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Trigonometry Lesson 12
Part I
Expanding SUM & Difference
Sum & Difference Identities: The following formulas are used to expand trigonometric
functions that have addition & subtraction in brackets.
sin(A + B) ≠ sinA + sinB , so we must use these rules whenever we want to expand.
sin(A + B) = sinAcosB + cosAsinB
sin(A - B) = sinAcosB - cosAsinB
cos(A + B) = cosAcosB - sinAsinB
cos(A - B) = cosAcosB + sinAsinB
Example 1: Expand sin(60o - 45o )
sin(60D − 45D )
= sin 60D cos 45D − cos 60D sin 45D
⎛ 3 ⎞⎛ 2 ⎞ ⎛ 1 ⎞⎛ 2 ⎞
= ⎜⎜
⎟⎟ ⎜⎜
⎟⎟ − ⎜ ⎟ ⎜⎜
⎟⎟
2
2
⎝
⎠⎝
⎠ ⎝ 2 ⎠⎝ 2 ⎠
6
2
−
4
4
6− 2
=
4
=
⎛π π ⎞
Example 2: Expand cos ⎜ + ⎟
⎝4 6⎠
⎛π π ⎞
cos ⎜ + ⎟
⎝4 6⎠
= cos
⎛
= ⎜⎜
⎝
⎛
= ⎜⎜
⎝
=
π
cos
π
− sin
π
sin
π
4
6
4
6
2 ⎞⎛ 3 ⎞ ⎛ 2 ⎞⎛ 1 ⎞
⎟⎜
⎟−⎜
⎟⎜ ⎟
2 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎝ 2 ⎠
6⎞ ⎛ 2⎞
⎟−⎜
⎟
4 ⎟⎠ ⎜⎝ 4 ⎟⎠
6− 2
4
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Trigonometry Lesson 12
Part I
Expanding SUM & Difference
Find the exact value by expanding each of the following:
π
π
π
π
1) sin(45º + 60º)
5) sin ⎛⎜ − ⎞⎟
⎝2 3⎠
2) cos(45º - 30º)
6) sin ⎛⎜ + ⎞⎟
⎝4 6⎠
3) sin(60º - 135º)
7) cos ⎛⎜ 0 −
4) cos(150º + 45º)
8) cos ⎛⎜ + ⎞⎟
⎝2 3⎠
⎝
π
3π ⎞
⎟
4 ⎠
π
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Trigonometry Lesson 12
Part I
1.
4.
Expanding SUM & Difference
sin(45º + 60º)
=sin45ºcos60º+cos45ºsin60º
2.
⎛ 2 ⎞⎛ 1 ⎞ ⎛ 2 ⎞⎛ 3 ⎞
= ⎜⎜
⎟⎟ ⎜ ⎟ + ⎜⎜
⎟⎟ ⎜⎜
⎟⎟
⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎠⎝ 2 ⎠
⎛ 2 ⎞⎛ 3 ⎞ ⎛ 2 ⎞ ⎛ 1 ⎞
= ⎜⎜
⎟⎜
⎟⎟ ⎜ ⎟
⎟⎜
⎟⎟ + ⎜⎜
⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠
2
6
=
+
4
4
2+ 6
=
4
6
2
=
+
4
4
6+ 2
=
4
cos(150º + 45º)
=cos150ºcos45º - sin150ºsin45º
5.
⎛
3 ⎞⎛ 2 ⎞ ⎛ 1 ⎞ ⎛ 2 ⎞
= ⎜⎜ −
⎟⎜
⎟⎟
⎟⎜
⎟⎟ − ⎜ ⎟ ⎜⎜
⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠
2
2
2
π
2
3
cos
π
8.
sin(60º - 135º)
=sin60ºcos135º - cos60ºsin135º
⎛ 3 ⎞⎛
2 ⎞ ⎛ 1 ⎞⎛ 2 ⎞
= ⎜⎜
⎟⎟ ⎜⎜ −
⎟⎟ − ⎜ ⎟ ⎜⎜
⎟⎟
⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎠⎝ 2 ⎠
2
− 6
−
4
4
− 6− 2
=
4
6.
