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$$\sin^3 x=\frac{3}{4}\sin x - \frac{1}{4}\sin 3x$$

Is there any formula that tells this or why is it like that?

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Certainly‌​. –  J. M. Jan 9 '12 at 16:34
Expand $$\sin 3x = \sin(x+2x) = \sin x \cos 2x + \cos x\sin 2x = \sin x(\cos^2 x - \sin^2 x) + 2\sin x \cos^2 x$$ and set $\cos^2 x = 1 - \sin^2 x$ –  Dilip Sarwate Jan 9 '12 at 16:47
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8 Answers

up vote 7 down vote accepted


$\sin(\alpha+\beta)=\sin \alpha \cdot \cos \beta + \cos \alpha \cdot \sin \beta$

$\sin 2\alpha =2\cdot \sin \alpha \cdot \cos \alpha$

$\cos 2\alpha=\cos^2 \alpha - \sin^2 \alpha$

$\sin^2 \alpha + \cos^2 \alpha=1 $

If you apply all these formulas you should get :

$\sin(3x)=3\cdot \sin x -4\cdot \sin^3 x$

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It's not $sin(3x)$, it's $sin^3(x)$ –  Andrew Jan 9 '12 at 16:54
@Andrew,I am sure that you know how to express $\sin^3 x$ from the last equality . –  pedja Jan 9 '12 at 16:57
@Andrew: $sin(3x)=3sinx-4sin^3x => sin^3x = \frac{3}{4}sinx - \frac{1}{4}sin3x $ –  GSBabil Jan 10 '12 at 3:30
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$$\sin(3x)=\Im(\mathrm e^{3\mathrm ix})=\Im\left((\cos(x)+\mathrm i\sin(x))^3\right)$$

$$(\cos(x)+\mathrm i\sin(x))^3=\cos^3(x)+3\mathrm i\ \cos^2(x)\sin(x)-3\cos(x)\sin^2(x)-\mathrm i\ \sin^3(x)$$

$$\sin(3x)=3\ \cos^2(x)\sin(x)-\sin^3(x)=3\sin(x)-4\sin^3(x)$$

$$4\ \sin^3(x)=3\sin(x)-\sin(3x)$$

Exercise Likewise, show that

$$16\ \sin^5(x)=\sin(5x)-5\sin(3x)+10\sin(x)$$

Added The other way round (this one generalizes easily and explains the appearance of the binomial coefficients $(1,3)$ for $\sin^3$ and $(1,5,10)$ for $\sin^5$):

$$(2\mathrm i\sin(x))^3=(\mathrm e^{\mathrm ix}-\mathrm e^{-\mathrm ix})^3=\mathrm e^{3\mathrm ix}-3\mathrm e^{\mathrm ix}+3\mathrm e^{-\mathrm ix}-\mathrm e^{-3\mathrm ix}$$ $$ (-8\mathrm i)\ \sin^3(x)=(2\mathrm i)\ (\sin(3x)-3\sin(x)) $$ $$ -4\ \sin^3(x)=\sin(3x)-3\sin(x) $$

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It's not $sin(3x)$, it's $sin^3(x)$ –  Andrew Jan 9 '12 at 16:55
@Andrew Didier's last line gives the required equality. –  David Mitra Jan 9 '12 at 17:13
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Continuing from my comment to bgins's answer, one can start from the (defining) Chebyshev polynomial identity


and make the substitution $x=\frac{\pi}{2}-z$ to yield the identity


(use the addition formula and mind the values of cosine and sine at integer multiples of $\pi$.)

Now, there is the identity

$$T_{2n-1}(u)=\sum_{m=0}^{n-1} (-1)^{n+m+1}\frac{2n-1}{2m+1}\binom{n+m-1}{2m} 4^m u^{2m+1}$$

which can be used to derive the matrix-vector identity

$$\begin{pmatrix}\sin\,x\\\sin\,3x\\\vdots\\\sin((2n-1)x)\end{pmatrix}=\mathbf L\mathbf D\begin{pmatrix}\sin\,x\\\sin^3 x\\\vdots\\\sin^{2n-1}x\end{pmatrix}$$

where the diagonal matrix $\mathbf D$ has the diagonal entries $d_{k,k}=(-4)^{k-1}$ and the unit lower triangular matrix $\mathbf L$ has the entries $\ell_{j,k}=\dfrac{2j-1}{2k-1}\dbinom{j+k-2}{2k-2}$. Inverting this relation gives

