Exponentiating Sine

$\sin5 \theta$ = $\sin^{5} \theta$ , mas

$\sin5 \theta$ = $\sin(3\theta + 2\theta)$ , mas

$\sin(3\theta + 2\theta)$ = $\sin3 \theta$ $\cos2 \theta$ + $\sin2 \theta$ $\cos3 \theta$ , e

$\sin(3\theta + 2\theta)$ = $(3\sin \theta - 4\sin^{3} \theta)$ $(1 - 2\sin^{2} \theta)$ + $(2\sin \theta\cos \theta)$ $(4\cos^{3} \theta - 3\cos \theta)$ .

Resolvendo a equação, teremos:

$\sin5 \theta$ = $16\sin^{5} \theta - 20\sin^{3} + 5\sin \theta$

$\sin^{5} \theta$ = $16\sin^{5} \theta - 20\sin^{3} + 5\sin \theta$

$20\sin^{3} \theta - 5\sin \theta$ = $15\sin^{5} \theta$

$5\sin \theta(4\sin^{2} \theta - 1)$ = $15\sin^{5} \theta$ . Então,

$4\sin^{2} \theta - 1$ = $3\sin^{4} \theta$ .

$3\sin^{4} \theta - 4\sin^{2} \theta + 1$ = 0.

Resolvendo esta equação, teremos:

$sin \theta$ = $\frac{\sqrt{3}}{3}$

Calculando o $cos \theta$ , obteremos:

$\cos^{2} \theta$ = $1 - \sin^{2} \theta$

$\cos \theta$ = $\frac{\sqrt{6}}{3}$

Utilizando a definição de $\tan \theta$ , temos:

$\tan \theta$ = $\frac{\sin \theta}{\cos \theta}$

$\tan \theta$ = $\frac{\sqrt{2}}{2}$

Calculando a $\tan2 \theta$ :

$\tan2 \theta$ = $\frac{2\tan \theta}{1 - \tan^{2} \theta}$ , temos:

$\tan2 \theta$ = $2\sqrt{2}$ . Portanto,

$a + b = 2 + 2 = \boxed{4}$

By the de Moivre formula (or simple but tedious trigonometry) we can express $\sin 5\theta$ as follows: $\sin 5\theta = \sin ^5 \theta -10\cos ^2 \theta\sin ^3 \theta + 5\cos ^4 \theta \sin \theta$

Equating this with $\sin ^5 \theta$ , and dividing out by $-5 \cos^5 \theta$ , we get $2 \frac {\sin^3 \theta}{\cos^3 \theta} - \frac {\sin \theta}{\cos \theta}=0$ . The solution that we eliminate from this step is irrelevant, because if the cosine of an angle is zero, its tangent is undefined. [Not exactly true. It is irrelevant because it falls out of range - Calvin] This simplifies to $2 \tan^3 \theta - \tan \theta = 0$ , which yields $\tan \theta = 0, \pm \frac {1}{\sqrt{2}}$ . We ignore the roots $0$ and $-\frac {1}{\sqrt{2}}$ since they fall out of range.

Using the double angle formula, we get $\tan (2 \theta) = \frac {2 \tan \theta}{1 - \tan^2 \theta} = 2 \sqrt{2}$ . Hence, $a+b=4$ .

Seeing $\sin 5\theta = \sin^5 \theta$ , it looks like something similar to de Moivre's theorem, so let's try it out with that. ${ \cos 5\theta + i \sin 5\theta } = { (\cos \theta + i \sin \theta)^5 }$ $\cos 5\theta + i \sin 5\theta = c^5 +i5c^4s - 10c^3s^2 -10ic^2s^3 + 5cs^4 + is^5$ where $c = \cos \theta , s = \sin \theta$

Equating the imaginary parts, $\sin 5\theta = 5\cos^4 \theta \sin \theta -10\cos^2 \theta \sin^3 \theta + \sin^5 \theta$ For $\sin 5\theta = \sin^5 \theta$ , $5\cos^4 \theta \sin \theta -10\cos^2 \theta \sin^3 \theta = 0$ $\cos^2 \theta \sin \theta ( \cos^2 \theta -2\sin^2 \theta) =0$ $\cos^2 \theta = 0$ and $\sin \theta = 0$ is rejected since they give solutions of $\theta=\frac {\pi}{2}$ and $\theta = 0$ respectively. So we're left with $\cos^2 \theta - 2\sin^2 \theta = 0$ $\tan^2 \theta = \frac {1}{2}$ $\tan \theta = \frac {1}{ \sqrt 2}$ Since $\tan 2\theta = \frac {2\tan \theta}{1-\tan^2 \theta}$ , $\tan 2\theta = \frac {\sqrt 2}{1 - \frac {1}{2}}$ $\tan 2\theta = 2\sqrt 2$ $\Rightarrow a+b =2+2=4$

First we simplify $sin 5\theta$ into an expression containing only $sin \theta$ .