)
π
3
- cos
π
2
3⎞
⎟
2 ⎟⎠
sin
sin(
π
4
π
+
π
6
)
π
π
=sin
3
⎛ 2 ⎞⎛ 3 ⎞ ⎛ 2 ⎞ ⎛ 1 ⎞
= ⎜⎜
⎟⎜
⎟⎟ + ⎜⎜
⎟⎟ ⎜ ⎟
⎟⎜
⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠
4
cos
6
+ cos
4
sin
6
2
+
4
4
6+ 2
=
4
=
π
cos( + )
2 3
= cos
π
2
cos
π
3
− sin
⎛
⎛1⎞
= ( 0 ) ⎜ ⎟ − (1) ⎜⎜
⎝2⎠
⎝
=−
π
2
3⎞
⎟
2 ⎟⎠
sin
π
π
1
=
2
3π
)
4
3π
3π
= cos 0 cos
+ sin 0sin
4
4
⎛
⎞
⎛
⎞
2
2
= (1) ⎜⎜ −
⎟⎟ + ( 0 ) ⎜⎜
⎟⎟
2
2
⎝
⎠
⎝
⎠
=−
-
π
3.
=
⎛
⎛1⎞
= (1) ⎜ ⎟ − ( 0 ) ⎜⎜
⎝2⎠
⎝
2
− 6
−
4
4
− 6− 2
=
4
cos(0 −
sin(
π
=sin
=
7.
cos(45º - 30º)
=cos45ºcos30º+sin45ºsin30º
π
3
3
2
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6
Trigonometry Lesson 12
Part Ii
condensing SUM & Difference
Given an expanded form, we can work backwards to find a single trigonometric expression
that can be easily solved using the unit circle.
Example 1: Express sin85o cos5o + cos85o sin5o as a single trigonometric
expression and solve.
We know sin(A + B) = sinAcosB + cosAsinB
Looking at the question and matching to the sum formula, we can see that:
A = 85º and B = 5º
Now plug A & B into the left side of the formula:
sin(A + B)
= sin(85º + 5º)
= sin(90º)
=1
1
Example 2: Express
⎛ π ⎞ cos ⎛ π ⎞ - cos ⎛ π ⎞ sin ⎛ π ⎞
⎟
⎜ ⎟
⎜ ⎟
⎜ ⎟
⎝3⎠
⎝6⎠
⎝3⎠
⎝6⎠
as a single trigonometric
sin ⎜
expression and solve.
We know: sin(A - B) = sinAcosB - cosAsinB
⎛π ⎞
⎛π ⎞
⎛π ⎞
⎛π ⎞
We have from the denominator of the question: sin ⎜ ⎟ cos ⎜ ⎟ - cos ⎜ ⎟ sin ⎜ ⎟
⎝3⎠
⎝6⎠
⎝3⎠
⎝6⎠
Comparing the two, A =
π
3
Plug A & B into sin(A - B)
= sin(
π
π
6
π
6
π
)
3
6
2π
π
- )
= sin(
6
6
= sin(
-
&B=
Subtract radians by
finding a common denominator.
(Or think in terms of degrees.)
)
We now know the denominator is sin(
Thus, we have
⎛π ⎞
= csc ⎜ ⎟ .
π
⎝6⎠
sin( )
6
1
π
6
).
Solving,
1
sin(
π
6
=
)
1
=2
1
2
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Trigonometry Lesson 12
Part Ii
condensing SUM & Difference
For each of the following, express as a single trigonometric expression and solve using the
unit circle.
1) cos 60D cos15D + sin 60D sin15D
3)
2) cos 2
π
6
5π
1
⎛π ⎞
⎛π ⎞
⎛π ⎞ ⎛π ⎞
sin ⎜ ⎟ cos ⎜ ⎟ − cos ⎜ ⎟ sin ⎜ ⎟
⎝2⎠
⎝3⎠
⎝2⎠ ⎝3⎠
− sin 2
π
6
π
5π
π
4) sin ⎛⎜ ⎞⎟ cos ⎛⎜ ⎞⎟ + cos ⎛⎜ ⎞⎟ sin ⎛⎜ ⎞⎟
⎝ 12 ⎠
⎝3⎠
⎝ 12 ⎠ ⎝ 3 ⎠
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Trigonometry Lesson 12
Part Ii
5)
7)
condensing SUM & Difference
1
cos(−15 ) cos(30 ) + sin(−15D ) sin(30D )
D
D
1
⎛π ⎞
⎛π ⎞
⎛π ⎞ ⎛π ⎞
cos ⎜ ⎟ cos ⎜ ⎟ + sin ⎜ ⎟ sin ⎜ ⎟
⎝2⎠
⎝4⎠
⎝2⎠ ⎝4⎠
π
π
π
π
6) sin ⎛⎜ ⎞⎟ cos ⎛⎜ ⎞⎟ − sin ⎛⎜ ⎞⎟ cos ⎛⎜ ⎞⎟
⎝3⎠
⎝6⎠
2π
π
⎝6⎠
2π
⎝3⎠
π
8) cos ⎛⎜ ⎞⎟ cos ⎛⎜ ⎞⎟ + sin ⎛⎜ ⎞⎟ sin ⎛⎜ ⎞⎟
⎝ 3 ⎠
⎝2⎠
⎝ 3 ⎠ ⎝2⎠
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Trigonometry Lesson 12
Part Ii
1.
cos 60D cos15D + sin 60D sin15D
A = 60
& B = 15
D
D
condensing SUM & Difference
2.
cos
cos(60 − 15 )
D
D
A=
D
cos(45 )
=
π
6
cos
π
π
6
− sin
& B=
6
π
π
6
sin
π
3.
