5.11 Proofs with trigonometric functions and identities — Practise with help

Choose one side to work on and stay with it. At each line, name the formula you used. Finish with the two sides identical.

Warm-up: remember

Expand $(x + y)(x - y)$ .
Write $\sin 2A$ in terms of $\sin A$ and $\cos A$ .
Write $\cos 2A$ in each of its three forms.

Answers

$(x + y)(x - y) = x^2 - y^2$ .
$\sin 2A = 2\sin A\cos A$ .
$\cos 2A = \cos^2 A - \sin^2 A = 1 - 2\sin^2 A = 2\cos^2 A - 1$ .

If question 3 was hard, revisit the double-angle formulae before you continue.

Practise with help

1. Prove that $(\sin A + \cos A)^2 \equiv 1 + \sin 2A$ .

Hint

Expand the square on the left. Look for $\sin^2 A + \cos^2 A$ and for $2\sin A\cos A$ .

Next step

$(\sin A + \cos A)^2 = \sin^2 A + 2\sin A\cos A + \cos^2 A$ . Group the two squares, then replace $2\sin A\cos A$ .

Full working

(\sin A + \cos A)^2 = \sin^2 A + 2\sin A\cos A + \cos^2 A = (\sin^2 A + \cos^2 A) + 2\sin A\cos A = 1 + \sin 2A.

2. Prove that $\sin(A + B)\sin(A - B) \equiv \sin^2 A - \sin^2 B$ .

Hint

Use the two addition formulae, then recognise a difference of two squares.

Next step

The product is $(\sin A\cos B + \cos A\sin B)(\sin A\cos B - \cos A\sin B)$ , which equals $\sin^2 A\cos^2 B - \cos^2 A\sin^2 B$ .

Full working

\sin(A + B)\sin(A - B) = \sin^2 A\cos^2 B - \cos^2 A\sin^2 B.

Replace the cosines with $1 - \sin^2$ :

= \sin^2 A(1 - \sin^2 B) - (1 - \sin^2 A)\sin^2 B = \sin^2 A - \sin^2 A\sin^2 B - \sin^2 B + \sin^2 A\sin^2 B = \sin^2 A - \sin^2 B.

3. Prove that $\cos 2A \equiv \dfrac{1 - \tan^2 A}{1 + \tan^2 A}$ .

Hint

Work on the right-hand side. Replace tangent with sine over cosine.

Next step

Multiply the top and the bottom of the fraction by $\cos^2 A$ . The denominator becomes $\cos^2 A + \sin^2 A$ .

Full working

\frac{1 - \tan^2 A}{1 + \tan^2 A} = \frac{\cos^2 A - \sin^2 A}{\cos^2 A + \sin^2 A} = \frac{\cos^2 A - \sin^2 A}{1} = \cos 2A.

4. Prove that $\tan A + \cot A \equiv 2\operatorname{cosec} 2A$ for $\sin A\cos A \neq 0$ .

Hint

Write both terms as fractions over the common denominator $\sin A\cos A$ .

Next step

$\dfrac{\sin A}{\cos A} + \dfrac{\cos A}{\sin A} = \dfrac{\sin^2 A + \cos^2 A}{\sin A\cos A} = \dfrac{1}{\sin A\cos A}$ . Now use $\sin 2A = 2\sin A\cos A$ .

Full working

\tan A + \cot A = \frac{\sin^2 A + \cos^2 A}{\sin A\cos A} = \frac{1}{\sin A\cos A} = \frac{2}{2\sin A\cos A} = \frac{2}{\sin 2A} = 2\operatorname{cosec} 2A.

5. Prove that $\cos^4 A - \sin^4 A \equiv \cos 2A$ .

Hint

The left-hand side is a difference of two squares.

Next step

$\cos^4 A - \sin^4 A = (\cos^2 A - \sin^2 A)(\cos^2 A + \sin^2 A)$ . Replace both brackets.

Full working

\cos^4 A - \sin^4 A = (\cos^2 A - \sin^2 A)(\cos^2 A + \sin^2 A) = (\cos 2A)(1) = \cos 2A.

6. Prove that $\sin 3A \equiv 3\sin A - 4\sin^3 A$ .

Hint

Write $3A = 2A + A$ and use the addition formula for sine, then the double-angle formulae.

Next step

$\sin 3A = \sin(2A + A) = \sin 2A\cos A + \cos 2A\sin A$ . Replace $\sin 2A = 2\sin A\cos A$ and $\cos 2A = 1 - 2\sin^2 A$ .

Full working

\sin 3A = \sin 2A\cos A + \cos 2A\sin A = (2\sin A\cos A)\cos A + (1 - 2\sin^2 A)\sin A.

= 2\sin A\cos^2 A + \sin A - 2\sin^3 A.

Replace $\cos^2 A = 1 - \sin^2 A$ :

= 2\sin A(1 - \sin^2 A) + \sin A - 2\sin^3 A = 2\sin A - 2\sin^3 A + \sin A - 2\sin^3 A = 3\sin A - 4\sin^3 A.