Proving trigonometric identities , practice

Identity-proving is one of the highest-mark areas of this chapter. The previous topics introduced the toolkit; here we drill it.

Strategies summarised

1. Pick the messier side. Usually the LHS is more complicated. Transform it into the RHS.
2. Convert to $\sin$ and $\cos$. When stuck, rewrite everything as $\sin$ and $\cos$. This is the always-works approach.
3. Combine fractions. Sums of fractions $\dfrac{1}{1 + \sin\theta} + \dfrac{1}{1 - \sin\theta}$ become $\dfrac{2}{1 - \sin^2\theta} = \dfrac{2}{\cos^2\theta} = 2 \sec^2\theta$.
4. Use $\sin^2 + \cos^2 = 1$ and its rearrangements. Especially $1 - \sin^2 = \cos^2$ and $\cos^2 - 1 = -\sin^2$.
5. Multiply by a clever form of $1$. For instance, $\dfrac{\sin\theta}{1 + \cos\theta}$: multiply top and bottom by $1 - \cos\theta$ to get $\dfrac{\sin\theta(1 - \cos\theta)}{1 - \cos^2\theta} = \dfrac{1 - \cos\theta}{\sin\theta}$.

More worked examples

Example 1. Prove $\dfrac{\sec\theta - 1}{\sec\theta + 1} = \dfrac{1 - \cos\theta}{1 + \cos\theta}$.

LHS $= \dfrac{1/\cos\theta - 1}{1/\cos\theta + 1} = \dfrac{(1 - \cos\theta)/\cos\theta}{(1 + \cos\theta)/\cos\theta} = \dfrac{1 - \cos\theta}{1 + \cos\theta}$. ✓

Example 2. Prove $\dfrac{1 + \cos\theta - \sin^2\theta}{\sin\theta(1 + \cos\theta)} = \cot\theta$.

Numerator: $1 + \cos\theta - \sin^2\theta = 1 + \cos\theta - (1 - \cos^2\theta) = \cos\theta + \cos^2\theta = \cos\theta(1 + \cos\theta)$.

Expression: $\dfrac{\cos\theta(1 + \cos\theta)}{\sin\theta(1 + \cos\theta)} = \dfrac{\cos\theta}{\sin\theta} = \cot\theta$. $\blacksquare$

Example 3. Prove $\dfrac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1} = \dfrac{1}{\sec\theta - \tan\theta}$.

Divide numerator and denominator of LHS by $\cos\theta$: $\text{LHS} = \frac{\tan\theta - 1 + \sec\theta}{\tan\theta + 1 - \sec\theta} = \frac{(\sec\theta + \tan\theta) - 1}{1 - (\sec\theta - \tan\theta)}.$

Use the identity $\sec^2\theta - \tan^2\theta = 1 \Rightarrow (\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = 1 \Rightarrow \sec\theta + \tan\theta = \dfrac{1}{\sec\theta - \tan\theta}$.

Let $u = \sec\theta - \tan\theta$. Then $\sec\theta + \tan\theta = 1/u$.

LHS $= \dfrac{1/u - 1}{1 - u} = \dfrac{(1 - u)/u}{1 - u} = \dfrac{1}{u} = \dfrac{1}{\sec\theta - \tan\theta}$. $\blacksquare$

Example 4. Prove $\sec^2\theta + \csc^2\theta = \sec^2\theta \cdot \csc^2\theta$.

LHS $= \dfrac{1}{\cos^2\theta} + \dfrac{1}{\sin^2\theta} = \dfrac{\sin^2\theta + \cos^2\theta}{\sin^2\theta \cos^2\theta} = \dfrac{1}{\sin^2\theta \cos^2\theta} = \csc^2\theta \sec^2\theta$. $\blacksquare$

Example 5. Prove $(\sin\theta + \cos\theta)^2 + (\sin\theta - \cos\theta)^2 = 2$.

Expand: $(\sin^2 + 2\sin\cos + \cos^2) + (\sin^2 - 2\sin\cos + \cos^2) = 2(\sin^2 + \cos^2) = 2$. $\blacksquare$

Try it yourself

1. Prove $\sin^4\theta - \cos^4\theta = \sin^2\theta - \cos^2\theta$.
2. Prove $\sin^4\theta + \cos^4\theta = 1 - 2 \sin^2\theta \cos^2\theta$.
3. Prove $(1 - \sin\theta)/(1 + \sin\theta) = (\sec\theta - \tan\theta)^2$.
4. Prove $\dfrac{1}{1 + \sin\theta} + \dfrac{1}{1 - \sin\theta} = 2 \sec^2\theta$.
5. Prove $(\tan\theta + \cot\theta)^2 = \sec^2\theta \csc^2\theta$.
6. Prove $\dfrac{1 + \sin\theta}{\cos\theta} + \dfrac{\cos\theta}{1 + \sin\theta} = 2 \sec\theta$.
7. Prove $\cos^2\theta - \sin^2\theta = 1 - 2\sin^2\theta$.
8. Prove $\dfrac{\cot^2\theta}{1 + \csc\theta} + 1 = \csc\theta$.
9. Prove $(1 - \tan\theta)^2 + (1 + \tan\theta)^2 = 2 \sec^2\theta$.
10. Prove $\sec\theta(1 - \sin\theta)(\sec\theta + \tan\theta) = 1$.

Pitfalls / Insight

• Don't move terms across an equality during a proof. Work each side independently or transform one side into the other.
• Watch denominators. A factor $(1 - \cos\theta)$ in the denominator excludes $\theta = 0^\circ$; mention domain if asked.
• If LHS = RHS is shown only for one value, you have not proved an identity.

Insight. A trig identity is true because of algebraic manipulation of the basic identities. Practice gives you a sense for which trick to try first.