Question

Which of the following is correct?

I : ${\cos }^2{76}^0+{\cos }^2{16}^0-\cos {76}^{0}cos{16}^0=\frac{3}{4}$
II : $\cos 6^0\sin {24}^0\cos {72}^0=\frac{1}{8}$

• A. only I is true
• B. only II is true
• C. Both I and II are true
• D. Neither I nor II are true

Answer 1

Answer: (C)

Both I and II are true

1. $(\cos 76{)}^2+(\cos 16{)}^2-\cos 76\cos 16$

$\Rightarrow (\cos 76+\cos 16{)}^2-3\cos 76\cos 16$

Using formula 1 for $\cos C+\cos D$, we get

$\Rightarrow (2\cos 46\cos 30{)}^2-3\cos 76\cos 16$

Putting value of $\cos 30$ and solving we get

$=3(\cos 46{)}^2-3\cos 76\cos 16$

Using formula 2 for $2\cos A\cos B$, we get

$=3(\cos 46{)}^2-\frac{3(\cos 60+\cos 92)}{2}$

Putting value of $\cos 60$ and solving, we get

$=3(\cos 46{)}^2-\frac{3\left(\cos 92\right)}{2}-\frac{3}{4}$

Using formula for $\cos 2A=2(\cos A{)}^2-1$, we get

$=3(\cos 46{)}^2-3[2(\cos 46{)}^2-1]/2-3/4$

$=3(\cos 46{)}^2-3(\cos 46{)}^2+\frac{3}{2}-\frac{3}{4}$

$=\frac{3}{4}$

2. $\cos 6\sin 24\cos 72$

$=\cos 6.\sin 24.\cos (90-18)$

$=1/2\left[2cos6.sin24\right]\left(sin18\right)$

Using formula 2 for $2\cos A\sin B$, we get

$=\frac{1}{2}\left[\sin \right(6+24)-\sin (6-24\left)\right]\sin 18$

$=\frac{1}{2}[\sin 30.\sin 18+\sin 18.\sin 18]$

Using formula 4 for $\sin 18$, we get

$=\frac{1}{2}[\frac{1}{2}\times \frac{(\sqrt{5}-1)}{2})+{\frac{(\sqrt{5}-1)}{2^2}}^2]$

After simplyfing we get,

$=\frac{1}{8}$

Hence, the correct answer is option C.