Determine of Cosine (A – B )

We can prove that cos (A B) = cos A – cos B is not an identity by finding one pair of values of A and B for which each side of the equation is defined and the equation is false. For example, if, in degree measure, A = 90° and B = 60°,

In order to write an identity that expresses cos (A B) in terms of function values of A and B,we will use the relationship between the unit circle and the sine and cosine of an angle. Let A and B be any two angles in standard position. The terminal side of \angle A intersects the unit circle at P(cos A, sin A) and the terminal side of \angle B intersects the unit circle at Q(cos B, sin B). Use the distance formula to express PQ in terms of sin A, cos A, sin B, and cos B:

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Now rotate \DeltaOQP through an angle of –B, that is, an angle of B units in the clockwise direction so that the image of P is P’ and the image of Q is Q’. Q’ is a point on the x-axis whose coordinates are (1, 0). Angle QOP’ is an angle in standard position whose measure is (A – B). Therefore, the coordinates of P’ are (cos (A B), sin(A B)). Use the distance formula to find PQ’.

Distance is preserved under a rotation. Therefore,

At the beginning of this section we showed that:

Does the identity that we proved make it possible to find cos (90° 2 60°)? We can check.

Example 1:

Use (60° – 45°) = 15° to find the exact value of cos 15°.


cos (A B) = cos A cos B + sin A sin B

cos (60° – 45°) = cos 60° cos 45° + sin 60° sin 45°

Note: Example 1 shows that cos 15° is an irrational number.

Also Read : Proving the identity of trigonometric

Cosine of (90° B)

The cofunction relationship between cosine and sine can be proved using cos (A B). Use the identity for cos (A B) to express cos (90° – B) in terms of a function of B.

Let A = 90°.

cos (A B) = cos A cos B + sin A sin B

cos (90° – B) = cos 90° cos B + sin 90° sin B

cos (90° – B) = 0 cos B + 1 sin B

cos (90° – B) = sin B

This is an identity, a statement true for all values of B.

Example 2:

Use the identity cos (90° = B) = sin B to find sin (90° – B).


Let B = (90° – A).

cos (90° – B) = sin B

cos (90° – (90° – A)) = sin (90° – A)

cos (90° – 90° + A) = sin (90° – A)

cos A = sin (90° – A)


We have proved the following identities:

cos (A B) = cos A cos B + sin A sin B

cos (90° B) = sin B

sin (90° A) = cos A


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