Question

If $\sin {47}^o+\sin {61}^o-\sin {11}^o-\sin {25}^o=$

• A. $\sin 7^o$
• B. $\cos 7^o$
• C. $\tan 7^o$
• D. $\sin {14}^o$

Answer 1

The correct option is B $\cos 7^o$

$(sin{47}^o-sin{25}^o)+(sin{61}^o-sin{11}^o)$

= (2 cos 36 x sin 11) + (2 cos 36 x sin 25)

= 2 cos 36 ( sin + sin 25)

= 4 cos 36 x sin 18 x cos 7 ..........(i)

sin 18 = $\frac{\sqrt{5}-1}{4}$

and cos 36 = $\frac{\sqrt{5}+1}{4}$

Put the values in eq (i)

= 4 $\left(\frac{\sqrt{5}+1}{4}\right)$ x ($\frac{\sqrt{5}-1}{4}$) x Cos $7^o$

By solving, we get

= 1 x cos 7

= Cos 7