Multiple-Angle Trigonometric Formulas

Multiple-Angle Trigonometric Formulas

Multiple-angle formulas allow us to express functions like $\sin (nx)$ and $\cos (nx)$ in terms of powers of $\sin (x)$ and $\cos (x)$. These identities are useful in algebraic simplification, solving equations, and signal analysis.

Common Multiple-Angle Identities

Sine:

• $\sin (2x)=2\sin (x)\cos (x)$
• $\sin (3x)=3\sin (x)-4{\sin }^3(x)$
• $\sin (4x)=4\sin (x)\cos (x)(1-2{\sin }^2(x))$

Cosine:

• $\cos (2x)={\cos }^2(x)-{\sin }^2(x)$
• $\cos (3x)=4{\cos }^3(x)-3\cos (x)$
• $\cos (4x)=8{\cos }^4(x)-8{\cos }^2(x)+1$

These can be derived using angle addition formulas or De Moivre’s Theorem.

General Formula for $\sin (nx)$?

For any positive odd integer $n$, $\sin (nx)$ can be written as a sum of powers of $\sin (x)$ and $\cos (x)$:

$$\sin (nx)=\sum _{k=0}^{\lfloor n/2\rfloor }(-1{)}^k(\genfrac{}{}{0ex}{}{n}{2k+1}){\cos }^{n-(2k+1)}(x){\sin }^{2k+1}(x)$$

WARNING

This identity selects only odd-power terms in the expansion of $\sin (nx)$, yielding a pure polynomial in $\sin (x)$ when expressed using the Pythagorean identity.

Deriving $\sin (3x)$ Example (also see Trigonometric identities: Problem 01)

Start with:

$$\sin (3x)=\sin (2x+x)=\sin (2x)\cos (x)+\cos (2x)\sin (x)$$

Use:

• $\sin (2x)=2\sin (x)\cos (x)$
• $\cos (2x)={\cos }^2(x)-{\sin }^2(x)$

Substitute:

$$\sin (3x)=(2\sin (x)\cos (x))\cos (x)+({\cos }^2(x)-{\sin }^2(x))\sin (x)$$ $$=2\sin (x){\cos }^2(x)+{\cos }^2(x)\sin (x)-{\sin }^3(x)$$ $$=(3{\cos }^2(x)-{\sin }^2(x))\sin (x)$$

Now, using the Pythagorean identity ${\cos }^2(x)=1-{\sin }^2(x)$, we can rewrite it purely in terms of $\sin (x)$:

$$\begin{array}{rl}\sin (3x)& =(3(1-{\sin }^2(x))-{\sin }^2(x))\sin (x)\\ & =(3-4{\sin }^2(x))\sin (x)\\ & =3\sin (x)-4{\sin }^3(x)\end{array}$$

This is a more compact polynomial form of $\sin (3x)$, often more convenient in algebraic manipulation or solving equations.

Generalized Multiple Angle Formulae

These identities express $\sin (n\theta )$, $\cos (n\theta )$, and $\tan (n\theta )$ using combinations of $\sin (k\theta )$ and $\cos (k\theta )$:

$$\begin{array}{rl}& \sin (n\theta )\\ & =\boxed{\sum _{k=0}^n(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k}){\cos }^{n-k}(\theta ){\sin }^k(\theta )\cdot \sin \left[\frac{1}{2}(n-k)\pi \right]}\\ & =\boxed{2\sin \left[(n-1)\theta \right]\cos (\theta )-\sin \left[(n-2)\theta \right]}\end{array}$$ $$\begin{array}{rl}& \cos (n\theta )\\ & =\boxed{\sum _{k=0}^n(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k}){\cos }^{n-k}(\theta ){\sin }^k(\theta )\cdot \cos \left[\frac{1}{2}(n-k)\pi \right]}\\ & =\boxed{2\cos \left[(n-1)\theta \right]\cos (\theta )-\cos \left[(n-2)\theta \right]}\end{array}$$ $$\begin{array}{rl}& \tan (n\theta )\\ & =\frac{\sin (n\theta )}{\cos (n\theta )}\\ & =\boxed{\frac{\tan \left[(n-1)\theta \right]+\tan (\theta )}{1-\tan \left[(n-1)\theta \right]\tan (\theta )}}\end{array}$$

These identities are useful in trigonometric polynomial expansion, harmonic analysis, and symbolic computation involving angular multiples.

