Mathematics presentation trigonometry and circle

Mathematics

Trigonometry and

Circle

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

Proudly Present

Some problems and solutions from

Trigonometry and Circle

Trigonometri

We are going to solve some

problems from Trigonometry and

the solutionsNumber : 4 , 9, 14, 19,

24, 29. 34

Given that and then

Find !

2yxtan 2

1tan x

ytan

Trigonometri

Problem i : Number 4

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

ytan22

12

y tan 22

3

2

1.

2

3ytan

4

3ytan

So, the value of is equal

ytan 4

3

C0ntinue . . .

Prove that

x2

1cotxcot xcosec

Trigonometri

Problem ii : Number 9

x2

1cotcotx xcosec

x2

1.

tan

1x

2

1cot

xcos1sin x

1

sin x

xcos1

Solution

Prove the right side

sin x

xcos1

sin x

xcos

sin x

1

cot x xcosec

cotx xcosecx2

1cot

x2

1cotcot x xcosec PROVED

Continue . . .

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

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 :

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 . . .

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 :

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 . . .

Evaluate each of the following :

a.

b.

oooo 90sin270csc.3180cos.2180tan

oooo 270cos.4180sec.590cot.30sin

Trigonometri

Problem iv : Number 19

oooo 90sin270csc.3180cos.2180tan

11.31.20

132

0

090sin270csc.3180cos.2180tan oooo

a.

Solution i

So,

b.

So,

oooo 270cos4180sec.590cot.30sin

041.50.30

0500

5

5270cos4180sec.590cot.30sin oooo

Solution ii

Prove that

2 xcoseccot x1 xsec

1 xsec

xcos1

xcos1

Trigonometri

Problem v : Number 25

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 that

xcos1

xcos.

xcos

xcos1

xcos1

xcos1

1 xsec

1 xsec

xcos1

xcos1

xcos xcos1

xcos xcos1

proved

Continue . . .

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 that

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 . . .

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 . . .

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

Prove that

1xtanxsec

xtanxsec22

44

Trigonometri

Problem vi : Number 29

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

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

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

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

CircleSome problems circle and the solutions

Number : 4, 9, 11

Show that the circle

and

are

tangent internally

(without a graphing)

054y2x6yx 22

0112y8x22yx 22

Problem i : Number 4Circle

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

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

Find the equation of the

circle through the point (5,1),

(4,6) and (2,-2)

Problem ii : Number 9

Circle

(4,6)

0CByAxyx 22 0CB6A464 22

0CB6A43616 52CB64A

…….Iii

SOLUTION

Trough Point

(5,1)


0CBA5125 …….Iii

Trough Point

26CBA5

26CB32A

(2,-2)


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 . . .

…….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

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

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

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

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

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

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

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

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 . . .

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 :

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 . . . .

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 . . . .

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 . . . .

Thank you For Your Attention

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Mathematics presentation trigonometry and circle