Question

Cos 60° • Cos 42° • Cos 66° • Cos 78° = $\frac{1}{16}$

Solve it.......


Help me to solve this question .......​

Answer 1

Answer:

LHS

$= \cos(60) \cos(42) \cos(66) \cos(78)$

$= \frac{1}{2} \times \frac{1}{2} \cos(42) ({{2 \cos(66) \cos(78) }})$

$= \frac{1}{4} \cos(42) ( \cos(144) + \cos(12)$

Answer 2

Answer:

LHS

= \cos(60) \cos(42) \cos(66) \cos(78)=cos(60)cos(42)cos(66)cos(78)

= \frac{1}{2} \times \frac{1}{2} \cos(42) ({{2 \cos(66) \cos(78) }})=

2

1

×

2

1

cos(42)(2cos(66)cos(78))

= \frac{1}{4} \cos(42) ( \cos(144) + \cos(12)=

4

1

cos(42)(cos(144)+cos(12)