5.11 Proofs with trigonometric functions and identities — Practise

Work through these on your own. Prove each identity by changing one side only. Name the formula you use at each line, and write down any angle where the identity is not defined.

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

Prove that $(\sin A + \cos A)^2 \equiv 1 + \sin 2A$ .
Prove that $\sin(A + B)\sin(A - B) \equiv \sin^2 A - \sin^2 B$ .
Prove that $\cos 2A \equiv \dfrac{1 - \tan^2 A}{1 + \tan^2 A}$ .
Prove that $\tan A + \cot A \equiv 2\operatorname{cosec} 2A$ for $\sin A\cos A \neq 0$ .
Prove that $\cos^4 A - \sin^4 A \equiv \cos 2A$ .
Prove that $\sin 3A \equiv 3\sin A - 4\sin^3 A$ .
Prove that $\cos 3A \equiv 4\cos^3 A - 3\cos A$ .
Prove that $\dfrac{\sin 2A}{1 + \cos 2A} \equiv \tan A$ for $1 + \cos 2A \neq 0$ .
Prove that $\dfrac{1 - \cos 2A}{\sin 2A} \equiv \tan A$ for $\sin 2A \neq 0$ .
Prove that $2\sin A\cos A \equiv \sin 2A$ .

Answers and concise solutions

1. The identity

Answer: Expand the square: $\sin^2 A + 2\sin A\cos A + \cos^2 A$ . Group the two squares and use $\sin^2 A + \cos^2 A = 1$ : this is $1 + 2\sin A\cos A$ . The double-angle formula gives $2\sin A\cos A = \sin 2A$ , so the left-hand side is $1 + \sin 2A$ .

2. The identity

Answer: Use the two addition formulae and the difference of two squares: $\sin(A + B)\sin(A - B) = \sin^2 A\cos^2 B - \cos^2 A\sin^2 B$ . Replace each cosine squared with $1 - \sin^2$ and collect. The cross terms cancel, leaving $\sin^2 A - \sin^2 B$ .

3. The identity

Answer: Multiply the top and the bottom by $\cos^2 A$ : $\dfrac{\cos^2 A - \sin^2 A}{\cos^2 A + \sin^2 A} = \dfrac{\cos^2 A - \sin^2 A}{1} = \cos 2A$ .

4. The identity

Answer: Write both terms over the common denominator: $\dfrac{\sin^2 A + \cos^2 A}{\sin A\cos A} = \dfrac{1}{\sin A\cos A}$ . Then multiply top and bottom by $2$ and use $\sin 2A = 2\sin A\cos A$ : $\dfrac{2}{\sin 2A} = 2\operatorname{cosec} 2A$ .

5. The identity

Answer: The left-hand side is a difference of two squares: $(\cos^2 A - \sin^2 A)(\cos^2 A + \sin^2 A)$ . The first bracket is $\cos 2A$ and the second is $1$ , so the product is $\cos 2A$ .

6. The identity

Answer: Write $3A = 2A + A$ , so $\sin 3A = \sin 2A\cos A + \cos 2A\sin A$ . Replace $\sin 2A = 2\sin A\cos A$ and $\cos 2A = 1 - 2\sin^2 A$ . Then replace $\cos^2 A = 1 - \sin^2 A$ and collect the sine terms.

7. The identity

Answer: Write $3A = 2A + A$ , so $\cos 3A = \cos 2A\cos A - \sin 2A\sin A$ . Replace $\cos 2A = 2\cos^2 A - 1$ and $\sin 2A = 2\sin A\cos A$ . Then replace $\sin^2 A = 1 - \cos^2 A$ and collect the cosine terms.

8. The identity

Answer: Replace both double angles: $\dfrac{2\sin A\cos A}{1 + 2\cos^2 A - 1} = \dfrac{2\sin A\cos A}{2\cos^2 A} = \dfrac{\sin A}{\cos A} = \tan A$ .

9. The identity

Answer: Use $\cos 2A = 1 - 2\sin^2 A$ , so the numerator is $2\sin^2 A$ . Use $\sin 2A = 2\sin A\cos A$ for the denominator. Then $\dfrac{2\sin^2 A}{2\sin A\cos A} = \dfrac{\sin A}{\cos A} = \tan A$ . The verifier reduces the left-hand side with a trigonometric simplification and matches the tangent.

10. The identity

Answer: This is the double-angle formula for sine. It follows from the addition formula $\sin(A + B) = \sin A\cos B + \cos A\sin B$ by putting $B = A$ .