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Math Help - A few questions

  1. #1
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    A few questions

    Hi
    Here are some of the questions I'm stuck on.
    1)Simply the expression: cos(y)cos(-2y)-sin(y)sin(-2y)
    this is what i have done and i cannot see where i have gone wrong.
    =cos((y)+(-2y))
    =cos(y) answers says its cos(-y).why?

    2)express sin(3x) in terms of sin(x). This questions follows on from the previous question which was asking to expand sin(x+2x) answer to this is sin(x)cos(2x)+cos(x)sin(2x).

    3)Use the compound angle formulas and appropriate angles to find the exact value of tan(\frac{5\pi}{12}).

    This is what i have done:
    tan(\frac{5\pi}{12}=tan(\pi-\frac{7\pi}{12})

    =tan(\pi-(\frac{\pi}{3}+\frac{\pi}{4})

    =\frac{tan(\pi)-tan(\frac{\pi}{3})-tan(\frac{\pi}{4})-tan(\pi)tan(\frac{\pi}{3})tan(\frac{\pi}{4})}{1+ta  n(\pi)tan(\frac{\pi}{3})+tan(\pi)tan(\frac{\pi}{4}  )+tan(\frac{\pi}{3})tan(\frac{\pi}{4})}

    =\frac{0-\sqrt3-1-0*\sqrt3*1}{1+0*\sqrt3+0*1+\sqrt3*1}

    =\frac{\sqrt3-1}{1+\sqrt3}

    book's answer is  2+\sqrt3
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  2. #2
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    Quote Originally Posted by Paymemoney View Post
    Hi
    Here are some of the questions I'm stuck on.
    1)Simply the expression: cos(y)cos(-2y)-sin(y)sin(-2y)
    this is what i have done and i cannot see where i have gone wrong.
    =cos((y)+(-2y))
    =cos(y) answers says its cos(-y).why?
    You're correct, and the answer's correct too, because cos(-x)=cos(x)

    2)express sin(3x) in terms of sin(x). This questions follows on from the previous question which was asking to expand sin(x+2x) answer to this is sin(x)cos(2x)+cos(x)sin(2x).
    Yes, but you can write cos(2x)=1-2sinČ(x) and sin(2x)=2sin(x)cos(x)
    and then use cosČ(x)=1-sinČ(x) to simplify and get something with respect to sin(x) only.

    3)Use the compound angle formulas and appropriate angles to find the exact value of tan(\frac{5\pi}{12}).

    This is what i have done:
    tan(\frac{5\pi}{12}=tan(\pi-\frac{7\pi}{12})

    =tan(\pi-(\frac{\pi}{3}+\frac{\pi}{4})

    =\frac{tan(\pi)-tan(\frac{\pi}{3})-tan(\frac{\pi}{4})-tan(\pi)tan(\frac{\pi}{3})tan(\frac{\pi}{4})}{1+ta  n(\pi)tan(\frac{\pi}{3})+tan(\pi)tan(\frac{\pi}{4}  )+tan(\frac{\pi}{3})tan(\frac{\pi}{4})}

    =\frac{0-\sqrt3-1-0*\sqrt3*1}{1+0*\sqrt3+0*1+\sqrt3*1}

    =\frac{\sqrt3-1}{1+\sqrt3}

    book's answer is  2+\sqrt3
    Multiply your answer by \frac{1-\sqrt{3}}{1-\sqrt{3}} (multiplying the numerator and the denominator by the conjugate of the denominator - it's similar to 'rationalizing the denominator')
    and use the identities : (a-b)(a+b)=aČ-bČ and (a-b)Č=aČ+bČ-2ab
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  3. #3
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    ok thanks
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  4. #4
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    So in question 2 does that mean i need to substitute values into?

    cos^2(x)=1-sin^2(x)

    This is what i done for question 3 but i still don't get the answer, what have a done wrong.

    =\frac{(\sqrt3-1)(1+\sqrt3)}{(1+\sqrt3)(1+\sqrt3)}

    =\sqrt3^2-1^2

    =\sqrt3^2+1^2-2\sqrt3

    =4-2\sqrt3
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  5. #5
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    For question 2, no. Follow the steps and use this identity to get the final form.

    For the other one... you're not multiplying by what I said

    \frac{\sqrt3-1}{1+\sqrt3}=\frac{\sqrt3-1}{1+\sqrt3}\cdot\frac{1-\sqrt3}{1-\sqrt3}=\frac{-(1-\sqrt3)^2}{(1+\sqrt3)(1-\sqrt3)}=\frac{-\dots}{1-3}=\dots
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  6. #6
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    if you don;t mind could you do question 2 for me coz i still can't get the answer -_-.
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  7. #7
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     sin(3x) = sin (2x +x)
     = sin2xcosx + cos2xsinx
     = cosx(2sinxcosx) + sinx(cos^2x - sin^2x)
     = 2sinx cos^2x + sinx(1- 2sin^2x)
     = 2sinxcos^2x + sinx - 2sin^3x
     = 2sinx(1-sin^2x) + sinx - 2sin^3x
     = 2sinx - 2sin^3x + sinx - 2sin^3x
     = 3sinx - 4sin^3x
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