Question

$\sin 47°+\sin 61°-\sin 11°-\sin 25°$ is equal to:

• A. $\sin 7°$
• B. $\cos 7°$
• C. $\sin 36°$
• D. $\cos 36°$

Answer 1

The correct option is B

$\cos 7°$

Explanation for correct option:

Simplifying the equations using trigonometric identities:

The given expression is: $\sin 47°+\sin 61°-\sin 11°-\sin 25°$

Simplifying the above expression:

$$\begin{gathered} =\sin 47°+\sin 61°-\sin 11°-\sin 25° \\ =(\sin 47°–\sin 25°)+(\sin 61°–\sin 11°) \end{gathered}$$

Applying the formula $\mathbf{sin}\mathbf{}\mathbf{x}\mathbf{}\mathbf{–}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{y}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{}\mathbf{sin}\left(\frac{x–y}{2}\right)\mathbf{cos}\left(\frac{x+y}{2}\right)$:

$$\begin{gathered} =2\left[\sin \left(\frac{47°-25°}{2}\right)\cos \left(\frac{47°+25°}{2}\right)\right]+2\left[\sin \left(\frac{61°-11°}{2}\right)\cos \left(\frac{61°+11°}{2}\right)\right] \\ =2\sin 11°\cos 36°+2\sin 25°\cos 36° \\ =2\cos 36°(\sin 11°+\sin 25°) \end{gathered}$$

Applying the formula $\mathbf{sin}\mathbf{}\mathbf{x}\mathbf{}\mathbf{+}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{y}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{}\mathbf{sin}\left(\frac{x+y}{2}\right)\mathbf{cos}\left(\frac{x-y}{2}\right)$:

$$\begin{gathered} =2\cos 36°\left(2\sin \left(\frac{11°+25°}{2}\right)\cos \left(\frac{11°-25°}{2}\right)\right) \\ =2\cos 36°(2\sin 18°\cos 7°) \\ =4\cos 36°\sin 18°\cos 7° \\ =4\times \frac{\sqrt{5}–1}{4}\times \frac{\sqrt{5}+1}{4}\times \cos 7°\left[\because \cos 36°=\frac{\sqrt{5}–1}{4}\&\sin 18°=\frac{\sqrt{5}+1}{4}\right] \\ =\frac{5-1}{4}\times \cos 7° \\ =\frac{4}{4}\times \cos 7° \\ =\cos 7° \end{gathered}$$

Therefore, the correct answer is option (B).