Trigonometric Identities: Application Problems (1)

easy • 2 min read
Tags Trigonometric Identities

Question

Prove that sin⁡θ1−cos⁡θ+1−cos⁡θsin⁡θ=2sin⁡θ\dfrac{\sin\theta}{1 - \cos\theta} + \dfrac{1 - \cos\theta}{\sin\theta} = \dfrac{2}{\sin\theta}.

Solution — Step by Step

LHS =sin⁡2θ+(1−cos⁡θ)2sin⁡θ(1−cos⁡θ)= \dfrac{\sin^2\theta + (1 - \cos\theta)^2}{\sin\theta(1 - \cos\theta)}.

sin⁡2θ+(1−cos⁡θ)2=sin⁡2θ+1−2cos⁡θ+cos⁡2θ\sin^2\theta + (1 - \cos\theta)^2 = \sin^2\theta + 1 - 2\cos\theta + \cos^2\theta

=(sin⁡2θ+cos⁡2θ)+1−2cos⁡θ= (\sin^2\theta + \cos^2\theta) + 1 - 2\cos\theta

=1+1−2cos⁡θ=2−2cos⁡θ=2(1−cos⁡θ)= 1 + 1 - 2\cos\theta = 2 - 2\cos\theta = 2(1 - \cos\theta).

LHS =2(1−cos⁡θ)sin⁡θ(1−cos⁡θ)=2sin⁡θ= \dfrac{2(1 - \cos\theta)}{\sin\theta(1 - \cos\theta)} = \dfrac{2}{\sin\theta}.

Final answer: Proved. LHS = 2sin⁡θ\dfrac{2}{\sin\theta} = RHS. ■\blacksquare

Why This Works

The Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 is the workhorse here. Whenever you see sin⁡2θ\sin^2\theta and cos⁡2θ\cos^2\theta on the same side, look for a chance to apply this identity — usually it collapses one of them or simplifies the expression dramatically.

The expansion of (1−cos⁡θ)2(1 - \cos\theta)^2 is straightforward, and the factor of (1−cos⁡θ)(1 - \cos\theta) that emerges in the numerator cancels with the same factor in the denominator. This kind of factor-cancellation is the soul of trig identity proofs.

Alternative Method

Multiply numerator and denominator of the first fraction by (1+cos⁡θ)(1 + \cos\theta):

sin⁡θ(1+cos⁡θ)(1−cos⁡θ)(1+cos⁡θ)=sin⁡θ(1+cos⁡θ)sin⁡2θ=1+cos⁡θsin⁡θ\dfrac{\sin\theta(1 + \cos\theta)}{(1 - \cos\theta)(1 + \cos\theta)} = \dfrac{\sin\theta(1 + \cos\theta)}{\sin^2\theta} = \dfrac{1 + \cos\theta}{\sin\theta}.

Then LHS =1+cos⁡θsin⁡θ+1−cos⁡θsin⁡θ=2sin⁡θ= \dfrac{1 + \cos\theta}{\sin\theta} + \dfrac{1 - \cos\theta}{\sin\theta} = \dfrac{2}{\sin\theta}. Cleaner — no expansion needed.

Common Mistake

Students often try to cancel terms across the addition: e.g., cancel sin⁡θ\sin\theta from the first numerator with sin⁡θ\sin\theta in the second denominator. That’s invalid — cancellation only works within a single fraction or across multiplied terms, not across additive ones. Always combine into one fraction first, then cancel.

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