Question

(sin 79° cos 11° + cos 79° sin 11°) = ?
(a) $\frac{1}{\sqrt{2}}$
(b) $\frac{1}{2}$
(c) 0
(d) 1

Answer 1

$$\begin{gathered} (d)\; 1 \\ \text{We have:}(\sin {79}^0\cos {11}^0+\cos {79}^0\sin {11}^0) \\ =\sin {79}^0\cos ({90}^0-{79}^0)+\cos {79}^0\sin ({90}^0-{79}^0) \\ =\sin {79}^0\sin {79}^0\text{+}\cos {79}^0\cos {79}^0[\because \cos ({90}^0-\theta )=\sin \theta \text{and}\sin ({90}^0-\theta )\text{=cos}\theta ] \\ ={\sin }^2{79}^0+{\cos }^2{79}^0 \\ =1 \end{gathered}$$