Question

PLEASE HELP MEOWT!!! Rewrite sin^(4)xtan^(2)x in terms of the first power of cosine.

Answer 1

Step-by-step explanation:

${ \sin(x) }^{4} { \tan(x) }^{2}$

${ \sin(x) }^{4} \frac{ { \sin(x) }^{2} }{ { \cos(x) }^{2} }$

$\frac{ ({ {1 - \cos(x) }^{2} })^{3} }{ { \cos(x) }^{2} }$

hopefully this helps, I'm rusty with my trig identities

Answer 2

Answer:

  sin⁴(x)tan²(x) = (10 -15cos(2x) +6cos(4x) -cos(6x))/(16(1 +cos(2x))

Step-by-step explanation:

The relevant identities are ...

$\sin^4{x}=\dfrac{3-4\cos{(2x)}+\cos{(4x)}}{8}\\\\\tan^2{x}=\dfrac{1-\cos{(2x)}}{1+\cos{(2x)}}\\\\\cos{(a)}\cos{(b)}=\dfrac{\cos{(a+b)}+\cos{(a-b)}}{2}$

Then your product is ...

$\sin^4{(x)}\tan^2{(x)}=\dfrac{3-4\cos{(2x)}+\cos{(4x)}}{8}\cdot\dfrac{1-\cos{(2x)}}{1+\cos{(2x)}}\\\\=\dfrac{3-4\cos{(2x)}+\cos{(4x)}-3\cos{(2x)}+4\cos^2{(2x)}-\cos{(4x)}\cos{(2x)}}{8(1+\cos{(2x)})}$

Collecting terms and using the identity for the product of cosines, we get ...

$=\dfrac{3-7\cos{(2x)}+\cos{(4x)}+4\dfrac{1+\cos{(4x)}}{2}-\dfrac{\cos{(6x)}+\cos{(2x)}}{2}}{8(1+\cos{(2x)})}\\\\=\dfrac{10-15\cos{(2x)}+6\cos{(4x)}-\cos{(6x)}}{16(1+\cos{(2x)})}$