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Trigonometry - grade 12

Compound angle identities

Derivation of sin ( α + β )

We have, for any angles α and β , that

sin ( α + β ) = sin α cos β + sin β cos α

How do we derive this identity? It is tricky, so follow closely.

Suppose we have the unit circle shown below. The two points L ( a , b ) and K ( x , y ) are on the circle.

We can get the coordinates of L and K in terms of the angles α and β . For the triangle L O K , we have that

sin β = b 1 b = sin β cos β = a 1 a = cos β

Thus the coordinates of L are ( cos β ; sin β ) . In the same way as above, we can see that the coordinates of K are ( cos α ; sin α ) . The identity for cos ( α - β ) is now determined by calculating K L 2 in two ways. Using the distance formula (i.e. d = ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2 or d 2 = ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2 ), we can find K L 2 :

K L 2 = ( cos α - cos β ) 2 + ( sin α - sin β ) 2 = cos 2 α - 2 cos α cos β + cos 2 β + sin 2 α - 2 sin α sin β + sin 2 β = ( cos 2 α + sin 2 α ) + ( cos 2 β + sin 2 β ) - 2 cos α cos β - 2 sin α sin β = 1 + 1 - 2 ( cos α cos β + sin α sin β ) = 2 - 2 ( cos α cos β + sin α sin β )

The second way we can determine K L 2 is by using the cosine rule for K O L :

K L 2 = K O 2 + L O 2 - 2 · K O · L O · cos ( α - β ) = 1 2 + 1 2 - 2 ( 1 ) ( 1 ) cos ( α - β ) = 2 - 2 · cos ( α - β )

Equating our two values for K L 2 , we have

2 - 2 · cos ( α - β ) = 2 - 2 ( cos α cos β + sin α · sin β ) cos ( α - β ) = cos α · cos β + sin α · sin β

Now let α 90 - α . Then

cos ( 90 - α - β ) = cos ( 90 - α ) cos β + sin ( 90 - α ) sin β = sin α · cos β + cos α · sin β

But cos ( 90 - ( α + β ) ) = sin ( α + β ) . Thus

sin ( α + β ) = sin α · cos β + cos α · sin β

Derivation of sin ( α - β )

We can use

sin ( α + β ) = sin α cos β + cos α sin β

to show that

sin ( α - β ) = sin α cos β - cos α sin β

We know that

sin ( - θ ) = - sin ( θ )

and

cos ( - θ ) = cos θ

Therefore,

sin ( α - β ) = sin ( α + ( - β ) ) = sin α cos ( - β ) + cos α sin ( - β ) = sin α cos β - cos α sin β

Derivation of cos ( α + β )

We can use

sin ( α - β ) = sin α cos β - sin β cos α

to show that

cos ( α + β ) = cos α cos β - sin α sin β

We know that

sin ( θ ) = cos ( 90 - θ ) .

Therefore,

cos ( α + β ) = sin ( 90 - ( α + β ) ) = sin ( ( 90 - α ) - β ) ) = sin ( 90 - α ) cos β - sin β cos ( 90 - α ) = cos α cos β - sin β sin α

Derivation of cos ( α - β )

We found this identity in our derivation of the sin ( α + β ) identity. We can also use the fact that

sin ( α + β ) = sin α cos β + cos α sin β

to derive that

cos ( α - β ) = cos α cos β + sin α sin β

As

cos ( θ ) = sin ( 90 - θ ) ,

we have that

cos ( α - β ) = sin ( 90 - ( α - β ) ) = sin ( ( 90 - α ) + β ) ) = sin ( 90 - α ) cos β + cos ( 90 - α ) sin β = cos α cos β + sin α sin β

Derivation of sin 2 α

We know that

sin ( α + β ) = sin α cos β + cos α sin β

When α = β , we have that

sin ( 2 α ) = sin ( α + α ) = sin α cos α + cos α sin α = 2 sin α cos α = sin ( 2 α )

Derivation of cos 2 α

We know that

cos ( α + β ) = cos α cos β - sin α sin β

When α = β , we have that

cos ( 2 α ) = cos ( α + α ) = cos α cos α - sin α sin α = cos 2 α - sin 2 α = cos ( 2 α )

However, we can also write

cos 2 α = 2 cos 2 α - 1

and

cos 2 α = 1 - 2 sin 2 α

by using

sin 2 α + cos 2 α = 1 .

The cos 2 α Identity

Use

sin 2 α + cos 2 α = 1

to show that:

cos 2 α = 2 cos 2 α - 1 1 - 2 sin 2 α

Problem-solving strategy for identities

The most important thing to remember when asked to prove identities is:

Trigonometric Identities

When proving trigonometric identities, never assume that the left hand side is equal to the right hand side. You need to show that both sides are equal.

A suggestion for proving identities: It is usually much easier simplifying the more complex side of an identity to get the simpler side than the other way round.

Prove that sin 75 = 2 ( 3 + 1 ) 4 without using a calculator.

  1. We only know the exact values of the trig functions for a few special angles ( 30 , 45 , 60 , etc.). We can see that 75 = 30 + 45 . Thus we can use our double-angle identity for sin ( α + β ) to express sin 75 in terms of known trig function values.

  2. sin 75 = sin ( 45 + 30 ) = sin ( 45 ) cos ( 30 ) + cos ( 45 ) sin ( 30 ) = 1 2 · 3 2 + 1 2 · 1 2 = 3 + 1 2 2 = 3 + 1 2 2 × 2 2 = 2 ( 3 + 1 ) 4

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Source:  OpenStax, Math 1508 (lecture) readings in precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11354/1.1
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