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Math Help - general solution

  1. #1
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    general solution

    Gievn that tan A =3 and sin (A-B)=2 cos (A+B) , find tan B

    i got tan B=-1/5

    Find general solution , in rad cos b + cos 3b +cos 5b =0

    cos b + 2 cos 4b cos b =0

    cos b (1+2cos 4b )=0

    cos b =0

    and cos b = -1/2

    I can only reach this far , i am not sure about the general solution >

    Deduce \cos^2 b+\cos^2 3b+\cos^2 5b=\frac{3}{2}

    no idea bout this .
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  2. #2
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    Hello, thereddevils!

    Your answer is correct . . . Good work!


    \text{Given: }\:\tan A = 3\:\text{ and }\:\sin(A-B)\:=\:2\cos (A+B),\;\text{ find }\tan B

    We are given: . \sin(A - B) \;=\;2\cos(A + B)

    Then: . \sin A\cos B - \sin B\cos A \;=\;2(\cos A\cos B - \sin A\sin B)

    . . . . . \sin A\cos B - \sin B\cos A \;=\;2\cos A\cos B - \sin A\sin B


    Divide by \cos A\!:\quad \frac{\sin A\cos B}{\cos A} - \frac{\sin B\cos A}{\cos A} \;=\;\frac{2\cos A\cos B}{\cos A} - \frac{\sin A\sin B}{\cos A}

    . . . . . . . . . . . . . . . . \underbrace{\tan A}_3\cos B - \sin B \;=\;2\cos B - 2\underbrace{\tan A}_3\sin B


    And we have: . 3\cos B - \sin B \;=\;2\cos B - 6\sin B

    . . . . . . . . . . . . . . . 5\sin B \;=\;-\cos B

    . . . . . . . . . . . . . . . . \frac{\sin B}{\cos B} \;=\;-\frac{1}{5}

    . . . . . . . . . . . . . . . . \boxed{\tan B \;=\;-\frac{1}{5}}

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  3. #3
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    Hello again, thereddevils!

    Find the general solution in radians: . \cos x + \cos3x +\cos5x \:=\:0

    \cos x + 2\cos4x\cos x \:=\:0

    \cos x (1+2\cos4x) \:=\:0 . . . . Correct!

    We have: . \cos x \:=\:0 \quad\Rightarrow\quad \boxed{ x \:=\:\frac{\pi}{2} + \pi n}

    And: . 1+2\cos4x\:=\:0 \quad\Rightarrow\quad \cos4x \:=\:-\tfrac{1}{2}

    . . . 4x \:=\:\pm\frac{2\pi}{3} + 2\pi n \quad\Rightarrow\quad\boxed{ x \:=\:\pm\frac{\pi}{6} + \frac{\pi}{2}n }

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  4. #4
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    Hello thereddevils
    Quote Originally Posted by thereddevils View Post
    Deduce \cos^2 b+\cos^2 3b+\cos^2 5b=\frac{3}{2}
    I think we have a problem here if this result is supposed to apply to all the solutions of the equation \cos b+\cos 3b+\cos 5b=0.

    Look at the solution b = \frac{\pi}{2}+n\pi = (2n+1)\frac{\pi}{2}. In other words, b is any odd multiple of \frac{\pi}{2}.

    If b is an odd multiple of \frac{\pi}{2}, then so are 3b and 5b, and at these values \cos b = \cos 3b = \cos 5b = 0.

    So, obviously, \cos^2 b+\cos^2 3b+\cos^2 5b=0 also.

    However, when we consider the other set of values of b (those for which \cos 4b = -\tfrac12) we do get the required result. Here's why:

    \cos^2 b+\cos^2 3b+\cos^2 5b=\tfrac12(1+\cos2b)+\tfrac12(1+\cos6b)+\tfrac12(  1+\cos10b)

    =\tfrac32+\tfrac12(\cos2b+\cos6b+\cos10b)

    =\tfrac32+\tfrac12(2\cos6b\cos4b+\cos6b)

    =\tfrac32+\tfrac12\cos6b(2\cos4b+1)

    =\tfrac32, when \cos4b = -\tfrac12

    So: are you sure you posted the complete question?

    Grandad
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