5.7 Trigonometric identities — Practise

Work through these on your own. Prove each identity by changing one side only. Start with the more complicated side, and show every step.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

Prove that $\dfrac{\sin^2\theta}{\sin\theta} \equiv \sin\theta$ for $\sin\theta \neq 0$ .
Prove that $\cos\theta\tan\theta \equiv \sin\theta$ for $\cos\theta \neq 0$ .
Prove that $1 + \tan^2\theta \equiv \sec^2\theta$ .
Prove that $(\sin\theta - \cos\theta)^2 \equiv 1 - 2\sin\theta\cos\theta$ .
Prove that $\dfrac{\cos^2\theta}{1 - \sin\theta} \equiv 1 + \sin\theta$ for $\sin\theta \neq 1$ .
Prove that $\sec^2\theta + \operatorname{cosec}^2\theta \equiv \sec^2\theta\,\operatorname{cosec}^2\theta$ for $\sin\theta\cos\theta \neq 0$ .
Prove that $\dfrac{1 - \cos^2\theta}{\sin\theta} \equiv \sin\theta$ for $\sin\theta \neq 0$ .
Prove that $\sin\theta\cos\theta(\tan\theta + \cot\theta) \equiv 1$ for $\sin\theta\cos\theta \neq 0$ .
Prove that $\tan\theta\cos\theta \equiv \sin\theta$ by simplifying the left-hand side.
Simplify $\sin^2\theta + \cos^2\theta$ using the basic identity.

Answers and concise solutions

1. The identity

Answer: Write the numerator as a product: $\dfrac{\sin^2\theta}{\sin\theta} = \dfrac{\sin\theta \times \sin\theta}{\sin\theta} = \sin\theta$ . The left-hand side becomes the right-hand side, so the identity holds wherever $\sin\theta \neq 0$ .

2. The identity

Answer: Replace tangent: $\cos\theta\tan\theta = \cos\theta \times \dfrac{\sin\theta}{\cos\theta} = \sin\theta$ .

3. The identity

Answer: Use a common denominator: $1 + \tan^2\theta = 1 + \dfrac{\sin^2\theta}{\cos^2\theta} = \dfrac{\cos^2\theta + \sin^2\theta}{\cos^2\theta} = \dfrac{1}{\cos^2\theta} = \sec^2\theta$ .

4. The identity

Answer: Expand the square: $\sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta$ . Group the two squares and use $\sin^2\theta + \cos^2\theta = 1$ . The result is $1 - 2\sin\theta\cos\theta$ .

5. The identity

Answer: Replace $\cos^2\theta$ with $1 - \sin^2\theta$ and factorise: $\dfrac{(1 - \sin\theta)(1 + \sin\theta)}{1 - \sin\theta} = 1 + \sin\theta$ .

6. The identity

Answer: Write both terms over $\cos^2\theta\sin^2\theta$ : $\dfrac{\sin^2\theta + \cos^2\theta}{\cos^2\theta\sin^2\theta} = \dfrac{1}{\cos^2\theta\sin^2\theta} = \sec^2\theta\,\operatorname{cosec}^2\theta$ .

7. The identity

Answer: From $\sin^2\theta + \cos^2\theta = 1$ , the numerator is $\sin^2\theta$ . Then $\dfrac{\sin^2\theta}{\sin\theta} = \sin\theta$ .

8. The identity

Answer: Expand the bracket: $\sin\theta\cos\theta\left(\dfrac{\sin\theta}{\cos\theta} + \dfrac{\cos\theta}{\sin\theta}\right) = \sin^2\theta + \cos^2\theta = 1$ .

9. The identity

$tan(x)*cos(x)$
$sin(x)$

Answer: Write tangent as a fraction: $\tan\theta\cos\theta = \dfrac{\sin\theta}{\cos\theta} \times \cos\theta = \sin\theta$ . The step from the product to the sine is the cancellation of the cosine, which the verifier checks as a trigonometric simplification.

10. The simplified expression

Answer: This is the first identity, true for every angle. On the unit circle it says that the point $(\cos\theta, \sin\theta)$ is always distance $1$ from the origin.