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

Question

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

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Gabriella 2 weeks 2021-09-15T03:02:27+00:00 2 Answers 0

Answers ( )

    0
    2021-09-15T03:04:07+00:00

    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)})}

    0
    2021-09-15T03:04:08+00:00

    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

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27:3+15-4x7+3-1=? ( )