Multi-angle identities (up to 4 theta)
Master multi-angle identities for Year 11 Specialist Mathematics in Queensland (QCAA). Starting from the double-angle identities for sine and cosine, you rewrite the trigonometric ratios of double, triple and quadruple angles using only functions of the original angle.
You will learn to prove and apply these identities up to four theta, choose the right form of the cosine double angle, and build results in stages — the identity foundation for solving trigonometric equations and sketching graphs later in the course.
Theory
Multi-angle identities express a trigonometric ratio of \(2t\), \(3t\) or \(4t\) in terms of the ratios of the single angle \(t\). In Year 11 Specialist Mathematics (QCAA, Queensland) you start from the double-angle identities for sine and cosine and apply them repeatedly to reach \(3t\) and \(4t\), for example \(\cos 4t=8\cos^4 t-8\cos^2 t+1\).
An angle such as \(2t\), \(3t\) or \(4t\) is a multiple of the base angle \(t\). A multi-angle identity rewrites its sine, cosine or tangent using only functions of \(t\).
The three double-angle identities are the foundation: \(\sin 2t=2\sin t\cos t\), \(\cos 2t=\cos^2 t-\sin^2 t\), and \(\tan 2t=\dfrac{2\tan t}{1-\tan^2 t}\). The cosine form has two useful rewrites, \(\cos 2t=2\cos^2 t-1\) and \(\cos 2t=1-2\sin^2 t\), found by substituting \(\sin^2 t+\cos^2 t=1\).
Applying these through the angle-sum identity \(\cos(A+B)=\cos A\cos B-\sin A\sin B\) reaches the triple angle: \(\cos 3t=4\cos^3 t-3\cos t\) and \(\sin 3t=3\sin t-4\sin^3 t\).
Treating \(4t\) as \(2(2t)\) gives the quadruple angle: \(\sin 4t=2\sin 2t\cos 2t\) and \(\cos 4t=2\cos^2 2t-1\), which expands fully to \(\cos 4t=8\cos^4 t-8\cos^2 t+1\). Every result is built by repeated application of the double-angle rules.
The double-angle identities, with the two extra forms of \(\cos 2t\):
Building up to \(3t\) and \(4t\):
How to evaluate or expand a multi-angle expression
- Identify the target angle: is it \(2t\), \(3t\) or \(4t\)? That fixes how many times a double-angle identity is applied.
- Choose the identity: for \(\cos 2t\) pick the form (\(2\cos^2 t-1\) or \(1-2\sin^2 t\)) that matches whether you are given \(\cos t\) or \(\sin t\).
- Substitute the known ratios and simplify one line at a time, keeping surds and fractions exact.
- Build up in stages: reach \(3t\) via \(2t+t\), and \(4t\) by doubling \(2t\) — find \(\cos 2t\) first, then feed it into the \(4t\) identity.
Apply the double-angle identity \(\sin 2t=2\sin t\cos t\):
| \(\sin 2t\) | \(=\) | \(2\sin t\cos t\) |
| \(=\) | \(2\times\dfrac{5}{13}\times\dfrac{12}{13}\) | |
| \(=\) | \(2\times\dfrac{60}{169}\) | |
| \(=\) | \(\dfrac{120}{169}\) |
\(\sin 2t=\dfrac{120}{169}\).
Only \(\sin t\) is given, so use the form \(\cos 2t=1-2\sin^2 t\):
| \(\cos 2t\) | \(=\) | \(1-2\sin^2 t\) |
| \(=\) | \(1-2\times\left(\dfrac{2}{5}\right)^2\) | |
| \(=\) | \(1-2\times\dfrac{4}{25}\) | |
| \(=\) | \(1-\dfrac{8}{25}\) | |
| \(=\) | \(\dfrac{17}{25}\) |
\(\cos 2t=\dfrac{17}{25}\).
Substitute \(\cos t=\dfrac{1}{4}\) into the triple-angle identity:
| \(\cos 3t\) | \(=\) | \(4\cos^3 t-3\cos t\) |
| \(=\) | \(4\times\left(\dfrac{1}{4}\right)^3-3\times\dfrac{1}{4}\) | |
| \(=\) | \(4\times\dfrac{1}{64}-\dfrac{3}{4}\) | |
| \(=\) | \(\dfrac{1}{16}-\dfrac{12}{16}\) | |
| \(=\) | \(-\dfrac{11}{16}\) |
\(\cos 3t=-\dfrac{11}{16}\).
Substitute \(\cos t=\dfrac{1}{2}\) into the quadruple-angle identity:
| \(\cos 4t\) | \(=\) | \(8\cos^4 t-8\cos^2 t+1\) |
| \(=\) | \(8\times\left(\dfrac{1}{2}\right)^4-8\times\left(\dfrac{1}{2}\right)^2+1\) | |
| \(=\) | \(8\times\dfrac{1}{16}-8\times\dfrac{1}{4}+1\) | |
| \(=\) | \(\dfrac{1}{2}-2+1\) | |
| \(=\) | \(-\dfrac{1}{2}\) |
\(\cos 4t=-\dfrac{1}{2}\).
Common pitfalls
Frequently asked questions
What is the double-angle identity for sine?
It is \(\sin 2t=2\sin t\cos t\). You multiply twice the sine of the angle by its cosine; it is not simply \(2\sin t\).
Why does cos 2t have three different forms?
Starting from \(\cos 2t=\cos^2 t-\sin^2 t\) and substituting \(\sin^2 t+\cos^2 t=1\) gives \(2\cos^2 t-1\) and \(1-2\sin^2 t\). All three are equal; you pick whichever matches the ratio you are given.
How do you get the identity for cos 3t?
Write \(3t=2t+t\) and expand \(\cos(2t+t)\) with the angle-sum identity, then substitute the double-angle forms. Simplifying with \(\sin^2 t=1-\cos^2 t\) gives \(\cos 3t=4\cos^3 t-3\cos t\).
How do you reach cos 4t = 8cos^4 t - 8cos^2 t + 1?
Treat \(4t\) as \(2(2t)\), so \(\cos 4t=2\cos^2 2t-1\). Substitute \(\cos 2t=2\cos^2 t-1\) and expand the square to get \(8\cos^4 t-8\cos^2 t+1\).
Is sin 4t = 2 sin 2t cos 2t correct?
Yes. It is the double-angle identity applied to the angle \(2t\): with \(A=2t\), \(\sin 2A=2\sin A\cos A\) gives \(\sin 4t=2\sin 2t\cos 2t\).
Do these identities work in radians and degrees?
Yes. The identities are true for any angle, so they hold whether \(t\) is measured in radians or degrees; just keep the mode consistent when you evaluate.