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Thread: simplifying identities

  1. #1
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    simplifying identities

    Simplify:

    \frac{sin6x}{sin2x} - \frac{cos6x}{cos2x}



    How do i simplify sin6x and cos6x?
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  2. #2
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    Hint: \sin (4x)=\sin (6x-2x)=\sin (6x)\cos (2x)-\cos (6x)\sin (2x), and \sin(4x)=2\sin(2x)\cos(2x).
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  3. #3
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    Hello, shawli!


    Simplify: . \frac{\sin6x}{\sin2x} - \frac{\cos6x}{\cos2x}

    \text{Combine the fractions: }\;\frac{\overbrace{\sin6x\cos2x - \cos6x\sin2x}^{\text{This is }\sin(6x-2x)}}{\sin2x\cos2x} \;=\;\frac{\sin4x}{\sin2x\cos2x}


    . . . . . . . . . . . . . =\frac{\sin4x}{\frac{1}{2}(\underbrace{2\sin2x\cos  2x)}_{\text{This is }\sin4x}} \;=\;\frac{\sin4x}{\frac{1}{2}\sin4x} \;=\;2

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  4. #4
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    You screwed out my Hint.

    Next time let them to proceed with the Hint.
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  5. #5
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    thank you both!
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