Mathematics presentation trigonometry and circle
Transcript of Mathematics presentation trigonometry and circle
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Mathematics
Trigonometry and
Circle
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Andriana Lisnasari
(01)Bramantya Nugraha
(06)Candra Dista Kusuma
(08)Dewi Setiyani Putri
(09)Khoirunnisa Ronaa Fairuuz
(16)
Okiana Wahyu K. (22)
XI.A2
We are from Fourth Group XI.A2
2011/2012
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Proudly Present
Some problems and solutions from
Trigonometry and Circle
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Trigonometri
We are going to solve some
problems from Trigonometry and
the solutionsNumber : 4 , 9, 14, 19,
24, 29. 34
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Given that and then
Find !
2yxtan 2
1tan x
ytan
Trigonometri
Problem i : Number 4
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ytan x.tan 1
ytan tan xyxtan
ytan.xtan1
ytan tan x2
Given that :
2y)(xtan
2
1tan x
Asked :
? .....y tan
ytan tan xy tan x.tan22
ytan 2
1y.tan
2
12.2
ytan 2
1ytan 2
ytan22
12
Solution
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ytan22
12
y tan 22
3
2
1.
2
3ytan
4
3ytan
So, the value of is equal
ytan 4
3
C0ntinue . . .
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Prove that
x2
1cotxcot xcosec
Trigonometri
Problem ii : Number 9
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x2
1cotcotx xcosec
x2
1.
tan
1x
2
1cot
xcos1sin x
1
sin x
xcos1
Solution
Prove the right side
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sin x
xcos1
sin x
xcos
sin x
1
cot x xcosec
cotx xcosecx2
1cot
x2
1cotcot x xcosec PROVED
Continue . . .
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Let
a.) Prove that
b.) Solve the equation ,
giving your answer for A in
the interval
o
o
o
o
f
cos
sin1
sin1
cos)(
of sec2)(
4f
o3600
Trigonometri
Problem iii : Number 14
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o
o
o
o
f
cos
sin1
sin1
cos)(
a. prove of sec2)(
oo
o2oo2
o
o
o
o
A cos . Asin 1
AsinA2sin Acos
Α cos
Αsin 1
Αsin 1
Α cos)A(
f
oo
oo2o2
A cos . Asin 1
Asin 21AsinAcos
oo
o
A cos . Asin 1
Asin 211
Solution i
Given that :
Asked :
Answer :
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oo
o
A cos . Asin 1
Asin 211
oo
o
A cos . Asin 1
Asin 22
oo
o
A cos . Asin 1
)Asin 1(2
oA cos
2
oA sec 2
xsec 2Α cos
Αsin 1
Αsin 1
Α cos)A(
o
o
o
o
f
proved
So,
Continue . . .
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b. A in the interval 4fo3600
of sec2)(
4f
4A 2sec o
2A sec o
2A cos
1o
2
1A cos o
o60 cosA cos o
Solution ii
Given that :
Asked :
Answer :
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o60 cosA cos o in the interval o3600
oo k.360αx
0k
oo k.36060A o60A
oo k.360αx oo k.36060A
0k o300A
So, the value A in the interval are
o3600 o300 and 60o
Continue . . .
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Evaluate each of the following :
a.
b.
oooo 90sin270csc.3180cos.2180tan
oooo 270cos.4180sec.590cot.30sin
Trigonometri
Problem iv : Number 19
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oooo 90sin270csc.3180cos.2180tan
11.31.20
132
0
090sin270csc.3180cos.2180tan oooo
a.
Solution i
So,
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b.
So,
oooo 270cos4180sec.590cot.30sin
041.50.30
0500
5
5270cos4180sec.590cot.30sin oooo
Solution ii
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Prove that
2 xcoseccot x1 xsec
1 xsec
xcos1
xcos1
Trigonometri
Problem v : Number 25
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1 xsec
1 xsec
xcos1
xcos1
2 xcoseccot x1 xsec
1 xsec
xcos1
xcos1
xcos xcos
xcos1
xcos xcos
xcos1
1 xsec
1 xsec
xcos xcos1
xcos xcos1
first step
Prove the right side
Prove that
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xcos1
xcos.
xcos
xcos1
xcos1
xcos1
1 xsec
1 xsec
xcos1
xcos1
xcos xcos1
xcos xcos1
proved
Continue . . .