6
π
6
π
π
7.
1
⎛π ⎞
⎛π ⎞
⎛π ⎞ ⎛π ⎞
cos ⎜ ⎟ cos ⎜ ⎟ + sin ⎜ ⎟ sin ⎜ ⎟
⎝2⎠
⎝4⎠
⎝2⎠ ⎝4⎠
A=
π
2
1
π
& B=
π
cos( − )
2 4
π
π
= sec( − )
2 4
2π π
= sec( − )
4 4
π
4
5.
1
cos(−15D ) cos(30D ) + sin(−15D ) sin(30D )
A = −15D & B = 30D
1
cos(−15D − 30D )
1
=
cos ( −45D )
1
2
2
2
=
2
=
=
= 2
8.
3
π
π
π
Getting A Common Denominator
π
= csc( )
6
=2
6.
⎛π ⎞
⎛π ⎞
⎛π ⎞
⎛π ⎞
sin ⎜ ⎟ cos ⎜ ⎟ − sin ⎜ ⎟ cos ⎜ ⎟
⎝3⎠
⎝6⎠
⎝6⎠
⎝3⎠
A=
π
3
π
& B=
π
π
6
sin( − )
3 6
2π π
= sin( − )
6 6
= sin
2
2
×
2
2
π
= csc( − )
2 3
3π 2π
= csc( − )
6
6
= cos( )
3
1
=
2
⎛ 5π ⎞
⎛π ⎞
⎛ 5π ⎞ ⎛ π ⎞
sin ⎜
⎟ cos ⎜ ⎟ + cos ⎜ ⎟ sin ⎜ ⎟
⎝ 12 ⎠
⎝3⎠
⎝ 12 ⎠ ⎝ 3 ⎠
5π
π
A=
& B=
12
3
5π π
sin( + )
12 3
5π 4π
= sin( + )
12 12
9π
= sin( )
12
3π
= sin( )
4
2
=
2
& B=
2
1
sin( − )
2 3
π
4.
π
A=
cos( + )
6 6
2π
= cos( )
6
2
2
1
⎛π ⎞
⎛π ⎞
⎛π ⎞ ⎛π ⎞
sin ⎜ ⎟ cos ⎜ ⎟ − cos ⎜ ⎟ sin ⎜ ⎟
⎝2⎠
⎝3⎠
⎝2⎠ ⎝3⎠
π
6
1
=
2
⎛ 2π ⎞
⎛π ⎞
⎛ 2π ⎞ ⎛ π ⎞
cos ⎜
⎟ cos ⎜ ⎟ + sin ⎜
⎟ sin ⎜ ⎟
⎝ 3 ⎠
⎝2⎠
⎝ 3 ⎠ ⎝2⎠
2π
π
A=
& B=
3
2
2π π
cos( − )
3 2
4π 3π
= cos( − )
6
6
= cos
π
6
3
=
2
π
= sec( )
4
2
=
2
= 2
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Trigonometry Lesson 12
Part Iii
non unit circle angles
The sum & difference formulas are useful in determining the exact values of sine & cosine
for angles not on the unit circle.
Example 1: Find the exact value of sin15º
First, think of how you can get 15º by using angles on the unit circle:
15º = 60º - 45º
15º = 45º - 30º
15º = 135º - 120º
15º = -30º + 45º
As you can see, there are many possibilities, you can choose any of them and
still get the right answer.
We’ll use the top one, 15º = 60º - 45º
sin15D
= sin(60D − 45D )
= sin 60D cos 45D − cos 60D sin 45D
⎛ 3 ⎞⎛ 2 ⎞ ⎛ 1 ⎞⎛ 2 ⎞
= ⎜⎜
⎟⎟ ⎜⎜
⎟⎟ − ⎜ ⎟ ⎜⎜
⎟⎟
⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎠⎝ 2 ⎠
=
6− 2
4
⎛ 5π ⎞
Example 2: Find the exact value of sec ⎜ −
⎟
⎝ 12 ⎠
5π 180D
It’s easiest to think in terms of degrees, so convert: −
×
= −75D (-75º = -45º - 30º)
π
12
sec(−75D )
= sec(-45º - 30º)
=
1
cos(-45º - 30º)
Now rationalize the denominator of your answer.
