Question

Prove that cos6.cos42.cos66.cos78=1/16.
Someone please help me out.

Answer 1

I hope it is clear ...........

Answer 2

Answer:

L. H.S = R. H.S

Step-by-step explanation:

In this question

$cos6\times cos42\times cos66\times cos78$ = 1/16

Let, R.H.S

$cos6\times cos42\times cos66\times cos78$

Multiple and divided by 2

$\frac{1}{2}\times (2cos42\times cos78)\times cos6\times cos66$

$\frac{1}{2}\times (cos120 + cos36)\times cos6\times cos66$

$\frac{1}{2}\times (-sin30\ + cos36)\times cos6\times cos66$

$\frac{1}{2}\times (\frac{-1}{2} + \frac{\sqrt5 + 1}{4} )\times cos6\times cos66$

Again multiple and divided by 2

$\frac{1}{4}\times (\frac{-1}{2} + \frac{\sqrt5 + 1}{4} )\times (2cos6\times cos66)$

$\frac{1}{4}\times (\frac{-1}{2} + \frac{\sqrt5 + 1}{4} )\times (cos72 + cos60)$

$\frac{1}{4}\times (\frac{-1}{2} + \frac{\sqrt5 + 1}{4} )\times ( \frac{\sqrt5 - 1}{4} + \frac{1}{2})$

$\frac{1}{4}\times (\frac{-2 + \sqrt5 + 1}{4} )\times ( \frac{\sqrt5 - 1 + 2}{4})$

$\frac{1}{4}\times (\frac{-1 + \sqrt5}{4} )\times ( \frac{\sqrt5 + 1}{4})$

$\frac{1}{4\times 4\times 4}\times (\sqrt5)^2 - 1^2$

$\frac{1}{4\times 4\times 4}\times (5 - 1)$

$\frac{1}{4\times 4\times 4}\times 4$

$\frac{1}{16}$

L.H.S