This can be done using the formula: >- $sin (A+B)$ = $cos (A)$ $sin (B)$ + $cos (B)$ $sin (A)$

We write $sin 5\theta$ as $sin (4\theta + \theta)$ and then use the above formula to expand and simplify to expressions containing only $sin \theta$ .

Simplified expression:

• $sin 5\theta$ = 5 $sin \theta$ - 20 $sin^3 \theta$ + 16 $sin^5 \theta$

Substituting the simplified expression into given equation $sin 5\theta$ = $sin^5 \theta$ , we get:

• 5 $sin \theta$ - 20 $sin^3 \theta$ + 16 $sin^5 \theta$ = $sin^5 \theta$

  5 $sin \theta$ - 20 $sin^3 \theta$ + 15 $sin^5 \theta = 0$

Substituting k for $sin \theta$ we get:

$15k^5 -20k^3 +5k = 0$

If we simliply the expression we get:

$k ( k^2 -1 ) ( 3k^2 - 1 ) = 0$

Hence, $k = 0, k = (+/-)1 \ or \ k = (+/-) \frac{1}{\sqrt{3}}$

Since $\theta$ is in the first quadrant, all the values of $sin \theta$ will be positive.

Hence, $k = 0,\ 1,\ or\ \frac{1}{\sqrt{3}}$

Replacing k for $sin \theta$ and using $sin^-1 (\theta)$ , we find that:

• $\theta = 0^{\circ} \ , \ 90^{\circ} \ or (\ sin^-1 (\theta))^{\circ}$

Since $\ tan 0^{\circ} = 0$ which cannot be expressed in the a\sqrt{b} format, and $\ tan 90^{\circ} = \infty$ , $\theta = 0^{\circ}$ and $\theta = 90^{\circ}$ cannot be a solution.

Hence, $\ \sin \theta \ = \frac{1}{\sqrt{3}}$ is our required value. Now, we will simplify $tan \theta$ . We can write $tan \theta$ as:

$tan \theta = \frac{\sin 2\theta}{\cos 2\theta}$

Using the trigonometric identitities:

• $sin 2\theta = 2sin \theta cos \theta$

• $cos 2\theta = cos^2 \theta -sin^2 \theta$

We get:

$tan 2\theta = \frac{2sin \theta cos \theta}{cos^2 \theta -sin^2 \theta}$

Now using the triangle concept, if $sin \theta = \frac{1}{\sqrt{3}}$ , then for angle $\theta$ the opposite side is 1 unit and the hypotenuse is $\sqrt{3}$ .

Then by using the Pythagoras Theorem, the adjacent side will be : $\sqrt{ \ 3 \ - \ 1} = \sqrt{2}$ .

Since, $\ cos \theta \$ is $\frac{adjacent side}{hypotenuse}$ , Therefore, $\ cos \theta \ = \frac{\sqrt{2}}{\sqrt{3}}$ .

Substituting the values of $sin \theta$ and $cos \theta$ into the expression for $tan \theta$ , we get the answer 4.

We know that ${sin5\theta}=5cos^4\theta\sin\theta-10cos^2\theta\sin^3\theta+sin^5\theta$ and we are given that ${sin5\theta}=sin^5\theta$ i.e. ${sin5\theta}=5cos^4\theta\sin\theta-10cos^2\theta\sin^3\theta+sin^5\theta=sin^5\theta$ which simplifies to $5cos^4\theta\sin\theta=10cos^2\theta\sin^3\theta$ dividing both sides by ${cos^2\theta\sin\theta}$ we get ${5cos^2\theta=10sin^2\theta}$ using the identity ${\sin^2\theta+\cos^2\theta={1}}$ the above equation simplifies to ${\sin^2\theta}=\frac{1}{3}$ from which ${\sin\theta=\frac{1}{\sqrt{3}}}$ using identities and the above value of sine ${cos\theta=\frac{\sqrt{6}}{3}}$ and ${tan\theta=\frac{1}{\sqrt{2}}}$ We know that ${tan2\theta}=\frac{2tan\theta}{1-tan^2\theta}$ substituting for tan we get ${\frac{2\times\frac{1}{\sqrt{2}}}{1-\frac{1}{2}}}$ from which we get ${tan2\theta}=2\sqrt{2}$ *but ${a\sqrt{b}}=2\sqrt{2}$ Hence ${a}={2},{b}={2}$ and ${a+b}={2+2}=\boxed{4}$

