Post on 27-Jun-2015
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 the right side
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 the right side
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)
0CByAxyx 22 0CB1A515 22
0CBA5125 …….Iii
Trough Point
26CBA5
26CB32A
(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 . . .
…….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