Question

31
B. Prove that (1+cosπ/8) (1+cos3π/8) (1+cos5π/8) (1+cos7π/8) =1/8
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Answer 1

Proof:

Let us derive each of the factor terms in its simplest form before we procced to the proof.

• 1 + cos(π/8)

• 1 + cos(3π/8)

= 1 + cos(π/2 - π/8)

= 1 + sin(π/8)

• 1 + cos(5π/8)

= 1 + cos(π/2 + 5π/8)

= 1 - sin(π/8)

• 1 + cos(7π/8)

= 1 + cos(π - π/8)

= 1 - cos(π/8)

Now, L.H.S.

= {1 + cos(π/8)} {1 + cos(3π/8)} {1 + cos(5π/8)} {1 + cos(7π/8)}

= {1 + cos(π/8)} {1 + sin(π/8)} {1 - sin(π/8)} {1 - sin(π/8)}

= {1 - cos²(π/8)} {1 - sin²(π/8)}

= sin²(π/8) cos²(π/8)

= 1/4 * {2 sin(π/8) cos(π/8)}²

= 1/4 * sin²(2 * π/8)

= 1/4 * sin²(π/4)

= 1/4 * (1/√2)²

= 1/4 * 1/2

= 1/8

= R.H.S.

Hence, proved.

Answer 2

L.H.S = R.H.S

Step-by-step explanation:

In this qyestion

We have given that

(1 + $cos[\frac{\pi}{8}]$)(1 + $cos[\frac{3\pi}{8}]$)(1 + $cos[\frac{5\pi}{8}]$)(1 + $cos[\frac{7\pi}{8}]$) = $\frac{1}{8}$

Let L.H.S

(1 + $cos[\frac{\pi}{8}]$)(1 + $cos[\frac{3\pi}{8}]$)(1 + $cos[\frac{5\pi}{8}]$)(1 + $cos[\frac{7\pi}{8}]$)

(1 + $cos[\frac{\pi}{8}]$)(1 + $cos[\frac{\pi}{2} - \frac{\pi}{8}]$)(1 + $cos[\frac{\pi}{2} + \frac{\pi}{8}]$)(1 + $cos[\pi - \frac{\pi}{8}]$)

(1 + $cos[\frac{\pi}{8}]$)(1 + $sin[\frac{\pi}{8}]$)(1 - $sin[\frac{\pi}{8}]$)(1 - $cos[\frac{\pi}{8}]$)

1 - $cos^2[\frac{\pi}{8}]$)(1 - $sin^2[\frac{\pi}{8}]$)

$sin^{2}\frac{\pi}{8}\times cos^{2}(\frac{\pi}{8})$  

$\frac{1}{4}[(2sin\frac{\pi}{8})\times cos\frac{\pi}{8}]^2$

$\frac{1}{4}sin^2\frac{2\pi}{8}$

$\frac{1}{4}sin^2\frac{\pi}{4}$

[tex]\frac{1}{4}\times [\frac{1}{\sqrt{2}}]^2

[tex]\frac{1}{8}[/frac]

R.H.S

Hense, L.H.S = R.H.S