NECO 2023 · Paper 1 · Q11

Given that $\cos A = \frac45$ and $\cos B = \frac{12}{13}$ , find $\sin(A + B)$ if $A$ and $B$ are both acute.

Worked solution (try it first)

$A$ is acute and $\cos A = \frac45$ : a 3-4-5 triangle gives $\sin A = \frac35$ .
Likewise a 5-12-13 triangle gives $\sin B = \frac{5}{13}$ .
Use the compound-angle formula: $\sin(A + B) = \sin A\cos B + \cos A\sin B$ .
That is $\frac35 \times \frac{12}{13} + \frac45 \times \frac{5}{13} = \frac{36}{65} + \frac{20}{65}$ .
So $\sin(A + B) = \frac{56}{65}$ , option E.