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2 xcoseccot x1 xsec
1 xsec
2 xcoseccot x1 xsec
1 xsec
xcos1
xcos1
cxcot x.cose 2xcosecxcot xcoseccot x 222
sin x
1.
sin x
xcos2
xsin
1
xsin
xcos22
2
xsin
xcos2
xsin
1
xsin
xcos222
2
xsin
xcos 21xcos2
2
second step
Prove the right side
Prove that
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xsin
xcos 21xcos2
2
xsinxsec
xsec2
xsecxsec
xsec1
2
22
2
2
xcos-1xsec
x2secxsec1
2
2
2
xsec1
xsecxsec
xsec xsec 2xsec1
22
2
2
2
Continue . . .
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xsec1
xsecxsec
xsec xsec 2xsec1
22
2
2
2
1xsec
xsec
xsec
xsec 2xsec12
2
2
2
1xsec
xsec 2xsec12
2
1 xsec1 xsec
1 xsec1) x(sec
1 xsec
1 xsec
1 xsec
1 xsec xcoseccot x 2
proved
Continue . . .
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1 xsec
1 xsec xcsccot x 2
2 xcoseccot x1 xsec
1 xsec
xcos1
xcos1
1 xsec
1 xsec
1 xsec
1 xsec
xcos1
xcos1
2 xcoseccot x1 xsec
1 xsec
xcos1
xcos1
From the first step we could prove that
And from the second step we could prove that
So, we can conclude that the
statement
Is proven
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Prove that
1xtanxsec
xtanxsec22
44
Trigonometri
Problem vi : Number 29
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1xtanxsec
xtanxsec22
44
xtanxsec
xtanxsecxtanxsec
xtanxsec
xtanxsec22
2222
22
44
xtanxsec 22
xcos
xsin
xcos
12
2
2
xcos
xsin12
2
xcos
xcos2
2
1
1xtanxsec
xtanxsec22
44
Solution
Prove the left side
proved
So, the statement is proved
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In a study of AC circuits, the
equation
sometimes arises,
Use a sum identity and algebra to
show this equation is equivalent
to
tsc
tsR
sin
cos.cos
tscR
tantan
1
Trigonometri
Problem vii : Number 34
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tsctsc
tsR
tantan
1
sin
cos.cos
tsctsc
ts
tantan
1
sin
cos.cos
)cos.sincos.(sin
cos.cos.
1
sin
cos.cos
ttts
ts
ctsc
ts
tstttsc
cos.cos)cos.sincos.(sin
1.
1
SolutionSHOW THAT
PROVE THE LEFT SIDE
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tstttsc
cos.cos)cos.sincos.(sin
1.
1
tsst
tsts
ccos.coscos.sin
cos.coscos.sin
1
tt
ss
ccossin
cossin
1
tsc tantan
1
tsctsc
ts
tantan
1
sin
cos.cos
continue
PROVED
So , the statement is proven
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CircleSome problems circle and the solutions
Number : 4, 9, 11
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Show that the circle
and
are
tangent internally
(without a graphing)
054y2x6yx 22
0112y8x22yx 22
Problem i : Number 4Circle
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Solution
054y2x6yx 22
0112y8x22yx 22
…….I
…….Ii
01666yx28 0833yx14
xy 14833
3
1483 xy
054y2x6yx 22
0543
14832x6
3
1483x
22
xx
Then substitution to the first equation
0543
28
3
1666
9
19623246889x
22
xx
xx
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0543
28
3
1666
9
19623246889x
22
xx
xx
09
84542324
9
4864986889
9
205 2
xxx
x
084542324)4864986889(205 2 xxxx
059052186205 2 xx
Then the value for the discriminant of 059052186205 2 xx
cabD ..42
5905.205.42186 2
4842100477856
4364244
0D The circle do not intersect
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Find the equation of the
circle through the point (5,1),
(4,6) and (2,-2)
Problem ii : Number 9
Circle
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(4,6)
0CByAxyx 22 0CB6A464 22
0CB6A43616 52CB64A
…….Iii
SOLUTION
Trough Point
(5,1)
0CByAxyx 22 0CB1A515 22
0CBA5125 …….Iii
Trough Point
26CBA5
26CB32A
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(2,-2)
0CByAxyx 22 0CB2A222 22
0CB2A244
8CB2A2 4CBA
26CBA5 …….I
…….Iii4CBA
22B26A 11B3A
26CB3A2 …….Ii
…….Iii4CBA
22B4A
…….Iii
CONTINUE
…….Iv…….v
Then, from I, and III equation we get . . .
