Trignometric equation - simplify LHS first.

Question

I know I have to begin by simplifying the LHS but it always becomes an algebraic mess. Please help

Answer 1

Assuming you are allow to use complex numbers, then we have \begin{align} \sin^8\frac{x}{2}+\cos^8\frac{x}{2} =&\ \left(\frac{e^{ix/2}+e^{-ix/2}}{2}\right)^8+\left(\frac{e^{ix/2}-e^{-ix/2}}{2i}\right)^8\\ =&\ \frac{1}{2^8}\sum^8_{k=0}\binom{8}{k}e^{ikx/2}e^{-i(8-k)x/2}+\frac{1}{2^8}\sum^8_{k=0}\binom{8}{k}(-1)^{8-k}e^{ikx/2}e^{-i(8-k)x/2}\\ =&\ \frac{1}{2^7}\left(\binom{8}{0}e^{-i4x}+\binom{8}{2}e^{-i2x}+\binom{8}{4}+\binom{8}{6}e^{i2x}+\binom{8}{8}e^{i4x} \right)\\ =&\ \frac{1}{2^6}\left(\cos4x+28\cos 2x+35 \right). \end{align} Then it follows \begin{align} 8\left(\sin^8\frac{x}{2}+\cos^8\frac{x}{2} \right) = 5 \ \ \implies& \ \ \cos4x+28\cos 2x+35 = 40\\ \implies&\ \cos 4x + 28 \cos 2x -5= 0 \\ \implies&\ \cos^2 2x+14\cos 2x -3=0\\ \implies&\ \cos 2x = \frac{-14+ \sqrt{14^2+12}}{2}. \end{align} Hence \begin{align} \cos 4 x = 2\left( \frac{-14+ \sqrt{14^2+12}}{2}\right)^2-1 = 201-56\sqrt{13} \end{align}

Answer 2

Hint:

$$ \cos ^8 (\frac{x}{2}) + \sin ^8 (\frac{x}{2})= \frac{5}{8}$$

$$ \implies \cos ^8 (\frac{x}{2}) +\sin ^8 (\frac{x}{2}) \pm 2\cos ^4 (\frac{x}{2}) \sin ^4 (\frac{x}{2}) = \frac{5}{8}$$

$$ \implies [\cos ^4 (\frac{x}{2}) - \sin ^4 (\frac{x}{2})]^2 + 2\cos ^4 (\frac{x}{2}) \sin ^4 (\frac{x}{2}) = \frac{5}{8}$$

$$ [\cos ^2 (\frac{x}{2}) + \sin ^2 (\frac{x}{2})]^2[\cos ^2 (\frac{x}{2}) - \sin ^2 (\frac{x}{2})]^2+ 2\cos ^4 (\frac{x}{2}) \sin ^4 (\frac{x}{2}) = \frac{5}{8}$$

$$ \cos ^2 (x) + \frac{1}{8} \sin ^4 (x) = \frac{5}{8}$$

$$\implies 1 - \sin ^2 (x) + \frac{1}{8} \sin ^4 (x) = \frac{5}{8}$$

So, the acceptable value from quadratic equation: $$ \implies \sin^2 (x) = {4-\sqrt{13}}.$$

Then: $$ \implies \cos(4x) = 1 - 2\sin^2 (2x) = 1 - 8\sin^2 (x)\cos^2 (x) = 201-56\sqrt{13}.$$

Answer 3

Hint -

$\cos^8 \frac x2 + \sin^8 \frac x2 = \frac 58$

$\left( \cos^4 \frac x2 + \sin^4 \frac x2 \right)^2 - 2 \sin^4 \frac x2 \cos^4 \frac x2 = \frac 58$

$\left[ \left( \cos^2 \frac x2 + \sin^2 \frac x2 \right)^2 - 2\sin^2 \frac x2 \cos^2 \frac x2 \right]^2 - 2 \sin^4 \frac x2 \cos^4 \frac x2 = \frac 58$

$\left[ (1)^2 - 2\sin^2 \frac x2 \cos^2 \frac x2 \right]^2 - 2 \sin^4 \frac x2 \cos^4 \frac x2 = \frac 58$

$\left[ 1 - 2\sin^2 \frac x2 \cos^2 \frac x2 \right]^2 - 2 \sin^4 \frac x2 \cos^4 \frac x2 = \frac 58$

$1 + 4 \sin^4 \frac x2 \cos^4 \frac x2 - 4 \sin^2 \frac x2 \cos^2 \frac x2 - 2 \sin^4 \frac x2 \cos^4 \frac x2 = \frac 58$

$1 + 2 \sin^4 \frac x2 \cos^4 \frac x2 - 4 \sin^2 \frac x2 \cos^2 \frac x2 = \frac 58$

Answer 4

HINT:

Using $\cos2A=1-2\sin^2A=2\cos^2A-1$

$$\left(2\sin^2\dfrac x2\right)^4+\left(2\cos^2\dfrac x2\right)^4=(1-\cos x)^4+(1+\cos x)^4$$

$$=2\left[1+\binom42\cos^2x+\binom44\cos^4x\right]$$

$$\implies5=1+6\cos^2x+\cos^4x$$

Again apply $2\cos^2x=1+\cos2x$ to form a Quadratic Equation in $\cos2x$