Find an expression for cos 3x in terms of cos x?
Answer 1:
It can be rewritten in terms of two addition identities:
\(\sin (u+v)=\sin u \cos v+\cos u \sin v\)
\(\cos (u+v)=\cos u \cos v-\sin u \sin v\)
From the identities above, we have:
\(\sin (2 x)=2 \sin x \cos x\)
\(\cos (2 x)=\cos ^{2} x-\sin ^{2} x\)
Hence, we have:
\(\sin (3 x)=(2 \sin x \cos x) \cos x+\left(\cos ^{2} x-\sin ^{2} x\right) \sin x\)
\(=2 \sin x \cos ^{2} x-\sin ^{3} x+\sin x \cos ^{2} x\)
\(=3 \sin x \cos ^{2} x-\sin ^{3} x\)
\(=3 \sin x\left(1-\sin ^{2} x\right)-\sin ^{3} x\)
\(=3 \sin x-4 \sin ^{3} x\)
Answer 2:
\(\sin 3 x=3 \sin x-4 \sin ^{3} x\)
Explanation:
Another approach to the excellent answer from @Truong-Son N. , which is less work should you need higher powers/multiples is to use de Moivre’s theorem, using complex numbers.
de Moivre’s theorem states that for any complex number \(z\) in trigonometric form \(z=\cos \theta+i \sin \theta\), then
\((\cos \theta+i \sin \theta)^{n}=\cos n \theta+i \sin n \theta\)With \(n=3\) we have:
And expanding the LHS using the binomial theorem we have:
\(\begin{aligned}&(\cos \theta)^{3}+3(\cos \theta)^{2}(i \sin \theta)+3(\cos \theta)(i \sin \theta)^{2}+(i \sin \theta)^{3}=\cos 3 \theta \\
&+i \sin 3 \theta
\end{aligned}\)
\(∴ \cos ^{3} \theta+i 3 \cos ^{2} \theta \sin \theta-3 \cos \theta \sin ^{2} \theta-i \sin ^{3} \theta=\cos 3 \theta+i \sin 3 \theta\)
\(∴ \left(\cos ^{3} \theta-3 \cos \theta \sin ^{2} \theta\right)+i\left(3 \cos ^{2} \theta \sin \theta-\sin ^{3} \theta\right)=\cos 3 \theta+i \sin 3 \theta\)
If we equate imaginary components we get:
\(\sin 3 \theta=3 \cos ^{2} \theta \sin \theta-\sin ^{3} \theta\)As we want the expression in terms of \(\sin \theta\) we can replace \(\cos ^{2} \theta\) using the fundamental identity \(\sin ^{2} A+\cos ^{2} A \equiv 1\) to get:
\sin 3 \theta &=3\left(1-\sin ^{2} \theta\right) \sin \theta-\sin ^{3} \theta \\
&=3 \sin \theta-3 \sin ^{3} \theta-\sin ^{3} \theta \\
&=3 \sin \theta-4 \sin ^{3} \theta
\end{aligned}\)
Or:
\(\sin 3 x=3 \sin x-4 \sin ^{3} x\)Incidentally, as an extension we also get an expression for \(\cos 3 x\) for free! Equating real components we get:
\(\begin{aligned}\cos 3 \theta &=\cos ^{3} \theta-3 \cos \theta \sin ^{2} \theta \\
&=\cos ^{3} \theta-3 \cos \theta\left(1-\cos ^{2} \theta\right) \\
&=\cos ^{3} \theta-3 \cos \theta+3 \cos ^{3} \theta \\
&=4 \cos ^{3} \theta-3 \cos \theta
\end{aligned}\)
Or:
\(\cos 3 x=4 \cos ^{3} x-3 \cos x\)