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If sin(A/3) = 1/2(a+1/a), prove that: sinA = -1/2(a^3 + 1/a^3)

please help me w/ this
dibas asked 2 years ago·

Here u go:

Solution:

Given: sin⁡A3=12(a+1a)\sin\frac{A}{3} = \frac{1}{2}\left(a+\frac{1}{a}\right)

To prove: sin⁡A=−12(a3+1a3)\sin A = -\frac{1}{2}\left(a^3 + \frac{1}{a^3}\right)

We'll use the triple angle formula for sine:

sin⁡(3θ)=3sin⁡θ−4sin⁡3θ\sin(3\theta) = 3\sin\theta - 4\sin^3\theta

Let θ=A3\theta = \frac{A}{3}, then:

sin⁡A=sin⁡(3×A3)=3sin⁡A3−4sin⁡3A3\sin A = \sin\left(3 \times \frac{A}{3}\right) = 3\sin\frac{A}{3} - 4\sin^3\frac{A}{3}

Substituting the given value sin⁡A3=12(a+1a)\sin\frac{A}{3} = \frac{1}{2}\left(a+\frac{1}{a}\right):

sin⁡A=3×12(a+1a)−4×[12(a+1a)]3\sin A = 3 \times \frac{1}{2}\left(a+\frac{1}{a}\right) - 4 \times \left[\frac{1}{2}\left(a+\frac{1}{a}\right)\right]^3

=32(a+1a)−4×18(a+1a)3= \frac{3}{2}\left(a+\frac{1}{a}\right) - 4 \times \frac{1}{8}\left(a+\frac{1}{a}\right)^3

=32(a+1a)−12(a+1a)3= \frac{3}{2}\left(a+\frac{1}{a}\right) - \frac{1}{2}\left(a+\frac{1}{a}\right)^3

Now, let's expand (a+1a)3\left(a+\frac{1}{a}\right)^3:

(a+1a)3=a3+3a2×1a+3a×1a2+1a3\left(a+\frac{1}{a}\right)^3 = a^3 + 3a^2 \times \frac{1}{a} + 3a \times \frac{1}{a^2} + \frac{1}{a^3}

=a3+3a+3a+1a3= a^3 + 3a + \frac{3}{a} + \frac{1}{a^3}

=a3+1a3+3(a+1a)= a^3 + \frac{1}{a^3} + 3\left(a + \frac{1}{a}\right)

Substituting back:

sin⁡A=32(a+1a)−12[a3+1a3+3(a+1a)]\sin A = \frac{3}{2}\left(a+\frac{1}{a}\right) - \frac{1}{2}\left[a^3 + \frac{1}{a^3} + 3\left(a + \frac{1}{a}\right)\right]

=32(a+1a)−12(a3+1a3)−32(a+1a)= \frac{3}{2}\left(a+\frac{1}{a}\right) - \frac{1}{2}\left(a^3 + \frac{1}{a^3}\right) - \frac{3}{2}\left(a + \frac{1}{a}\right)

=32(a+1a)−32(a+1a)−12(a3+1a3)= \frac{3}{2}\left(a+\frac{1}{a}\right) - \frac{3}{2}\left(a + \frac{1}{a}\right) - \frac{1}{2}\left(a^3 + \frac{1}{a^3}\right)

=−12(a3+1a3)= -\frac{1}{2}\left(a^3 + \frac{1}{a^3}\right)

Hence proved.