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Let \(\theta \) be an acute angle such that \(\sin 5 \theta = \sin^5 \theta\)
Compute \(\tan 2 \theta \)

 

this isn't my homework or anything, really curious how to do it

 Jun 6, 2021
 #1
avatar+32633 
+4

Here's a helping hand:

 

 

See if you can take it from here.

 Jun 6, 2021
 #4
avatar+114818 
0

Alan, can you please explain your very first line.

That is where I got lost.        frown sad

Melody  Jun 7, 2021
 #2
avatar+208 
+2

because of DeMoivre's Theorem:

\(\begin{align*} \operatorname{cis} n \theta &= (\operatorname{cis} \theta)^n \\ &= (\cos \theta + i \sin \theta)^n \\ &= \cos^n \theta + \binom{n}{1} i \cos^{n - 1} \theta \sin \theta - \binom{n}{2} \cos^{n - 2} \theta \sin^2 \theta - \binom{n}{3} i \cos^{n - 3} \theta \sin^3 \theta + \dotsb. \end{align*}\)

\(\begin{align*} \cos n \theta &= \cos^n \theta - \binom{n}{2} \cos^{n - 2} \theta \sin^2 \theta + \binom{n}{4} \cos^{n - 4} \theta \sin^4 \theta - \dotsb, \\ \sin n \theta &= \binom{n}{1} \cos^{n - 1} \theta \sin \theta - \binom{n}{3} \cos^{n - 3} \theta \sin^3 \theta + \binom{n}{5} \cos^{n - 5} \theta \sin^5 \theta - \dotsb. \end{align*}\)

\(\begin{align*} \sin 5 \theta &= \binom{5}{1} \cos^4 \theta \sin \theta - \binom{5}{3} \cos^2 \theta \sin^3 \theta + \binom{5}{5} \sin^5 \theta \\ &= 5 \cos^4 \theta \sin \theta - 10 \cos^2 \theta \sin^3 \theta + \sin^5 \theta. \end{align*}\)

\(\sin 5 \theta = \sin^5 \theta \) 

becomes 

\(5 \cos^4 \theta \sin \theta - 10 \cos^2 \theta \sin^3 \theta + \sin^5 \theta = \sin^5 \theta \)

\(5 \cos^4 \theta \sin \theta - 10 \cos^2 \theta \sin^3 \theta = 0 \)

\(\theta\) is an acute angle, so \(\cos^2 \theta - 2 \sin^2 \theta = 0 \)

\(\cos^2 \theta = 2 \sin^2 \theta\)

so, I know that \(\tan^2 \theta = \frac{1}{2}\)

\(\tan \theta = \frac{1}{\sqrt{2}}\)

\(\tan 2 \theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} = \frac{\sqrt{2}}{1 - \frac{1}{2}} = {2 \sqrt{2}} \)

 Jun 6, 2021
 #3
avatar+114818 
+1

That is an impressive answer JKP.

Did you do it yourself?

Melody  Jun 7, 2021

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