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anonymous

  • one year ago

Rewrite with only sin x and cos x. sin 2x - cos 2x

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  1. anonymous
    • one year ago
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    This is so confusing because I know that what I did was correct, but my answer is not one of the choices.

  2. campbell_st
    • one year ago
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    well what do you know about sin(2x) and cos(2x)

  3. anonymous
    • one year ago
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    So I kind of made the choices as the guide, so here's what I did.

  4. anonymous
    • one year ago
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    \[\sin2x-\cos2x\] \[2 \sin x \cos x-(1-2\sin^2x)\] \[2 \sin x \cos x -1+2\sin^2x\]

  5. anonymous
    • one year ago
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    Here are the choices btw. a.2 sinx cosx - 1 + 2 sin2x b. 2 sin x cos2x - 1 + 2 sin2x c. 2 sin x cos2x - sin x + 1 - 2 sin2x d. 2 sin x cos2x - 1 - 2 sin2x And my answer is A.

  6. campbell_st
    • one year ago
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    that works.... so looking at the answer options they have used cos(2x) = 1 - 2 sin^2(x) so as you have posted 2sin(x)cos(x) - (1 - 2sin^2(x)) = 2 sin(x)cos(x) -1+ 2sin^2(x)

  7. anonymous
    • one year ago
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    I know that sin^2x = 2 sin x cos x cos 2x = 1-2sin^2x = 2 cos^2x-1 =cos^2x-sin^2x

  8. campbell_st
    • one year ago
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    you have the correct solution

  9. anonymous
    • one year ago
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    So am I correct?

  10. anonymous
    • one year ago
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    Oh okay thanks! :)

  11. campbell_st
    • one year ago
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    yes \[2\sin(x)\cos(x) - 1 + 2 \sin^2(x)\]

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