1
cos(-45º)cos(30º)+sin(-45º)sin(30º)
1
=
⎛ 2 ⎞⎛ 3 ⎞ ⎛ − 2 ⎞⎛ 1 ⎞
⎜
⎟⎜
⎟+⎜
⎟⎜ ⎟
⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎠⎝ 2 ⎠
1
=
6
2
−
4
4
1
=
6− 2
4
4
=
6− 2
=
4
6− 2
=
4
6+ 2
×
6− 2
6+ 2
4( 6 + 2)
6−2
4( 6 + 2)
=
4
= 6+ 2
=
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304
Trigonometry Lesson 12
Part Iii
non unit circle angles
Find the exact value of each of the following:
Note that there are multiple ways of getting to the correct answer. Rationalize the denominator when necessary.
1) cos(-15º)
2) sec(105º)
3) csc(-105º)
5π
)
12
5) csc(165º)
6) sec(
4) sin( −
13π
)
12
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Trigonometry Lesson 12
Part Iii
1.
cos(30D − 45D )
= cos 30D cos 45D + sin 30D sin 45D
non unit circle angles
2.
⎛ 3 ⎞⎛ 2 ⎞ ⎛ 1 ⎞ ⎛ 2 ⎞
= ⎜⎜
⎟⎜
⎟⎟ + ⎜ ⎟ ⎜⎜
⎟⎟
⎟⎜
⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠
sec(60D + 45D )
1
=
cos(60D + 45D )
3.
=
1
cos 60D cos 45D − sin 60D sin 45D
1
=
⎛ 1 ⎞ ⎛ 2 ⎞ ⎛ 3 ⎞⎛ 2 ⎞
⎟+⎜
⎟⎜
⎟
⎜ ⎟⎜
⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠⎝ 2 ⎠
1
=
2
6
+
4
4
1
=
2+ 6
4
4
=
2+ 6
6
2
+
4
4
6+ 2
=
4
=
=
=
4
2− 6
×
2+ 6
2− 6
=
4( 2 − 6)
2−6
4(− 6 + 2)
6−2
4(− 6 + 2)
=
4
=− 6+ 2
4( 2 − 6)
−4
= −1( 2 − 6)
= 6− 2
5π
)
12
= sin( −75D )
sin(−
5.
= sin( −45 − 30 )
D
D
= sin( −45D ) cos 30D − cos(−45D ) sin 30D
⎛
2 ⎞⎛ 3 ⎞ ⎛ 2 ⎞ ⎛ 1 ⎞
= ⎜⎜ −
⎟⎜
⎟⎟ ⎜ ⎟
⎟⎜
⎟⎟ − ⎜⎜
⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠
6
2
−
4
4
−( 6 + 2)
=
4
=−
csc(165D )
1
=
sin(165D )
1
=
sin(120D + 45D )
6.
1
sin(120D ) cos 45D + cos(120D ) sin 45D
1
=
⎛ 3 ⎞⎛ 2 ⎞ ⎛ 1 ⎞⎛ 2 ⎞
⎜
⎟⎜
⎟ + ⎜ − ⎟⎜
⎟
⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎠⎝ 2 ⎠
1
=
6
2
−
4
4
1
=
6− 2
4
4
=
6− 2
=
=
− 6+ 2
4
×
− 6− 2 − 6+ 2
=
=
4.
