equation:

4sinxcosx+2sinx+2cosx+1 = 0

Answer: Above equation can be written as (4sinxcosx+2sinx) + (2cosx+1 ) = 0

⇨ 2sinx (2cosx+1) + (2cosx+1) = 0

⇨ (2sinx+1) (2cosx+1) = 0

⇨ (2sinx+1) = 0 or 2cosx+1 = 0

⇨ sinx = -1/2 or cosx = -1/2

⇨ sinx = sin(π + π6 )

or cosx = cos (π - π3 )

⇨ x = nπ +(-1)n 7 π6

or x = 2nπ ± 2 π3 , ∀ n Є Z (Integers).

Question – 12 If cosx = -1213 and π <x <3 π

2 , find the

value of sin3x and cos3x.

Answer: Since x lies in 3rd quadrant ∴ sinx is negative.

Sinx = – √1−cos ² x = - 513 then sin3x = 3sinx

– sin3x = -20352197

Cos3x = 4cos3x – 3cosx = - 8282197 [by

putting the values of sinx & cosx]

Question – 13 (i) If cos(A+B) sin(C-D) = cos(A-B) sin(C+D) , then show that

tanA tanB tanC + tanD = 0

(ii) If sinθ = n sin(θ+2𝛂), prove that

tan(θ+𝛂) = 1+n1−n tan𝛂.

Answer: We can write above given result as cos (A+B)cos( A−B) = sin(C+D)

sin(C−D)

By C & D cos ( A+B )+cos ( A−B)cos ( A+B )−cos( A−B) =

sin (C+D )+sin(C−D)sin (C+D )−sin(C−D)

cosA . cosB−sinA . sinB = sinC .cosD

cosC . sinD

-cotA.cotB = tanC.cotD

- 1tanA . 1

tanB = tanC. 1tanD

⇨ - tanD = tanA tanB tanC ⇨ tanA tanB tanC + tanD = 0 .

(ii) sin(θ+2α )sinθ = 1

n , by C & D sin (θ+2α )+sinθsin (θ+2α)−sinθ = 1+n

1−n

⇨ 2 sin (θ+α )cosα2 sinα cos (θ+α) = 1+n

1−n ⇨ tan(θ+𝛂) = 1+n1−n tan𝛂.

Question – 14 Prove that

(i) cos 520+cos 680+cos 1720 = 0

(ii) sinA .sin(600 – A).sin(600 + A) = 14 sin3A.

(it lies in 2nd quad.)

Answer: (i) L.H.S. cos 520+(cos 680+cos 1720) = cos520 + 2 cos1200 cos520

= cos520 + 2 ׿ ) cos520 [∵ cos1200 = cos(1800-600)]

= 0

(ii) L.H.S. sinA.[sin(600 – A).sin(600 + A)] = sinA [sin2 600 – sin2 A]

=

sinA [34 - sin2 A] = 1

4 [3sinA – 4sin3A]

= 1

4 sin3A.

[We know that sin2A-sin2B = sin(A+B)sin(A-B) & cos2A – sin2B = cos(A+B)cos(A-B)]

Question – 15 If sinx + siny = a and cosx + cosy = b, find

the values (i) tan( x− y2

) (ii) tan¿.

Answer: sinx + siny = a ⇨ 2sin( x+ y2

¿ cos¿) =

a ..............1

cosx + cosy = b ⇨ 2cos( x+ y2

¿ cos¿) =

b ..............2

(i) By squaring & adding above results, we get

4 cos² ¿) [sin²( x+ y2

¿ + cos²( x+ y2

¿] = a2+b2

( 1 )

Sec2( x− y2

) = 4a ²+b ² ⇨ tan2( x− y

2) = 4

a ²+b ² - 1 ⇨ tan

( x− y2

) = √ 4−a ²−b ²a ²+b ²

.

(ii) dividing 1 by 2, we get tan( x+ y2

) = ab .

Question – 16 Prove that

(i) tan700 = 2 tan500 + tan200 .

(ii) Prove that: tan300+ tan150+ tan300. tan150 =1.

(iii) cos2 A + cos2 B – 2cosAcosBcos(A+B) = sin2 (A+B).

Solution: (i) We have tan700 = tan(500 + 200)

= tan 50 °+ tan 20 °1−tan50 ° . tan 20 °

⇨ tan700 [1−tan 50 ° . tan 20 °] = tan50 °+ tan 20°

⇨ tan700 - tan700. tan 50 ° . tan 20° = tan50 °+ tan 20°

⇨ tan700 – tan(900-200). tan 50 ° . tan 20° = tan50 °+ tan 20°

(cot200 = 1tan 20 ° )

⇨ tan700 – tan500 = tan50 °+ tan 20°

⇨ tan700 = 2 tan500 + tan200.

