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Math Help - Trigo Help!!!

  1. #1
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    Trigo Help!!!

    1.) Prove 6sin4xcos4x = 3sin8x

    2.) Derive sin( u + v) = sinucosv + cosusinv from the identity
    cos(u+v) =cosucosv - sinusinv
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  2. #2
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    Quote Originally Posted by ^_^Engineer_Adam^_^
    1.) Prove 6sin4xcos4x = 3sin8x
    Use the identity:

    sin(u + v) = sin(u) cos(v) + cos(u) sin(v)

    with v=u=4x, then multiply the whole thing by 3.

    RonL
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  3. #3
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    Quote Originally Posted by ^_^Engineer_Adam^_^
    2.) Derive sin( u + v) = sinucosv + cosusinv from the identity
    cos(u+v) =cosucosv - sinusinv
    cos(u+v)=cos(u)cos(v)-sin(u)sin(v)
    \sqrt{1-sin^{2}(u+v)}=cos(u)\sqrt{1-sin^{2}(v)}+sinu\sqrt{1-cos^{2}(v)}
    Solve for sin(u+v)
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  4. #4
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    Quote Originally Posted by ^_^Engineer_Adam^_^
    2.) Derive sin( u + v) = sinucosv + cosusinv from the identity
    cos(u+v) =cosucosv - sinusinv
    You have,
    \cos (u+v)=\cos u\cos v-\sin u\sin v
    Then, by co-funtions
    \sin (u+v)=\cos \left( \frac{\pi}{2}-(u+v) \right)
    Thus,
    \cos \left[ \left( \frac{\pi}{2}-u\right)-v\right]
    Thus,
    \cos \left( \frac{\pi}{2}-u\right)\cos u+\sin\left( \frac{\pi}{2}-u \right) \sin v
    Thus,
    \sin u\cos v+\cos u\sin v
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  5. #5
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    Hello, ^_^Engineer_Adam^_^!


    1) Prove: . 6\sin4x\cos4x \,= \,3\sin8x

    You're expected to know the identity: . 2\sin\theta\cos\theta \,=\,\sin2\theta

    We have: . 6\sin4x\cos4x \;=\:3(2\sin4x\cos4x) \:=\:3(\sin 8x)

    Therefore: . 6\sin4x\cos4x \:=\:3\sin8x

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  6. #6
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    Thanks!!!
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  7. #7
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    Quote Originally Posted by shubhadeep
    cos(u+v)=cos(u)cos(v)-sin(u)sin(v)
    \sqrt{1-sin^{2}(u+v)}=cos(u)\sqrt{1-sin^{2}(v)}+sinu\sqrt{1-cos^{2}(v)}
    Solve for sin(u+v)
    I see how you derived the irrationals, but can you please explain the next few steps?

    Thanks
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  8. #8
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    Quote Originally Posted by chancey
    I see how you derived the irrationals, but can you please explain the next few steps?

    Thanks
    Its much easier to do it using the method presented by ImPerfect
    Hacker, less algebra, less to go wrong.

    RonL
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  9. #9
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    Quote Originally Posted by CaptainBlank
    Its much easier to do it using the method presented by ImPerfect
    Hacker, less algebra, less to go wrong.

    RonL
    Also, I do not think it is entirely correct.
    Because,
    \sqrt{1-\sin^2 x}=\cos x
    Is not true for all x!
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  10. #10
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    Quote Originally Posted by ThePerfectHacker
    Also, I do not think it is entirely correct.
    You can make it correct by definiton \mbox{H4X}(x). It has the propery that,
    \mbox{H4X}(x)=\sqrt{1-\sin^2 x} when it should be and,
    \mbox{H4X}(x)=\sqrt{1-\sin^2 x} when it should be.
    But that is not ideal.
    ---
    Note: If you want to use the radical method you cannot write,
    \pm \sqrt{1-\sin^2 x}. Because the expression would not be well-definied, i.e. not a function. My way through the H4X function is ye only way.
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  11. #11
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    Hello, ^_^Engineer_Adam^_^!

    I found a way, but it's not worth the trouble . . .


    2) Derive \sin( u + v) \:= \:\sin u\cos v + \cos u\sin v
    from the identity: \cos(u+v) \:=\:\cos u\cos v - \sin u\sin v

    Square the given identity:
    . . \cos u\cos v - \sin u\sinv)^2" alt="\cos^2(u + v)\:=\\cos u\cos v - \sin u\sinv)^2" />
    . . \cos^2(u + v)\:= \:\cos^2\!u\cos^2\!v - 2\sin u\sin v\cos u\cos v  + \sin^2\!u\sin^2\!v

    Subtract from 1:
    . . 1 - \cos^2(u + v) \:=\:1 - \cos^2\!u\cos^2\!v +  2\sin y\sin v\cos u\cos v - \sin^2\!u\sin^2\!v

    Add and subtract \sin^2\!v and \cos^2\!v


    . . = \:1 + cosēv - \cos^2\!u\cos^2\!v + 2\sin u\sin v\cos u\cos v + sinēv   - \sin^2\!u\sin^2\!v - cosēv - sinēv

    Factor:
    . . = \:1 + \cos^2\!v(1 - \cos^2\!u) + 2\sin u\sin v\cos u\cos v +  \sin^2\!v(1 - \sin^2\!u) - (\sin^2\!v + \cos^2\!v)

    Note that: 1 - (\sin^2\!v + \cos^2\!v) \:=\:0 . . . and we have:
    . . \sin^2(u+v)\:=\:\sin^2\!u\cos^2\!v + 2\sin u\sin v\cos u\cos v  + \sin^2\!v\s\cos^2\!u

    Factor: . \sin^2(u+v) \:=\:\left(\sin u\cos v + \sin v\cos u)^2

    Therefore: . \boxed{\sin(u + v)\:=\:\sin u\cos v + \sin v\cos u}


    I need a nap . . .

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  12. #12
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    just another question... what if the case would be deriving sin(u - v) to cos(u-v)?

    Elow thanks people!!!
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  13. #13
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    Quote Originally Posted by ^_^Engineer_Adam^_^
    just another question... what if the case would be deriving sin(u - v) to cos(u-v)?

    Elow thanks people!!!
    Just stick with what you have, you will have derived

    <br />
\sin( u + v)=\sin (u)\cos (v) + \cos (u)\sin (v)<br />

    now use the symmetry properties of \sin and \cos to get what you want:

    <br />
\sin(u-v)=\sin( u + (-v))=\sin (u)\cos (-v) + \cos (u)\sin (-v)<br />

    but \sin(-x)=-\sin(x), and \cos(-x)=\cos(x), so:

    <br />
\sin(u-v)=\sin (u)\cos (v) - \cos (u)\sin (v)<br />

    RonL
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