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Thread: 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: $\displaystyle cos(y)cos(-2y)-sin(y)sin(-2y)$
    this is what i have done and i cannot see where i have gone wrong.
    $\displaystyle =cos((y)+(-2y))$
    $\displaystyle =cos(y)$ 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 $\displaystyle sin(x)cos(2x)+cos(x)sin(2x)$.

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

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

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

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

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

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

    book's answer is$\displaystyle 2+\sqrt3$
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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: $\displaystyle cos(y)cos(-2y)-sin(y)sin(-2y)$
    this is what i have done and i cannot see where i have gone wrong.
    $\displaystyle =cos((y)+(-2y))$
    $\displaystyle =cos(y)$ 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 $\displaystyle sin(x)cos(2x)+cos(x)sin(2x)$.
    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 $\displaystyle tan(\frac{5\pi}{12})$.

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

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

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

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

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

    book's answer is$\displaystyle 2+\sqrt3$
    Multiply your answer by $\displaystyle \frac{1-\sqrt{3}}{1-\sqrt{3}}$ (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?

    $\displaystyle cos^2(x)=1-sin^2(x)$

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

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

    $\displaystyle =\sqrt3^2-1^2$

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

    $\displaystyle =4-2\sqrt3$
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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

    $\displaystyle \frac{\sqrt3-1}{1+\sqrt3}=\frac{\sqrt3-1}{1+\sqrt3}\cdot\frac{1-\sqrt3}{1-\sqrt3}=\frac{-(1-\sqrt3)^2}{(1+\sqrt3)(1-\sqrt3)}=\frac{-\dots}{1-3}=\dots$
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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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    $\displaystyle sin(3x) = sin (2x +x) $
    $\displaystyle = sin2xcosx + cos2xsinx$
    $\displaystyle = cosx(2sinxcosx) + sinx(cos^2x - sin^2x)$
    $\displaystyle = 2sinx cos^2x + sinx(1- 2sin^2x)$
    $\displaystyle = 2sinxcos^2x + sinx - 2sin^3x$
    $\displaystyle = 2sinx(1-sin^2x) + sinx - 2sin^3x$
    $\displaystyle = 2sinx - 2sin^3x + sinx - 2sin^3x$
    $\displaystyle = 3sinx - 4sin^3x $
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