Question

If $sin A = \frac{1}{2}$ and $tan B = 3$ , then verify that $cos (A + B) = cos A cos B - sin A sin B$ .

Approach

Verify the identity $cos (A + B) = cos A cos B - sin A sin B$ with $A = 3 0^{\circ}, B = 6 0^{\circ}$ . LHS: $cos 9 0^{\circ} = 0$ . RHS: $\frac{3}{2} \cdot \frac{1}{2} - \frac{1}{2} \cdot \frac{3}{2} = 0$ . Both sides equal, identity verified.

Step-by-step working

1
$sin A = \frac{1}{2} \Rightarrow A = 3 0^{\circ}$ , $tan B = 3 \Rightarrow B = 6 0^{\circ}$ .
1 mark
2
LHS: $cos (3 0^{\circ} + 6 0^{\circ}) = cos 9 0^{\circ} = 0$ . RHS: $cos 3 0^{\circ} cos 6 0^{\circ} - sin 3 0^{\circ} sin 6 0^{\circ} = \frac{3}{2} \cdot \frac{1}{2} - \frac{1}{2} \cdot \frac{3}{2} = 0$ . LHS=RHS.
1 mark

Concepts used

trigonometric identitycompound angleverifycos

Approach

Step-by-step working

Concepts used

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