if sinθ = 1/2(x+1/x), show that sin 3θ + 1/2 (x^3 + 1/x^3) = 0

Question

it is related to trigonometric identities

Answer 1

Given SinѲ = 1/2(x+1/x)

We know the formula Sin3Ѳ = 3SinѲ-4Sin^3Ѳ

Now Sin3Ѳ = 3(1/2(x+1/x))-4(1/2(x+1/x))^3

= 3/2(x+1/x)-4(1/2)^3(x+1/x)^3

Recall the formula (a+b)^3 = a^3+b^3+3a^2b+3ab^2

Takecommon out 1/2 from above expression.

= 1/2[3x+3/x-4(1/2)^2(x^3+1/x^3+3x^2*1/x+3*1/x^2*x)

= 1/2[3x+3/x-4*1/4(x^3+1/x^3+3x+3/x)

= 1/2[3x+3/x-x^3+1/x^3-3x-3/x]

= 1/2[-x^3-1/x^3]

= -1/2[x^3+1/x^3]

Sin3Ѳ = -1/2[x^3+1/x^3]

Add 1/2[x^3+1/x^3] to each side.

Sin3Ѳ+1/2[x^3+1/x^3] = 0

 

Comments

Your solution is correct.

Typo mistake in 9 th line of above solution.

= 1/2[3x + 3/x - x^3 - 1/x^3 - 3x - 3/x].