Systematic way for proving trigonometric equations

Question

I am often asked to prove identities such as:

• $\sin^2(2t)=4\sin^2(t)-4\sin^4(t)$
• $\sin(3t)=3\sin(t)-4\sin^3(t)$

the proofs involve only elementary trigonometric identities. The method I use to prove these identities is very ad hoc and it often costs me too much time to execute. So, I would like to optimize my current method. Hence, I am wondering whether there is a systematic way of proving these identities. By systematic I mean a step-by-step strategy/algorithm.

Answer 1

There is at least two strategies:

• For identities like $\sin(3t)=3 \sin(t)-4\sin^3(t)$, i.e expressing $\sin(nt)$ as a sum of powers of $\sin(t)$, this is actually the Chebyshev polynomials.
• For expressing $\sin^n(t)$ as a sum of sinuses $\sin(kt)$ you can use Moivre formula and a bynomial expansion: $$\sin^3(t)=\left(\frac{e^{it}-e^{-it}}{2i} \right)^3=\frac{1}{-8i} \left( e^{3it}-3e^{(2i-i)t}+3e^{(i-2i)t}-e^{-3it}\right)=\frac{1}{4}\left(-\frac{e^{3it}-e^{-3it}}{2i}+3\frac{e^{it}-e^{-it}}{2i}\right)$$ so: $$\sin^3(t)=\frac{1}{4}(-\sin(3t)+3 \sin(t))$$

Answer 2

You can use the definition of sine and cosine with Euler's formula for a more "constructed" way of handling such identities, I'd not algorithmics. For example, sine can be defined as $sint =\frac{ e^{it} - e^{-it}}{2i}$. As such we can see that:

$\sin(3t) = \frac{e^{i3t} - e^{-i3t}}{2i} = \frac{(e^{it} - e^{-it})^3 + 3e^{i2t - it} - 3e^{it -i2t}}{2i} = -4 \frac{(e^{it} - e^{-it})^3}{(2i)^3} +3 \frac{ e^{it} - e^{-it}}{2i} = -4(\sin t)^3 + 3\sin t$

As seen by the identities you provided.

Answer 3

Without using higher level mathematics I am wondering whether a binary type algorithm might help where the fundamental forms are idenities like

$\sin(a\pm b)=\sin a \;\cos b \;\pm \cos a \;\sin b\;\;$ and $\;\;\cos(a\pm b)=\cos a \;\cos b \;\mp \sin a \;\sin b$

Since $$\sin (2^n a)=2 \sin (2^{n-1} a)\cos(2^{n-1} a)$$ and $$\cos (2^n a)= \cos^2 (2^{n-1} a)-\sin^2(2^{n-1} a)$$

For example in the case of proving $$\sin 6a=32 \cos^5 a \sin a-32 \cos^3 \sin a +6 \cos a \sin a \tag{1}$$

We can start with the left hand side of $(1)$ $$ \sin 6a=\sin 4a \cos 2a + \cos 4a \sin 2a$$

Quickly getting to $$ \sin 6a=\sin 2a \left(3\cos^2 2a - \sin^2 2a \right)=\sin 2a\left( 4 \cos^2 2a - 1 \right)$$

[or alternatively $ \sin 6a = \sin 2a\left( 3- 4 \sin^2 2a \right)$]

Then continue with the r.h.s. of $(1)$ $$32 \cos^5 a \sin a-32 \cos^3 \sin a +6 \cos a \sin a = \sin 2a \left( 16 \cos^4 a - 16 \cos^2 a +3 \right)$$

and we are then simply left to prove $$ 4 \cos^2 2a - 1 = 16 \cos^4 a - 16 \cos^2 a +3 $$

An identity involving $\sin 13a$ for example would decompose as

$$\sin 13a =\sin 12a \cos a + \cos 12a \sin a$$ then $$\sin 13a =\left(\sin 8a \cos 4a+ \cos 8a \sin 4a \right) \cos a + \left(\cos 8a \cos 4a- \sin 8a \sin 4a \right) \sin a$$ and so on...