(ii) [hint: take tan450 = tan(300+150)

(iii) L.H.S. cos2 A + cos2 B – (2cosAcosB)cos(A+B) = cos2 A + cos2 B – (cos(A+B)+cos(A-B))cos(A+B)

cos2 A + cos2 B – [cos2(A+B)+cos(A-B)cos(A+B)]

cos2 A + cos2 B – [cos2(A+B)+cos2(A) – sin2(B]

cos2 A + cos2 B - cos2(A+B)-cos2(A) + sin2(B]

1 - cos2(A+B) = sin2 (A+B).

Question – 17 If x+y = π4 , prove that (i) (1+tanx)

(1+tany) = 2 (ii) (cotx – 1)(coty – 1) = 2.

Answer: (i) tan(x+y) = tanπ4 ⇨ tanx+tany

1−tanx . tany = 1

⇨ tanx+tany+tanx.tany = 1

⇨ (1+ tanx) + tany(1 + tanx) = 1+1

⇨ (1+tanx)(1+tany) = 2

(ii) Similarly for second part by using cot(x+y) = cotx . coty−1cotx+coty .

Question – 18 Prove that (i) tan1890 = cos36 °−sin 36 °cos36 °+sin 36 °

(ii) Find the value of tan220 30’.

Answer: (i) we can take both sides to prove above result

L.H.S. tan1890 = tan(1800 +90) = tan9° =

tan(45°-36°) = tan 45 °+ tan36 °1−tan 45 ° . tan36 °

= 1−tan36 °

1+tan 36 ° = cos36 °−sin 36 °cos36 °+sin 36 ° .

(ii) Let x= 220 30’ then 2x = 450

We know that tanx = √ 1−cos2 x1+cos2 x ⇨ tanx

= √ 1−cos 45°1+cos45 °

= √ 1− 1√ 2

1+ 1√ 2

= √ (√2−1 )(√2−1)(√2+1 )(√2−1)

tan220 30’ = √2−1 [∵ it lies in 1st quad.]

Question: If tanx+tany = a and cotx+coty = b, prove that 1/a - 1/b = cot(x+y).

Answer: L.H.S. = 1tanx+tany - 1

coty+cotx = cosxcosy−sinxsinysinxcosy−cosxsiny

{after simplification}.

Question – 19 Prove that cos2 x + cos2 (x+π3 ) + cos2 (x-π

3

) = 32 .

Answer: L.H.S. cos2 x + cos2 (x+π3 ) +[1 - sin2 (x-π

3 ) ]

⇨ 1+ cos2 x + [ cos2 (x+π3 ) - sin2 (x-π

3 ) ]

⇨ 1+ cos2 x + [ cos (x+π3

+x− π3

¿ ) cos(x+π3

-x+π3 ) ] ⇨ 1+ cos2 x + [cos (2x) .cos(2 π

3 ) ]

( ∵cos(π - π3

¿ it lies in 2nd quad.)

⇨ 1+ cos2 x + [cos (2x) ׿ )] ⇨ 2+2 cos² x−cos2 x2 =

2+2cos² x−(2cos ² x−1)2

= 2+2 cos² x−2 cos² x+12 = 3

2 .

Question – 20 Prove that

(a) sin3x + sin3(2 π3 +x) + sin3(4 π

3 +x) = - 34

sin3x.

[Hint: Use sin3A = 3sinA – 4sin3A ⇨ 4sin3A = 3sinA - sin3A ⇨ sin3A = ¼[3sinA - sin3A]]

(b) If tanx + tan(x+π3 ) + tan¿+x) = 3 , then

show that tan3x = 1. [ use formula of tan(x+y)]

(c) Find in degrees and radians the angle subtended b/w the hour hand and the minute hand

Of a clock at half past three. [answer is 750 , 5π/12]

(d) 1tan 3 A−tanA - 1

cot 3 A−cotA = cot 2A. [Hint

put cot3A = 1/tan3A & cotA = 1/tanA then take l.c.m.

and use formula tan(A-B).]

(e) If α , 𝛃 are the distinct roots of acosθ +

bsinθ = c, prove that sin(𝛂+𝛃) = 2 aba ²+b ² .

Solution of (e) If α , 𝛃 are the distinct roots of acosθ + bsinθ = c, then

acos𝛂 + bsin𝛂 = c & acos𝛃 + bsin𝛃 = c

By subtracting , we get a(cos𝛂 – cos𝛃) + b(sin𝛂 – sin𝛃) = 0 ⇨ a(cos𝛂 – cos𝛃) = b(sin𝛂 – sin𝛃) ⇨ 2a sin α+β2 sin α−β

2 = 2b cosα+β2 sin

α−β2

⇨ tanα+β2 = b

a ⇨ sin(𝛂+𝛃) = 2 tan

α+β2

1+ tan ²α+β

2

⇨

sin(𝛂+𝛃) = 2aba ²+b ² .

