8.1 Q-6

Exercise 8.1 — Question 6

Problem

Simplify the following using trigonometric identities:

$\sin 138^{\circ }\cos 46^{\circ }-\cos 138^{\circ }\sin 46^{\circ }$

Solution

Recall the sine difference formula (fundamental law of trigonometry):

$\sin (\alpha -\beta )=\sin \alpha \cos \beta -\cos \alpha \sin \beta$

Comparing $\sin 138^{\circ }\cos 46^{\circ }-\cos 138^{\circ }\sin 46^{\circ }$ with the right-hand side, we identify:

$\alpha =138^{\circ },\beta =46^{\circ }$

Therefore:

$\sin 138^{\circ }\cos 46^{\circ }-\cos 138^{\circ }\sin 46^{\circ }=\sin (138^{\circ }-46^{\circ })=\sin 92^{\circ }$

Key Identity Used

$\boxed{\sin (\alpha -\beta )=\sin \alpha \cos \beta -\cos \alpha \sin \beta }$

This identity allows any expression of the form $\sin A\cos B-\cos A\sin B$ to be written as a single sine term $\sin (A-B)$.