Question

The value of $1+\cos 56°+\cos 58°-\cos 66°$is

• A. $4\cos 28°\cos 29°\sin 33°$
• B. $\cos 28°\cos 29°\sin 33°$
• C. $4\cos 28°\sin 29°\cos 33°$
• D. $4\cos 28°\sin 29°\sin 33°$

Answer 1

The correct option is A

$4\cos 28°\cos 29°\sin 33°$

Explanation for the correct option.

Find the value.

Given: $1+\cos 56°+\cos 58°-\cos 66°$

Formula used $1-\cos 2\mathrm{A}=2{\sin }^2\mathrm{A},\mathrm{cosA}+\mathrm{cosB}=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$

$$\begin{gathered} \Rightarrow 1-\cos 66°+\cos 58°+\cos 56° \\ =2{\sin }^{2}33°+2\cos \left(\frac{58°+56°}{2}\right)\cos \left(\frac{58°-56°}{2}\right) \\ =2{\sin }^{2}33°+2\cos 57°\cos 1° \\ =2{\sin }^{2}33°+2\cos (90-33)°\cos (90-89)° \\ =2{\sin }^{2}33+2\sin 33°\sin 89° \\ =2\sin 33°(\sin 33°+\sin 89°) \\ =2\sin 33°(2\sin 61°\cos 28°)\left[\mathrm{sinA}+\mathrm{sinB}=2\sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\right] \\ =2\sin 33°\left(2\sin (90-29)°\cos 28°\right) \\ =2\sin 33°\times 2\cos 29°\cos 28° \\ =4\sin 33°\cos 29°\cos 28° \end{gathered}$$

Hence $o\mathrm{ption}\left(\mathrm{A}\right)$ is the correct answer.