1
sin(−60D − 45D )
1
sin(−60D ) cos 45D − cos(−60D ) sin 45D
1
=
⎛
⎞⎛
3
2 ⎞ ⎛ 1 ⎞⎛ 2 ⎞
⎜−
⎟⎜
⎟ − ⎜ ⎟⎜
⎟
2
2
⎝
⎠⎝
⎠ ⎝ 2 ⎠⎝ 2 ⎠
1
=
6
2
−
−
4
4
1
=
− 6− 2
4
4
=
− 6− 2
=
=
csc(−60D − 45D )
4
6+ 2
×
6− 2
6+ 2
4( 6 + 2)
4
= 6+ 2
=
⎛ 13π ⎞
sec ⎜
⎟
⎝ 12 ⎠
= sec(195D )
= sec(150D + 45D )
=
1
cos(150D + 45D )
1
cos150D cos 45D − sin150D sin 45D
1
=
⎛
3 ⎞⎛ 2 ⎞ ⎛ 1 ⎞ ⎛ 2 ⎞
⎜−
⎟⎜
⎟ − ⎜ ⎟⎜
⎟
⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠
1
=
6
2
−
−
4
4
1
=
− 6− 2
4
4
=−
( 6 + 2)
=
=−
4
6− 2
×
( 6 + 2)
6− 2
4( 6 − 2)
4
= −( 6 − 2)
=−
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Trigonometry Lesson 12
Part Iv
sum & difference proofs
The following examples illustrate some basic proofs you can do with the sum & difference
identities:
π
Example 1: Prove that cos( - x) = sinx
2
We start with the formula: cos( A − B ) = cos A cos B + sin A sin B
Plug in A =
π
π
2
&B=x
cos( − x ) = cos A cos B + sin A sin B
2
= cos
π
cos x + sin
π
2
2
= ( 0 ) cos x + (1) sin x
sin x
= sin x
Example 2: Prove that csc(π + x) = -cscx
csc(π + x)
1
=
sin(π + x)
1
sin π cos x + cos π sin x
1
=
( 0 ) cos x + ( −1) sin x
=
1
− sin x
= − csc x
=
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Trigonometry Lesson 12
Part Iv
sum & difference proofs
Prove each of the following:
π
3π
1) cos( − x) = − sin x
2
5) sin( − x) =
2) sin(270D − x) = − cos x
6) csc( + x) =
π
2
π
2
3) cos( + x) = − sin x
7) sec(π + x)
4) cos(π − x) =
8) csc(π − x)
2
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Trigonometry Lesson 12
Part Iv
3π
1. cos( 2 − x)
sum & difference proofs
2.
3π
3π
cos x + sin
sin x
= cos
2
2
= ( 0 ) cos x + (−1) sin x
π
sin(270D − x)
= sin 270D cos x − cos 270D sin x
= (−1) cos x − (0) sin x
= − cos x
= − sin x
π
4. cos(π − x)
= cos π cos x + sin π sin x
= (−1) cos x + (0) sin x
= − cos x
π
= cos
π
cos x − sin
π
2
2
= (0) cos x − (1) sin x
= − sin x
sin x
π
5. sin( 2 − x)
= sin
3. cos( 2 + x)
cos x − cos
π
2
2
= (1) cos x − (0) sin x
= cos x
6. csc( 2 + x)
sin x
1
=
π
sin( + x)
2
=
sin
π
2
1
cos x + cos
π
2
1
(1) cos x + (0) sin x
1
=
cos x
= sec x
=
7. sec(π + x)
=
8. csc(π − x)
1
=
cos(π + x)
1
cos π cos x − sin π sin x
1
=
(−1) cos x − (0) sin x
1
=
− cos x
= − sec x
=
1
sin(π − x)
1
sin π cos x − cos π sin x
1
=
(0) cos x − ( −1) sin x
1
=
sin x
= csc x
=
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sin x
Trigonometry Lesson 12
Part v
triangle questions
In some questions, you will be given incomplete information, and must use triangles to find
all the trig ratios required by the sum & difference formulas.
Example 1: Given tan x = -
tan y =
5
(In quadrant II) and
12
3
(In quadrant III), find the exact value of sec(x + y)
5
Draw a triangle corresponding to
5 and find the unknown side.
tan x = 12
Draw a triangle corresponding to
3
tan y = and find the unknown side.
5
a 2 + b2 = c2
a 2 + b2 = c2
(−12) 2 + (5) 2 = c 2
(−5) 2 + (−3) 2 = c 2
169 = c 2
c = 13
34 = c 2
c = 34
Now find sin y & cos y
Now find sin x & cos x .