So, for example, $\cos (4x)$ could be written in the following ways:

• $\cos (4x)$
• $8{\cos }^4(x)-8{\cos }^2(x)+1$

  derived from $\cos (n\theta )={\sum }_{k=0}^n(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k}){\cos }^{n-k}(\theta ){\sin }^k(\theta )\cdot \cos \left[\frac{1}{2}(n-k)\pi \right]$

• $2\cos (3x)\cos (x)-\cos (2x)$

  derived from $\cos (n\theta )=2\cos \left[(n-1)\theta \right]\cos (\theta )-\cos \left[(n-2)\theta \right]$

Complete example of $\cos (4x)=8{\cos }^4(x)-8{\cos }^2(x)+1$

Using the sum formula for cosine:

$$\cos (4x)=\sum _{k=0}^4(-1{)}^k(\genfrac{}{}{0ex}{}{4}{k}){\cos }^{4-k}(x){\sin }^k(x)\cdot \cos \left[\frac{1}{2}(4-k)\pi \right]$$

Evaluate the cosine factor $\cos \left[\frac{1}{2}(4-k)\pi \right]$ for each $k$:

$k$	$4-k$	$\frac{1}{2}(4-k)\pi$	$\cos \left[\frac{1}{2}(4-k)\pi \right]$
0	4	$2\pi$	1
1	3	$\frac{3\pi }{2}$	0
2	2	$\pi$	$-1$
3	1	$\frac{\pi }{2}$	0
4	0	0	1

Only terms where the cosine factor is nonzero contribute, so:

$$\begin{array}{rl}\cos (4x)& =(-1{)}^0(\genfrac{}{}{0ex}{}{4}{0}){\cos }^4(x){\sin }^0(x)\cdot 1\\ & +(-1{)}^2(\genfrac{}{}{0ex}{}{4}{2}){\cos }^2(x){\sin }^2(x)\cdot (-1)\\ & +(-1{)}^4(\genfrac{}{}{0ex}{}{4}{4}){\cos }^0(x){\sin }^4(x)\cdot 1\\ & ={\cos }^4(x)-6{\cos }^2(x){\sin }^2(x)+{\sin }^4(x)\end{array}$$

Use the Pythagorean identity ${\sin }^2(x)=1-{\cos }^2(x)$ and rewrite ${\sin }^4(x)$:

$${\sin }^4(x)=(1-{\cos }^2(x){)}^2=1-2{\cos }^2(x)+{\cos }^4(x)$$

Substitute into the expression:

$$\begin{array}{rl}\cos (4x)& ={\cos }^4(x)-6{\cos }^2(x)(1-{\cos }^2(x))+1-2{\cos }^2(x)+{\cos }^4(x)\\ & ={\cos }^4(x)-6{\cos }^2(x)+6{\cos }^4(x)+1-2{\cos }^2(x)+{\cos }^4(x)\end{array}$$

Combine like terms:

$$\begin{array}{rl}\cos (4x)& =({\cos }^4(x)+6{\cos }^4(x)+{\cos }^4(x))+(-6{\cos }^2(x)-2{\cos }^2(x))+1\\ & =8{\cos }^4(x)-8{\cos }^2(x)+1\end{array}$$

Thus,

$$\boxed{\cos (4x)=8{\cos }^4(x)-8{\cos }^2(x)+1}$$

General Form via De Moivre’s Theorem

Using complex numbers:

$$(\cos x+i\sin x{)}^n=\cos (nx)+i\sin (nx)$$

Expanding with the binomial theorem gives a way to derive general formulas for $\sin (nx)$ and $\cos (nx)$.

Applications

• Trigonometric simplification
• Fourier analysis
• Vibrations and wave modeling
• Engineering and physics problems

See Also

• Trigonometric Identities