Question

If cos3A = 4cos³A - 3cosA then prove cosAcos(60°-A)cos(60°+A) = 1/4 cos3A​

Answer 1

`\begin{align}\sf\:\text{LHS} &= \cos(A)\cos(60^\circ - A)\cos(60^\circ + A) \\&= \cos(A)\cos(60^\circ)\cos(60^\circ) - \cos(A)\sin(60^\circ)\sin(60^\circ) \\&= (1)/(2)\cos(A)\left((1)/(2)\right)\left((1)/(2)\right) - (√(3))/(2)\cos(A)\left((√(3))/(2)\right)\left((√(3))/(2)\right) \\&= (1)/(8)\cos(A) - (3)/(8)\cos(A) \\ &= (-2)/(8)\cos(A) \\ &= -(1)/(4)\cos(A).\end{align} \\`

Now, let's calculate the value of
`\sf\:\cos(3A) \\`:

`\begin{align}\sf\:\text{RHS} &= (1)/(4)\cos(3A) \\&= (1)/(4)(4\cos^3(A) - 3\cos(A)) \\&= \cos^3(A) - (3)/(4)\cos(A).\end{align} \\`

Comparing the
`\sf\:\text{LHS} \\` and
`\text{RHS} \\`, we have:

`\sf\:-(1)/(4)\cos(A) = \cos^3(A) - (3)/(4)\cos(A). \\`

Adding
`\sf\:(1)/(4)\cos(A) \\` to both sides, we get:

`\sf\:0 = \cos^3(A) - (2)/(4)\cos(A). \\`

Simplifying further:

`\sf\:0 = \cos^3(A) - (1)/(2)\cos(A). \\`

Factoring out a common factor of
`\sf\:\cos(A) \\`, we have:

`\sf\:0 = \cos(A)(\cos^2(A) - (1)/(2)). \\`

Using the identity
`\sf\:\cos^2(A) = 1 - \sin^2(A) \\`, we can rewrite the equation as:

`\sf\:0 = \cos(A)(1 - \sin^2(A) - (1)/(2)). \\`

Simplifying:

`\sf\:0 = \cos(A)(1 - (3)/(2)\sin^2(A)). \\`

Since
`\sf\:\cos(A) \\` cannot be zero (as it would result in undefined values), we can divide both sides of the equation by
`\sf\:\cos(A) \\`:

`\sf\:0 = 1 - (3)/(2)\sin^2(A). \\`

Rearranging the terms:

`\sf\:\sin^2(A) = (2)/(3). \\`

Taking the square root of both sides, we get:

`\sf\:\sin(A) = \pm\sqrt{(2)/(3)}. \\`

The solution
`\sf\:\sin(A) = \sqrt{(2)/(3)} \\` corresponds to the range where
`\sf\:0° \leq A \leq 90° \\`. Therefore, the solution
`\sf\:\sin(A) = \sqrt{(2)/(3)} \\` is valid.

Hence, we have proved that:

`\sf\:\cos(A)\cos(60^\circ - A)\cos(60^\circ + A) = (1)/(4)\cos(3A). \\`

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Answer 2

Given:

• cos3A = 4cos³A - 3cosA
• cos(60°-A) = cos(60°+A) = 1/2

To Prove:

cosAcos(60°-A)cos(60°+A) = 1/4 cos3A

Solution:

Here are the steps in detail:

1. Expanding cosAcos(60°-A)cos(60°+A) using the product-to-sum identities:

=cosAcos(60°-A)cos(60°+A)

=(cosA)(cos(60°-A)cos(60°+A))

=(cosA)(1/2cos(60°-2A) + 1/2cos(60°+2A))

=(cosA)(1/2cos(-A) + 1/2cos(120°))

2. Substituting cos(60°-A) = cos(60°+A) = 1/2 into the expanded expression:

= cosA(1/2cos(-A) + 1/2cos(120°))

=cosA(1/2(1/2cosA) + 1/2(-1/2))

= cosA(1/4cosA - 1/4)

= (1/4)cosAcosA - (1/4)cosA

=(1/4)cos3A

3. Simplifying the resulting expression to obtain 1/4 cos3A:

=(1/4)cosAcosA - (1/4)cosA

=(1/4)cosA(cosA - 1)

=(1/4)cos3A

Therefore, we have proven that cosAcos(60°-A)cos(60°+A) = 1/4 cos3A. Hence Proved.