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Prove the trigonometric identity: sin(3θ)/sin θ - cos(3θ)/cos θ = 2

Mathematics

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To prove the trigonometric identity sin⁡(3θ)sin⁡θ−cos⁡(3θ)cos⁡θ=2\frac{\sin(3\theta)}{\sin\theta} - \frac{\cos(3\theta)}{\cos\theta} = 2, we will start with the left-hand side (LHS) and simplify it to the right-hand side (RHS).

  1. Write the left-hand side of the identity and substitute the triple angle formulas for sin⁡(3θ)\sin(3\theta) and cos⁡(3θ)\cos(3\theta).

    We know that sin⁡(3θ)=3sin⁡θ−4sin⁡3θ\sin(3\theta) = 3\sin\theta - 4\sin^3\theta and cos⁡(3θ)=4cos⁡3θ−3cos⁡θ\cos(3\theta) = 4\cos^3\theta - 3\cos\theta.

    LHS=3sin⁡θ−4sin⁡3θsin⁡θ−4cos⁡3θ−3cos⁡θcos⁡θ\text{LHS} = \frac{3\sin\theta - 4\sin^3\theta}{\sin\theta} - \frac{4\cos^3\theta - 3\cos\theta}{\cos\theta}

  2. Factor out sin⁡θ\sin\theta from the numerator of the first term and cos⁡θ\cos\theta from the numerator of the second term, then simplify.

    LHS=sin⁡θ(3−4sin⁡2θ)sin⁡θ−cos⁡θ(4cos⁡2θ−3)cos⁡θ\text{LHS} = \frac{\sin\theta(3 - 4\sin^2\theta)}{\sin\theta} - \frac{\cos\theta(4\cos^2\theta - 3)}{\cos\theta} LHS=(3−4sin⁡2θ)−(4cos⁡2θ−3)\text{LHS} = (3 - 4\sin^2\theta) - (4\cos^2\theta - 3)

  3. Expand the expression and group terms.

    LHS=3−4sin⁡2θ−4cos⁡2θ+3\text{LHS} = 3 - 4\sin^2\theta - 4\cos^2\theta + 3 LHS=6−4sin⁡2θ−4cos⁡2θ\text{LHS} = 6 - 4\sin^2\theta - 4\cos^2\theta LHS=6−4(sin⁡2θ+cos⁡2θ)\text{LHS} = 6 - 4(\sin^2\theta + \cos^2\theta)

  4. Apply the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1.

    LHS=6−4(1)\text{LHS} = 6 - 4(1) LHS=6−4\text{LHS} = 6 - 4 LHS=2\text{LHS} = 2

    Since LHS = RHS, the identity is proven.

    Hence, sin⁡(3θ)sin⁡θ−cos⁡(3θ)cos⁡θ=2\boxed{\frac{\sin(3\theta)}{\sin\theta} - \frac{\cos(3\theta)}{\cos\theta} = 2} is proven.

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