−3
34
−5
cos y =
34
5
sin x =
13
−12
cos x =
13
sin y =
1
cos( x + y )
1
=
cos x cos y − sin x sin y
1
=
⎛ −12 ⎞ ⎛ −5 ⎞ ⎛ 5 ⎞ ⎛ −3 ⎞
⎜
⎟⎜
⎟ − ⎜ ⎟⎜
⎟
⎝ 13 ⎠ ⎝ 34 ⎠ ⎝ 13 ⎠ ⎝ 34 ⎠
sec( x + y ) =
1
60
15
+
13 34 13 34
1
=
75
13 34
=
=
13 34
75
Final Answer
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Trigonometry Lesson 12
Part v
sin x = −
1)
triangle questions
1
(Quadrant III)
2
Find cos( x − y )
3
tan y =
(Quadrant I)
4
cos x =
2)
12
(Quadrant IV)
13
csc y = −
sec x =
3)
Find csc( x + y )
5
(Quadrant III)
2
7
(Quadrant I)
5
Find sin( x − y )
3
cot y = −
(Quadrant II)
4
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Trigonometry Lesson 12
Part v
tan x =
4)
triangle questions
−6
(Quadrant IV)
7
Find sec( x + y )
2
sin y = −
(Quadrant IV)
5
5)
cot x = −5 ( cos x < 0)
Find csc( x − y )
3
tan y =
( sin y > 0)
4
sec x = −
6)
8
( sin x > 0)
7
tan y = −3
Find sec( x − y )
( sin y < 0)
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Trigonometry Lesson 12
Part v
triangle questions
1) First draw out each triangle, then use Pythagoras to find the unknown side:
- 3
Now find the sine & cosine trig ratios for each triangle:
−1
2
− 3
cos x =
2
3
5
4
cos y =
5
sin x =
Use these to evaluate
sin y =
cos( x − y )
cos( x − y )
= cos x cos y + sin x sin y
⎛ − 3 ⎞ ⎛ 4 ⎞ ⎛ −1 ⎞⎛ 3 ⎞
= ⎜⎜
⎟⎟ ⎜ ⎟ + ⎜ ⎟⎜ ⎟
⎝ 2 ⎠ ⎝ 5 ⎠ ⎝ 2 ⎠⎝ 5 ⎠
=
−4 3 3
−
10
10
=
−4 3 − 3
10
2) First draw out each triangle, then use Pythagoras to find the unknown side:
− 21
Now state the sine cosine trig ratios for each triangle:
−5
13
12
cos x =
13
sin x =
Use these to evaluate
csc( x + y )
csc( x + y )
1
=
sin x cos y + cos x sin y
1
=
⎛ −5 ⎞ ⎛ − 21 ⎞ ⎛ 12 ⎞ ⎛ −2 ⎞
⎟ + ⎜ ⎟⎜ ⎟
⎜ ⎟⎜
⎝ 13 ⎠ ⎝ 5 ⎠ ⎝ 13 ⎠ ⎝ 5 ⎠
sin y =
−2
5
cos y =
− 21
5
1
⎛ −5 21 ⎞ ⎛ −24 ⎞
⎜
⎟+⎜
⎟
⎝ 65 ⎠ ⎝ 65 ⎠
1
=
−5 21 − 24
65
65
=
−5 21 − 24
=
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Trigonometry Lesson 12
Part v
triangle questions
3) First draw out each triangle, and use Pythagoras to find the unknown side:
2 6
Now find the sine & cosine trig ratios for each triangle:
4
5
−3
cos y =
5
2 6
7
5
cos x =
7
sin y =
sin x =
Use these to evaluate sin( x − y )
sin( x − y )
= sin x cos y − cos x sin y
⎛ 2 6 ⎞ ⎛ −3 ⎞ ⎛ 5 ⎞ ⎛ 4 ⎞
= ⎜⎜
⎟⎟ ⎜ ⎟ − ⎜ ⎟ ⎜ ⎟
⎝ 7 ⎠⎝ 5 ⎠ ⎝ 7 ⎠⎝ 5 ⎠
−6 6 20
−
35
35
−6 6 − 20
=
35
=
4) First draw out each triangle, and use Pythagoras to find the unknown side:
21
85
Now find the sine & cosine trig ratios for each triangle:
−6
85
7
cos x =
85
sin y =
sin x =
Use these to evaluate
cos y =
−2
5
21
5
sec( x + y )
sec( x + y )
1
cos x cos y − sin x sin y
1
=
⎛
⎞
⎛ 7 ⎞ 21 ⎛ −6 ⎞ ⎛ −2 ⎞
⎟−⎜
⎜
⎟⎜
⎟⎜ ⎟
⎝ 85 ⎠ ⎝ 5 ⎠ ⎝ 85 ⎠ ⎝ 5 ⎠
=
=
1
⎛ 7 21 ⎞ ⎛ 12 ⎞
⎜
⎟−⎜
⎟
⎝ 5 85 ⎠ ⎝ 5 85 ⎠
=
1
7 21 − 12
5 85
=
5 85
7 21 − 12
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Trigonometry Lesson 12
Part v
triangle questions
5) First draw out each triangle, and use Pythagoras to find the unknown side:
cot x < 0 in quadrants II & IV
cos x < 0 in quadrants II & III
Overlap in II
tan y > 0 in quadrants I & III
sin y > 0 in quadrants I & II
Overlap in I
26
Now find the sine & cosine trig ratios for each triangle:
1
26
−5
cos x =
26
3
5
4
cos y =
5
sin x =
Use these to evaluate csc( x − y )
csc( x − y )
1