Let's calculate $sin 5 \theta$ in terms of $sin \theta$ .

$sin 5 \theta = sin 5 \theta + sin \theta - sin \theta$

$= 2sin 3 \theta cos 2 \theta - sin \theta$

$= 2(3 sin \theta - 4 sin^{3} \theta )(1 - 2 sin^{2} \theta ) - sin \theta$

$= 16sin^{5} \theta - 20sin^{3} \theta + 5sin \theta$

Since $sin 5 \theta = sin^{5} \theta$ , we can write

$16sin^{5} \theta - 20sin^{3} \theta + 5sin \theta = sin^{5} \theta$

$15sin^{5} \theta - 20sin^{3} \theta + 5sin \theta = 0$

$3sin^{5} \theta - 4sin^{3} \theta + sin \theta = 0$

$sin \theta (3sin^{4} \theta - 4sin^{2} \theta + 1) = 0$

$sin \theta = 0$ or $3sin^{4} \theta - 4sin^{2} \theta + 1 = 0$

$\theta = 0$ or $sin^{2} \theta = \frac{1}{3}$ or $sin^{2} \theta = 1$

$\theta = 0$ or $sin^{2} \theta = \frac{1}{3}$ or $\theta = \pm{90^{o}}$

First and third conditions can't be applied as $tan 2 \theta = \frac{sin 2 \theta}{cos 2 \theta}$ is not of the form $a \sqrt{b}$ as mentioned in the question.

Hence, $sin^{2} \theta = \frac{1}{3}$

$sin \theta = \frac{1}{\sqrt{3}}$ .

$cos \theta = \frac{\sqrt{2}}{\sqrt{3}}$

Hence, $tan 2 \theta = \frac{sin 2 \theta}{cos 2 \theta} = \frac{2sin \theta cos \theta}{1 - 2sin^{2} \theta}$

$= 2\sqrt{2}$

So, $a + b = 2 + 2 = 4$

That's the answer!

We can write above equation as

$sinx((1-2sin^{2}2x) + (4(1-2sin^{2}2x)\cos^{2}x)sinx)=\sin ^{5}x$

Since sin x lies strictly lies between 0 to pie/2, we can simplify this as :

$3sin^{3}x-4sin^{2}x+1=0$ which yields $sin^{2}=1/3$ sin x lies between 0 and pie/2

from this we get $cos 2x= 1/3, sin 2x=2\sqrt{2}/3$ implies $tan 2x= 2\sqrt{2}$

What I did was restrict the amount of possible values. Because $0 \le \theta \le \frac{\pi}{2}$ , we know that $0 \le \sin \theta \le 1$ . Therefore, we can also state the following:

$0 \le \sin ^{5} \theta \le 1$

By the given equallity, we now know that $0 \le \sin 5\theta \le 1$ , which means that:

$2k\pi \le 5\theta \le (2k+1)\pi, \forall k\in N$

But, because $\theta \le \frac{\pi}{2}$ we're left with:

$0 \le \sin 5\theta \le 1 \Rightarrow 0 \le \theta \le \frac{\pi}{5}$

So, now, we can say that $0 \le \tan 2\theta \le (\tan (\frac{2\pi}{5}) \approx 3.0777)$ . And, then:

$0 < a\sqrt{b} < 3.0777$

This means that $0 < ba^{2} < 9.473$ . As $a$ and $b$ are integers: $0 < ba^{2} \le 9$ . We now have a total of 12 possible a's and b's. What you do next is just try every combination until you get a $\theta$ that works. In this case, a=2 and b=2. So a+b=4 (that's the answer to the problem).

Please, let me know if there is a better (perhaps more algebraic) way to solve it. Thank you very much!