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Math Help - Need 3 Homework problems

  1. #1
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    Need 3 Homework problems

    Verify these: [i need help getting started and other help you can throw at me]

    1) sin(x + y)cos y - cos (x + y)sin y = sin x

    2) cot x = (cos3x + cos x) / (sin 3x - sin x)

    3) tan(B/2) = sec B / (sec B csc B + csc B)
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  2. #2
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    Look up (and learn) the expansion rules for sin(a+b), cos(a+b), tan(a+b), sin (a-b),
    sin2a, etc. And apply them here. Have a go or you'll never never know.
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  3. #3
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    Hello Johnny Walker Black
    Quote Originally Posted by Johnny Walker Black View Post
    Verify these: [i need help getting started and other help you can throw at me]

    1) sin(x + y)cos y - cos (x + y)sin y = sin x

    2) cot x = (cos3x + cos x) / (sin 3x - sin x)

    3) tan(B/2) = sec B / (sec B csc B + csc B)
    Here's the method for each one:

    1) Use the identities below to expand the LHS:
    \sin(x+y) = \sin x \cos y +\cos x \sin y

    \cos(x+y) = \cos x \cos y - \sin x \sin y
    Simplify the result. Then take out a common factor of \sin x. Then use:
    \cos^2y + \sin^2y = 1
    and you're done.

    2) Use the identities:
    \cos 3x = 4\cos^3x - 3\cos x

    \sin 3x = 3\sin x - 4\sin^3 x
    Simplify the result, and factorise. Then use:
    \cos^2x = 1 - \sin^2 x
    in the numerator. Simplify; cancel, and you're there.

    3) Get rid of \sec B and \csc B (horrible things!) by multiplying top-and-bottom of the fraction by \sin B\cos B, using the fact that:
    \csc B\sin B = 1 and \sec B \cos B = 1
    Then use:
    \sin B = 2 \sin\tfrac12B\cos\tfrac12B

    \cos B = 2\cos^2\tfrac12B - 1
    Simplify. Then use \tan\tfrac12B = \frac{\sin\tfrac12B}{\cos\tfrac12B}, and you're there.

    Grandad
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