sin x cos y − cos x sin y
1
=
⎛ 1 ⎞ ⎛ 4 ⎞ ⎛ −5 ⎞ ⎛ 3 ⎞
⎜
⎟⎜ ⎟ − ⎜
⎟⎜ ⎟
⎝ 26 ⎠ ⎝ 5 ⎠ ⎝ 26 ⎠ ⎝ 5 ⎠
=
sin y =
1
⎛ 4 ⎞ ⎛ −15 ⎞
⎜
⎟−⎜
⎟
⎝ 5 26 ⎠ ⎝ 5 26 ⎠
1
=
19
5 26
=
=
5 26
19
6) First draw out each triangle, and use Pythagoras to find the unknown side:
sec x < 0 in quadrants II & III
sin x > 0 in quadrants I & II
Overlap in II
tan y < 0 in quadrants II & IV
sin y < 0 in quadrants III & IV
Overlap in IV
15
10
Now find the sine & cosine trig ratios for each triangle:
−3
10
1
cos y =
10
15
8
−7
cos x =
8
sin y =
sin x =
Use these to evaluate sec( x − y )
sec( x − y )
1
cos x cos y + sin x sin y
1
=
⎛ −7 ⎞ ⎛ 1 ⎞ ⎛ 15 ⎞ ⎛ −3 ⎞
⎟⎜
⎜ ⎟⎜
⎟+⎜
⎟
⎝ 8 ⎠ ⎝ 10 ⎠ ⎝ 8 ⎠ ⎝ 10 ⎠
=
=
1
⎛ −7 ⎞ ⎛ −3 15 ⎞
⎟
⎜
⎟+⎜
⎝ 8 10 ⎠ ⎝ 8 10 ⎠
=
1
−7 − 3 15
8 10
=
8 10
−7 − 3 15
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Trigonometry Lesson 12
Part vI
Double angle Identities
The following double angle identities are frequently used:
cos2A = cos 2 A - sin 2 A
sin2A = 2sinAcosA
cos2A = 2cos 2 A - 1
cos2A = 1 - 2sin 2 A
Example 1: Expand sin 60º using sin2A = 2sinAcosA :
sin 2 A = 2sin A cos A
sin 60D = 2sin 30D cos 30D
What goes here is
always half of the
angle on the left side.
Example 2: Expand cos 90º using cos2A = 2cos 2 A -1
cos 2 A = 2 cos 2 A − 1
cos 90D = 2 cos 2 45D − 1
Example 3: Expand cos
2π
using cos2A = cos 2 A - sin 2 A
3
cos 2 A = cos 2 A − sin 2 A
2π
π
π
= cos 2 − sin 2
cos
3
3
3
Example 4: Expand cos 8x using cos2A =1- 2sin 2 A
cos 2 A = 1 − 2sin 2 A
cos8 x = 1 − 2sin 2 4 x
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Trigonometry Lesson 12
Part vI
Double angle Identities
In the following examples, you must work backwards to condense the identity:
Example 5: Condense: 1- 2sin 2150o
cos 2 A = 1 − 2sin 2 A
cos 300D = 1 − 2sin 2 150D
What goes on the left
side is double the
angle on the right.
Example 6: Condense: 2sin3xcos3x
sin 2 A = 2sin A cos A
sin 6 x = 2sin 3 x cos 3 x
Example 7: Condense: cos 2π - sin 2π
cos 2 A = cos 2 A − sin 2 A
cos 2π = cos 2 π − sin 2 π
Questions:
1) Expand sin 180D using sin 2 A = 2sin A cos A
2) Expand cos
π
3
using cos 2 A = 1 − 2sin 2 A
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Trigonometry Lesson 12
Part vI
Double angle Identities
3) Expand cos16x using cos 2 A = cos 2 A − sin 2 A
4) Expand cos
π
2
using cos 2 A = 2 cos 2 A − 1
5) Condense: 2 cos 2
6) Condense: cos 2
π
6
1
2
2π
−1
3
− sin 2
π
6
1
2
7) Condense: cos 2 x − sin 2 x
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Trigonometry Lesson 12
Part vI
Double angle Identities
1) sin 180D
sin 2 A = 2sin A cos A
sin180D = 2sin 90D cos 90D
2) cos
π
3
cos 2 A = 1 − 2sin 2 A
cos
π
3
= 1 − 2sin 2
π
6
3) cos16x
cos 2 A = cos 2 A − sin 2 A
cos16 x = cos 2 8 x − sin 2 8 x
4) cos
π
2
cos 2 A = 2 cos 2 A − 1
cos
π
2
= 2 cos 2
π
4
−1
2π
−1
3
cos 2 A = 2 cos 2 A − 1
4π
2π
cos
= 2 cos 2
−1
3
3
5) 2 cos 2
6) cos 2
π
− sin 2
π
6
6
2
cos 2 A = cos A − sin 2 A
cos
π
3
= cos 2
π
6
− sin 2
π
6
1
1
2
2
2
cos 2 A = cos A − sin 2 A
1
1
cos x = cos 2 x − sin 2 x
2
2
7) cos 2 x − sin 2 x
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Trigonometry Lesson 12
Part vII
Double angle Proofs
In the following examples, the double-angle identities will be used in completing proofs.