$$\begin{pmatrix}\sin\,x\\\sin^3 x\\\vdots\\\sin^{2n-1}x\end{pmatrix}=\mathbf D^{-1}\mathbf W\begin{pmatrix}\sin\,x\\\sin\,3x\\\vdots\\\sin((2n-1)x)\end{pmatrix}$$

where $\mathbf W=\mathbf L^{-1}$ is also a unit lower triangular matrix with entries $w_{j,k}=(-1)^{j+k} \dbinom{2j-1}{j+k-1}$. (The proof that $\mathbf L\mathbf W=\mathbf I$ is not too hard, and is left as an exercise.) In particular, the second row yields the OP's desired identity.

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de Moivre's formula says $$ \begin{align} \cos(3x)+i\sin(3x) &=(\cos(x)+i\sin(x))^3\\ &=\left(\cos^3(x)-3\cos(x)\sin^2(x)\right)+i\left(3\cos^2(x)\sin(x)-\sin^3(x)\right)\\ &=\left(4\cos^3(x)-3\cos(x)\right)+i\left(3\sin(x)-4\sin^3(x)\right)\tag{1} \end{align} $$ Therefore, $$ \begin{align} \cos(3x)&=4\cos^3(x)-3\cos(x)\tag{2}\\ \sin(3x)&=3\sin(x)-4\sin^3(x)\tag{3} \end{align} $$ Solving $(3)$ for $\sin^3(x)$ yields $$ \sin^3(x)=\frac34\sin(x)-\frac14\sin(3x)\tag{4} $$

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Here is a mild generalization.

Let $X,Y\ $ be indeterminates, and let $f(X;Y)\in\mathbb C[X,Y\ ]$ be a polynomial.

The following observations give a purely algebraic method to decide if we have $$f(\cos t\,;\sin t)=0$$ for all $t$ in $\mathbb R$.

If $U,V,T$ are other indeterminates, then the following conditions are equivalent:

(a) $\ f(\cos t\,;\sin t)=0$ for all $t$ in $\mathbb R$,

(b) $\ X^2+Y^2-1$ divides $f(X;Y)$,

(c) $\ UV-1$ divides $$g(U,V):=f\left(\frac{U+V}{2}\ ;\ \frac{U-V}{2i}\right),$$

(d) $\ g(T,T^{-1})=0$.

The interpretation of the indeterminates $X,Y,U,V,T$ can be informally expressed by the equalities $$ X=\cos t,\quad Y=\sin t,\quad T=U=e^{it},\quad T^{-1}=V=e^{-it}. $$

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You might want to look at Chebyshev polynomials of the first and second kind, denoted $T_n$ and $U_n$ respectively, which are defined recursively by $$ \begin{matrix} T_0 = 1,& T_1(x) = x,& T_{n+1} = 2x T_n(x) - T_{n-1}(x) \\ U_0 = 1,& U_1(x) = 2x,& U_{n+1} = 2x U_n(x) - U_{n-1}(x) \end{matrix} $$ (hence each of degree $n$) and give the formulas $$ \cos(n\theta) = T_n(\cos\theta), \qquad \frac{\sin\left((n+1)\theta\right)}{\sin\theta} = U_n(\cos\theta), $$ and have various closed form expressions, for example involving binomial coefficients.

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In particular, $T_{2n-1}(\sin\,z)=(-1)^{n+1}\sin((2n-1)z)$ –  J. M. Jan 10 '12 at 3:46
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If you're ever unsure of a trig identity, just convert it to complex exponentials and expand. It's especially nice if you have a CAS that can do the expansion for you (or if you're just handy with such things).

$\sin^3(x) = \left (\frac{e^{i x} - e^{-i x}}{2i}\right )^3 = \frac{e^{3ix}-3e^{ix}+3e^{-ix}-e^{-3ix}}{-8i}$ and $\frac{3}{4}\sin(x) - \frac{1}{4}\sin(3x) = \frac{3}{4}\left (\frac{e^{i x} - e^{-i x}}{2i}\right ) - \frac{1}{4}\left (\frac{e^{3i x} - e^{-3i x}}{2i}\right ) = \frac{e^{3ix}-3e^{ix}+3e^{-ix}-e^{-3ix}}{-8i}$.

Alternately, we could derive the right hand side by converting the exponentials back to sines and cosines using $e^{ix} = \cos(x) + i\sin(x)$ and canceling.

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