Question

Trignometry problem help plz
no 3, 5 and 6​

Answer 1

3. First factor $\cos^6A-\sin^6A$ as a difference of cubes:

$\cos^6A-\sin^6A=\underbrace{(\cos^2A-\sin^2A)}_{\cos2A}(\cos^4A+\cos^2A\sin^2A+\sin^4A)$

For the remaining group, apply the double angle identity.

$\cos^2A=\dfrac{1+\cos2A}2$

$\sin^2A=\dfrac{1-\cos2A}2$

$\implies\begin{cases}\cos^4A=\left(\dfrac{1+\cos2A}2\right)^2=\dfrac{1+2\cos2A+\cos^22A}4\\\\\cos^2A\sin^2A=\dfrac{(1+\cos2A)(1-\cos2A)}4=\dfrac{1-\cos^22A}4\\\\\sin^4A=\left(\dfrac{1-\cos2A}2\right)^2=\dfrac{1-2\cos2A+\cos^22A}4\end{cases}$

$\implies4(\cos^6A-\sin^6A)=\cos2A[(1+2\cos2A+\cos^22A)+(1-\cos^22A)+(1-2\cos2A+\cos^22A)]$

$=\cos2A(3+\cos^22A)=\cos^32A+3\cos2A$

5. seems rather tricky. You might want to post another question for that problem alone...

6. Factorize the left side as a sum of cubes:

$\cos^320^\circ+\sin^310^\circ=(\cos20^\circ+\sin10^\circ)(\cos^220^\circ-\cos20^\circ\sin10^\circ+\sin^210^\circ)$

From here we have to prove that

$\cos^220^\circ-\cos20^\circ\sin10^\circ+\sin^210^\circ=\dfrac34$

We can write everything in terms of sine:

$\cos^220^\circ=(1-2\sin^210^\circ)^2=1-4\sin^210^\circ+4\sin^410^\circ$ (double angle identity)

$\cos20^\circ\sin10^\circ=\dfrac{\sin30^\circ-\sin10^\circ}2=\dfrac14-\dfrac12\sin10^\circ$ (angle sum identity)

After some simplifying, we're left with showing that

$4\sin^410^\circ-3\sin^210^\circ+\dfrac12\sin10^\circ=0$

or

$4\sin^310^\circ-3\sin10^\circ+\dfrac12=0$

This last equality follows from what you could the triple angle identity for sine,

$\sin3x=3\sin x-4\sin^3x$