# 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 intersects the unit circle at *P*(cos *A*, sin *A*) and the terminal side of 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*:

Also Read:

Now rotate *OQP *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 *Q*’*OP*’ 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 *P*’*Q*’.

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°.

Solution:

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*).

Solution:

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*)

**SUMMARY**

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*