Question

(sin 43°cos 47° + cos 43°sin 47°) = ?
(a) sin 4°
(b) cos 4°
(c) 1
(d) 0

Answer 1

$$\begin{gathered} \text{(c) 1} \\ \text{We have:} \\ (\sin {43}^0\cos {47}^0+\cos {43}^0\sin {47}^0) \\ =\sin {43}^0\cos ({90}^0-{43}^0)+\cos {43}^0\sin ({90}^0-{43}^0) \\ =\sin {43}^0\sin {43}^0+\cos {43}^0\cos {43}^0[\because \cos ({90}^0-\theta )=\sin \theta \text{and}\sin ({90}^0-\theta )=\cos \theta ] \\ ={\sin }^2{43}^0+{\cos }^2{43}^0 \\ =1 \end{gathered}$$