Multiple and Sub-Multiple Angle Formulas | NIMCET

Quick answer: Setting $B = A$ in the compound angle formulas gives the double angle formulas: $\sin 2A = 2\sin A\cos A$, $\cos 2A = \cos^2 A - \sin^2 A = 1 - 2\sin^2 A = 2\cos^2 A - 1$ and $\tan 2A = \frac{2\tan A}{1 - \tan^2 A}$. The triple angle formulas are $\sin 3A = 3\sin A - 4\sin^3 A$ and $\cos 3A = 4\cos^3 A - 3\cos A$. Replacing $A$ by $\frac{A}{2}$ gives the half-angle (sub-multiple) formulas.

"Multiple angles" means $2A$, $3A$ and so on; "sub-multiple angles" means $\frac{A}{2}$, $\frac{A}{3}$ and so on. All of them come from the compound angle formulas.

Double angle formulas

Put $B = A$ in $\sin(A + B)$, $\cos(A + B)$ and $\tan(A + B)$:

The three forms of $\cos 2A$ are all useful:

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

Rearranged, they give the power-reduction formulas used constantly in integration:

$\sin^2 A = \frac{1 - \cos 2A}{2}, \qquad \cos^2 A = \frac{1 + \cos 2A}{2}$

Triple angle formulas

Derivation of sin 3A: write $\sin 3A = \sin(2A + A) = \sin 2A\cos A + \cos 2A\sin A$, then substitute the double angle formulas and use $\cos^2 A = 1 - \sin^2 A$.

The rearranged forms $\sin^3 A = \frac{3\sin A - \sin 3A}{4}$ and $\cos^3 A = \frac{3\cos A + \cos 3A}{4}$ are handy for integrals and sums.

Half-angle (sub-multiple) formulas

Replace $A$ by $\frac{A}{2}$:

Taking square roots gives $\sin\frac{A}{2} = \pm\sqrt{\frac{1 - \cos A}{2}}$; the sign depends on the quadrant of $\frac{A}{2}$, not of $A$.

The t = tan(θ/2) substitution

With $t = \tan\frac{\theta}{2}$:

$\sin\theta = \frac{2t}{1 + t^2}, \qquad \cos\theta = \frac{1 - t^2}{1 + t^2}$

This turns equations like $a\cos\theta + b\sin\theta = c$ into quadratics in $t$.

Exact values: 18°, 36°, 22.5°

sin 18°: Let $\theta = 18^\circ$, so $5\theta = 90^\circ$ and $2\theta = 90^\circ - 3\theta$. Then $\sin 2\theta = \cos 3\theta$:

$2\sin\theta\cos\theta = 4\cos^3\theta - 3\cos\theta$

Divide by $\cos\theta$ (non-zero): $2\sin\theta = 4\cos^2\theta - 3 = 1 - 4\sin^2\theta$. So $4\sin^2\theta + 2\sin\theta - 1 = 0$, giving

$\sin 18^\circ = \frac{\sqrt{5} - 1}{4}$

(taking the positive root, since 18° is acute).

The value of $\cos 36^\circ$ follows from $\cos 36^\circ = 1 - 2\sin^2 18^\circ$. The value of $\tan 22.5^\circ$ comes from $\tan\frac{A}{2} = \frac{1 - \cos A}{\sin A}$ with $A = 45^\circ$.

Worked examples

Example 1: given tan A, find sin 2A and cos 2A

If $\tan A = \frac{3}{4}$, then

$\sin 2A = \frac{2 \cdot \frac{3}{4}}{1 + \frac{9}{16}} = \frac{3/2}{25/16} = \frac{24}{25}, \qquad \cos 2A = \frac{1 - \frac{9}{16}}{1 + \frac{9}{16}} = \frac{7}{25}$

Check: $\left(\frac{24}{25}\right)^2 + \left(\frac{7}{25}\right)^2 = 1$ ✓.

Example 2: a quick simplification

Simplify $\frac{1 - \cos 2\theta}{\sin 2\theta}$.

$\frac{2\sin^2\theta}{2\sin\theta\cos\theta} = \tan\theta$

Example 3: product of sines

Find $\sin 18^\circ \cos 36^\circ$.

$\frac{\sqrt{5} - 1}{4} \cdot \frac{\sqrt{5} + 1}{4} = \frac{5 - 1}{16} = \frac{1}{4}$

Example 4: cos 2A in an equation

Solve $\cos 2x + 3\sin x = 2$ for $0 \le x \le 2\pi$.

Step 1: Replace $\cos 2x$ by $1 - 2\sin^2 x$: $1 - 2\sin^2 x + 3\sin x = 2$.

Step 2: $2\sin^2 x - 3\sin x + 1 = 0$, so $(2\sin x - 1)(\sin x - 1) = 0$.

Step 3: $\sin x = \frac{1}{2}$ gives $x = \frac{\pi}{6}, \frac{5\pi}{6}$; $\sin x = 1$ gives $x = \frac{\pi}{2}$.

Summary table

Exam tips

• Choose the form of $\cos 2A$ that matches the rest of the expression: use $1 - 2\sin^2 A$ if the equation is in $\sin A$.
• Expressions like $1 \pm \cos A$ or $1 \pm \sin A$ usually simplify with half angles: $1 + \sin A = \left(\sin\frac{A}{2} + \cos\frac{A}{2}\right)^2$.
• Learn the 18° and 36° values; they appear in product and sum questions.
• In square-root half-angle results, decide the sign from the quadrant of the half angle.

See the NIMCET Trigonometry chapter for related topics.

Common mistakes

• Writing $\sin 2A = 2\sin A$. The formula is $2\sin A\cos A$.
• Mixing up the signs in $\sin 3A = 3\sin A - 4\sin^3 A$ and $\cos 3A = 4\cos^3 A - 3\cos A$.
• Dividing by $\cos x$ or $\sin x$ in an equation without checking whether it can be zero.
• Choosing the sign of $\sin\frac{A}{2}$ from the quadrant of $A$.

Frequently Asked Questions

What are the three forms of cos 2A?

cos 2A = cos²A − sin²A = 2cos²A − 1 = 1 − 2sin²A.

What is the formula for sin 3A?

sin 3A = 3 sin A − 4 sin³A.

What is the exact value of sin 18°?

sin 18° = (√5 − 1)/4.

What is tan 22.5°?

tan 22.5° = √2 − 1, from tan(A/2) = (1 − cos A)/sin A with A = 45°.

How do I express sin θ in terms of tan(θ/2)?

sin θ = 2t/(1 + t²), where t = tan(θ/2).