Trough Point
And from II, and III equation we get . . .
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…….Iv
…….v
2211A
11B3A 22B4A
4
1
44B412A
22B4A
2A
…….Iv11B3A 11B23
11B6 5B
4CBA 4C52 1C
CONTINUEFrom elimination IV and V equation, we can get the value of A
Then, we are going to find the
value of B from IV equation And the value of C can we get by
substitution of the III equation
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0CByAxyx 22 2A
5B
1C
0Cy5x2yx 22
0Cy5x2yx 22
After that, substitution the value of A, B, and C that we already got to the equation of the circle
CONTINUE
So, the equation of the circle is
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a. Determine the equation of a circle
that passes through points (-2,4)
and (7,7) and has a center on line
b. Determine the equation of tangent
line at points (7,7) on the circle
c. Determine the equation of the
tangent
line on a circle which is parallel to
line
7yx
01y2x
Problem iii : Number 14
Circle
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0CByAxyx 22
0CByAxyx 22
0CB4A2164 20CB4A2
Solution for number a.
Given that :Trough point A = (-2,4)
and B(7,7)Center on line x + y
= 7
Asked :
The equation of circle
Answer :Suppose that the equation
of the circle is
Trough point A (-2,4)
…... i
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0CByAxyx 22
0CB7A74949
98CB7A7
B
2
1A,
2
1C
7yx 7yx
7B2
1A
2
1
14BA
Then, substitution the center of the circle to the equation of center line circle
The center of circle
…... iii
Continue . . .
Trough point A (-2,4)
…... ii
x y
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20CB4A2 98CB7A7
78B3A9 26BA3
26BA3 14BA
122A 6A
14BA 14B6
8B
…... i…... ii
…... iv…... iii
…... iii
Then, elimination the I and the II equation
…... iV
And from elimination Iv and Iii equation we can get the value of A
Continue . . .
Then, the value of B can we get from Iii equation
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20CB4A2 20C8462
20C3212 0C
0C
6A 8B
0CByAxyx 22
0y86xyx 22
0y86xyx 22
…... i
The value of A and B are known, so we can get the value C from I equation
So, the equation of circle is
Continue . . .
After that, substitution the value of A, B, and C that we already got to the equation of the circle
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0y86xyx 22
CHECK point (7,7) 0y86xyx 22
0787677 22
056424949 00
SOLUTION FOR NUMBER B.Given
that :Asked :
Answer :
The equation of tangent line at points (7,7) on the circle
The equation of the circle
on the circle
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0Cy)(y B2
1x)(xA
2
1yyxx 1111
The equation of tangent line is
00y)(78 2
1x)(76
2
1y77x
00y)(74x)(73y77x
04y28x321y77x
0493y4x
So, the equation of tangent line is
0493y4x
Continue . . .
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The equation of the circle0y86xyx 22
the equation of the tangent line on a circle which is parallel to line
01y2x
Find the gradient of the equation 01y2x
1x2y 01y2x
2
1x
2
1y So,
m =2
1
Solution for number C.
Given that :
Asked :
Answer :
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Find the center and radius of the circle which has equation 0y86xyx 22
CB4
1A
4
1r 22
0)8(4
16
4
1 22
169
25
5 So, r = 5
B
2
1A,
2
1C
)8(
2
1,6
2
1C
C(3,4) So, the center is (3,4)
Continue . . . .
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21)( mraxmby
The equation of the tangent line on a
circle
which is parallel to line
01y2x
0y86xyx 22 L
2
1m
5r
C(3,4)
2
2
1153
2
14
)(xy
ba,
4
115
2
3
2
14 xy
04
554
2
3
2
1 xy
052
5
2
5
2
1 xy
05552 xy
Continue . . . .
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The equation of the tangent line on a circlewhich is parallel to line are
01y2x 0y86xyx 22 L
And
And
05552 xy
05552 xy 05552 xy
05552 xy05552 xy
Continue . . . .
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Thank you For Your Attention