Example 1: Prove sin 2x + cos x = cos x(2sin x + 1)
sin 2 x + cos x
= 2sin x cos x + cos x
= cos x(2sin x + 1)
Example 2: Prove: cos2x = cos 2 x - sin 2 x
Hint: Write cos 2x as cos (x + x)
cos( x + x) = cos x cos x − sin x sin x
cos( x + x) = cos 2 x − sin 2 x
Questions:
1) cos 2 x + cos x = (2 cos x − 1)(cos x + 1)
2) 2 cos 2 x − sin x + 1 = −(4sin x − 3)(sin x + 1)
3) cos 2 x = 2 cos 2 x − 1
4) cos 2 x = 1 − 2sin 2 x
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Trigonometry Lesson 12
Part vII
5)
1 + cos 2 x
= cot x
sin 2 x
Double angle Proofs
6)
7) sin 2 x = 2sin x cos x
9) (sin x + cos x) 2 = 1 + sin 2 x
2
= sec 2 x
1 + cos 2 x
8) cos 2 x − 1 + 2sin x = 2sin x(1 − sin x)
10) sin( x − y ) sin( x + y ) = cos 2 y − cos 2 x
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Trigonometry Lesson 12
Part vII
Double angle Proofs
1) cos 2 x + cos x
= 2 cos 2 x − 1 + cos x
6)
2
= sec 2 x
1 + cos 2 x
2
=
1 + (2 cos 2 x − 1)
2
=
2 cos 2 x
1
=
cos 2 x
= sec 2 x
7)
sin 2 x = sin( x + x)
sin( x + x) = sin x cos x + cos x sin x
= 2 cos 2 x + cos x − 1
= (2 cos x − 1)(cos x + 1)
2) 2 cos 2 x − sin x + 1
= 2(1 − 2sin 2 x) − sin x + 1
= 2 − 4sin 2 x − sin x + 1
= −4sin 2 x − sin x + 3
= −(4sin x + sin x − 3)
= −(4sin x − 3)(sin x + 1)
2
= 2sin x cos x
3) cos 2 x = cos( x + x)
cos( x + x) = cos x cos x − sin x sin x
8) cos 2 x − 1 + 2sin x
= (1 − 2sin 2 x) − 1 + 2sin x
= −2sin 2 x + 2sin x
cos 2 x = cos 2 x − sin 2 x
= 2sin x(− sin x + 1)
cos 2 x = cos 2 x − (1 − cos 2 x)
= 2sin x(1 − sin x)
cos 2 x = cos x − 1 + cos x
2
2
cos 2 x = 2 cos 2 x − 1
9)
(sin x + cos x) 2
= sin 2 x + 2sin x cos x + cos 2 x
4) cos 2 x = 1 − 2sin x
cos( x + x) = cos x cos x − sin x sin x
2
= sin 2 x + cos 2 x + 2sin x cos x
= 1 + 2sin x cos x
cos 2 x = cos 2 x − sin 2 x
cos 2 x = (1 − sin 2 x) − sin 2 x
cos 2 x = 1 − sin 2 x − sin 2 x
10)
cos 2 x = 1 − 2sin x
sin( x − y )sin( x + y) = cos2 y − cos2 x
2
1 + cos 2 x
5)
sin 2 x
1 + (2 cos 2 x − 1)
=
2sin x cos x
2 cos 2 x
=
2sin x cos x
cos x
=
sin x
= cot x
= [sin x cos y − cos x sin y ][sin x cos y + cos x sin y ]
= ( sin x cos y ) + sin x cos y cos x sin y − cos x sin y sin x cos y − (cos x sin y) 2
2
= sin 2 x cos 2 y − cos2 x sin 2 y
= sin 2 x cos 2 y − (1 − sin 2 x)(1 − cos 2 y )
= sin 2 x cos 2 y − [1 − cos 2 y − sin 2 x + sin 2 x cos2 y]
= sin 2 x cos 2 y − 1 + cos2 y + sin 2 x − sin 2 x cos 2 y
= cos 2 y + sin 2 x − 1
= cos 2 y − (1 − sin 2 x )
= cos 2 y − cos 2 x
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