Question

Prove that sin $({40}^{\circ }+\theta )$ cos$({10}^{\circ }+\theta )$ - cos $({40}^{\circ }+\theta )$ sin $({10}^{\circ }+\theta )$ = $\frac{1}{2}$

Answer 1

We know that
$\text{sin x cos y - cos x sin y = sin (x - y)}$

Putting x = $({40}^{\circ }+\theta )$ and y = $({10}^{\circ }+\theta )$ in this identity, we get

=$sin({40}^{\circ }+\theta )$ $cos({10}^{\circ }+\theta )$ - $cos({40}^{\circ }+\theta )sin({10}^{\circ }+\theta )$

= $sin\left[\right({40}^{\circ }+\theta )$ - $({10}^{\circ }+\theta )$]

= $sin{30}^{\circ }=\frac{1}{2}$