Question

1fA is not an integral multiple of pi , prove that cos A.cos2A.cos 4A.cos 8A=
sin 16A/16sinA​

Answer 1

Answer:

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Answer 2

Let's start by using the following trigonometric identity:

sin 2θ = 2 cos θ sin θ

Using this identity, we can write:

cos A cos 2A = (1/2)(cos 3A + cos A)

cos 4A cos 8A = (1/2)(cos 12A + cos 4A)

Now, substituting these expressions into the given expression, we get:

cos A cos 2A cos 4A cos 8A = [(1/2)(cos 3A + cos A)][(1/2)(cos 12A + cos 4A)]

Expanding this expression, we get:

cos A cos 2A cos 4A cos 8A = (1/4)[cos 3A cos 12A + cos A cos 12A + cos 3A cos 4A + cos A cos 4A]

Using the identity cos x cos y = (1/2)(cos(x+y) + cos(x-y)), we can simplify this expression further:

cos A cos 2A cos 4A cos 8A = (1/8)[cos 15A + cos 9A + cos 7A + cos A]

Now, let's use the following identity:

sin 2θ sin 4θ sin 8θ sin 16θ = (1/16) sin 32θ

We can rewrite the denominator of the given expression as:

16 sin A = 2(2 sin A) = 2 sin 2A

= 2(2 sin 2A) = 4 sin 4A

= 4(2 sin 4A) = 8 sin 8A

= 8(2 sin 8A) = 16 sin 16A

Therefore, the given expression can be rewritten as:

cos A cos 2A cos 4A cos 8A = (1/8)[cos 15A + cos 9A + cos 7A + cos A]

= (1/16)[2(cos 15A + cos 9A + cos 7A + cos A)](16 sin A)

= (1/16) [2(cos 15A + cos 9A + cos 7A + cos A)](2 sin 2A)(4 sin 4A)(8 sin 8A)

= (1/16) sin 16A


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