Question: If sin2A = k sin2B, prove that tan( A+B)tan (A−B) =

k+1k−1

[ Hint by c & d sin 2 A+sin 2 B

sin 2 A−sin 2 B = k+1k−1 , use formula of sinx+siny= 2sin( x+ y

2¿ cos¿)]

Question: If cos(x+2A) = n cosx, show that cotA = n+11−n tan(x+A). Question: Prove that tan(∝−β ¿ = sin 2 β

5−cos 2 β , if 2tan∝ = 3tanβ. [Hint: L.H.S. = tanα−tanβ1+ tanα . tanβ , put the value of tan∝ and simplify it]. Question: If sinx + siny = a and cosx + cosy = b,find the

value of cos(x-y).

Answer: squaring and adding above results, we will get cos(x-y) = ½[a2+b2 – 2]

Question: Show that 1sin 10° - √ 3

cos10 ° =4

[Hint: 2 [ 1/2sin 10° - √3/2

cos10 ° ] = 2 [ sin 30°sin 10° - cos30 °

cos10 ° ]

Question: Prove that : 1+sinx−cosx1+sinx+cosx = tan(x/2)

[ Hint: use 1 – cosx = 2sin2x/2 , 1+cosx = 2cos2x/2 and sinx = 2sinx/2.cosx/2]

** Question Solve: √2 sec∅ + tan∅ = 1

Solution √2 + sin∅ = cos∅ ⇨ cos∅ - sin∅ = √2 dividing by √a2+b2 =√2 ∵ a=1,b=1 (cos∅ - sin∅ )/ √2 = 1 ⇨ cos( /4)п cos∅ - sin( /4) п sin∅ =1 ⇨ cos(∅+ /4) = cos0п 0 ∅+ /4 = 2nп п±0, n∈Z ⇨ ∅ = 2n –п /4.п

** Question: If tan2A = 2tan2B + 1, prove that cos2A + sin2B =0.

Answer: L.H.S. 1−tan ² A1+ tan ² B + sin2B =

−2 tan ² B

2(1+tan2 B) + sin2B , by

putting above result and simplify it.

** Question: Find the maximum and minimum values of sinx+cosx.

Answer: maximum value of (asinx+bcosx) = √a2+b2 = √ 2

Minimum value of (asinx+bcosx) = - √a2+b2 = - √ 2

Or

√ 2[ 1√ 2 sinx + 1

√ 2 cosx] = √ 2 sin(x+п/4), as -1 ≤

sin(x+п/4) ≤1 ∀ x

**Question: Find the minimum value of 3cosx+4sinx+8. [ answer is 3 as above result]

**Question: ∀ x in (0,п/2), show that cos(sinx) > sin(cosx).

Answer: п/2> √ 2 [∵п/2 = 22/7=1.57 and √ 2 =1.4]

We know that п/2 > √ 2 ≥ sinx+cosx ⇨ п/2 - sinx > cosx ⇨ cos(sinx) > sin(cosx).

** Question: If A = cos2x + sin4x ∀ x, prove that ¾ ≤ A ≤ 1

Answer: A = cos2x + sin2x. sin2x ≤ cos2x + sin2x ⇨ A ≤ 1 , A = (1 - sin2x) + sin4x = [sin2x – (½)]2 + (¾) ≥ ¾.

** Question: (i) find the greatest value of sinx.cosx [ ½.(2sinx.cosx) ≤ ½]

(ii) If sinx and cosx are the roots of ax2 – bx+c = 0, then a2 – b2 +2ac = 0.

[ Hint: sum and product of roots ⇨ (sinx+cosx)2 = 1+2(c/a)]

** Question : If f(x) = cos2x+sec2x , then find which is correct f(x)<1, f(x)=1, 2< f(x)<1, f(x)≥2.

[A.M. ≥ G.M.⇨ f(x)/2= (cos2x+sec2x)/2 ≥√ (cos2x.sec2x) , ANSWER IS f(x)≥2].

** Question: solve 3sin 2x+2 cos ² x+31−sin2 x +2sin² x = 28.

[hint: use 2cos2x = 1+ cos2x , 2sin2x = 1 – cos2x and let sin2x+cos2x = y

Above equation becomes 31+ y+32− y = 28. Then put t = 3y ⇨ 3t2 – 28t +9=0 ⇨ t=1/3 , 9

When y=-1 then sin2x+cos2x = -1 ⇨ cosx = 0 or tanx = -1

When y = 2 then sin2x+cos2x =2 ⇨ 1√ 2 sin2x+ 1

√ 2 cos2x = √ 2 ⇨ sin(2x+π /4 ¿=√ 2 >1 which is not possible