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Thread: inverse

  1. #1
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    inverse

    If sin^-1(a)=%pi/6 and cos^-1(%pi/4)=b, then the exact value of sin(%pi*a+b) is:

    i think you get rid of the inverse by going

    sin %pi/6 = a so then a = 1/2

    cosb = %pi/4

    but im not sure how you solve for b or get the value the question is asking for.....
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  2. #2
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    Quote Originally Posted by samtheman17 View Post
    If sin^-1(a)=%pi/6 and cos^-1(%pi/4)=b, then the exact value of sin(%pi*a+b) is:

    i think you get rid of the inverse by going

    sin %pi/6 = a so then a = 1/2

    cosb = %pi/4

    but im not sure how you solve for b or get the value the question is asking for.....
    Good start ,

    $\displaystyle \sin^{-1} a=\frac{\pi}{6}$

    $\displaystyle a=\frac{1}{2}$

    $\displaystyle \cos^{-1} \frac{\pi}{4}=b$

    $\displaystyle \cos b=\frac{\pi}{4}$

    $\displaystyle =\sin (\pi\cdot a+b)=\sin (\frac{\pi}{2}+b)$

    $\displaystyle =\sin \frac{\pi}{2}\cos b+\cos \frac{\pi}{2}\sin b$

    $\displaystyle
    =\cos b=\frac{\pi}{4}
    $
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  3. #3
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    ahh okay thanks,

    but then how do you put it into the form of


    sin(%pi*a + b)?

    i think you have to use

    sin(a+b)=cos(a)sin(b)+sin(a)cos(b)

    but i can't seem to get it right.....
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  4. #4
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    Quote Originally Posted by samtheman17 View Post
    ahh okay thanks,

    but then how do you put it into the form of


    sin(%pi*a + b)?

    i think you have to use

    sin(a+b)=cos(a)sin(b)+sin(a)cos(b)

    but i can't seem to get it right.....
    $\displaystyle \sin (\frac{1}{2}\pi+b)$ ... substitute the a with 1/2

    after that , that's what i did , expand using the sin formulas .
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