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Math Help - sinus, cosinus expression

  1. #1
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    sinus, cosinus expression

    sinx = -2/(5^0.5)

    3pi/2 < x < 2pi

    the question that i donīt understand is that i shuold use this values and decide the exacly value of this expression: cos(2x-3pi/4)

    so i did this way

    arcsin(-2/(5^0.5) = x = -63.4

    cos(2*(-63.4)-3pi/4)

    are i doing right ?
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by paulaa View Post
    sinx = -2/(5^0.5)

    3pi/2 < x < 2pi

    the question that i donīt understand is that i shuold use this values and decide the exacly value of this expression: cos(2x-3pi/4)

    so i did this way

    arcsin(-2/(5^0.5) = x = -63.4

    cos(2*(-63.4)-3pi/4)

    are i doing right ?
    But... 3pi/2 < x < 2pi (270* < x < 360*)
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  3. #3
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    sinx = -2/(5^0.5)
    3pi/2 < x < 2pi

    x = 2*pi*n-sin^(-1)(2/sqrt(5)), där n = Z

    n=1.

    x = 2*pi*1-sin^(-1)(2/sqrt(5))= ~ 5.17604

    is this the answer for this question ?
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  4. #4
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    \displaystyle \cos{\left(2x - \frac{3\pi}{4}\right)} = \cos{2x}\cos{\frac{3\pi}{4}} + \sin{2x}\sin{\frac{3\pi}{4}}

    \displaystyle = \cos{2x}\cos{\left(\pi - \frac{\pi}{4}\right)} + \sin{2x}\sin{\left(\pi - \frac{\pi}{4}\right)}

    \displaystyle = -\cos{\frac{\pi}{4}}\cos{2x} + \sin{\frac{\pi}{4}}\sin{2x}

    \displaystyle = -\frac{\sqrt{2}}{2}\cos{2x} + \frac{\sqrt{2}}{2}\sin{2x}

    \displaystyle = -\frac{\sqrt{2}}{2}\left(1 - 2\sin^2{x}\right) + \frac{\sqrt{2}}{2}( 2\sin{x}\cos{x})

    \displaystyle = -\frac{\sqrt{2}}{2}\left(1 - 2\sin^2{x}\right) + \sqrt{2}\sin{x}\sqrt{1 - \sin^2{x}}

    \displaystyle = -\frac{\sqrt{2}}{2}\left[1 - 2\left(-\frac{2}{\sqrt{5}}\right)^2\right] + \sqrt{2}\left(-\frac{2}{\sqrt{5}}\right)\sqrt{1 - \left(-\frac{2}{\sqrt{5}}\right)^2}

    \displaystyle = -\frac{\sqrt{2}}{2}\left[1 - 2\left(\frac{4}{5}\right)\right] - \frac{2\sqrt{2}}{\sqrt{5}}\sqrt{1 - \frac{4}{5}}

    \displaystyle = -\frac{\sqrt{2}}{2}\left(-\frac{3}{5}\right) - \frac{2\sqrt{2}}{\sqrt{5}}\left(\frac{1}{\sqrt{5}}  \right)

    \displaystyle = \frac{3\sqrt{2}}{10} - \frac{2\sqrt{2}}{5}

    \displaystyle = \frac{3\sqrt{2}}{10} - \frac{4\sqrt{2}}{10}

    \displaystyle = -\frac{\sqrt{2}}{10}.
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  5. #5
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    Quote Originally Posted by paulaa View Post
    sinx = -2/(5^0.5)

    3pi/2 < x < 2pi

    the question that i donīt understand is that i should use this value and decide the exact value of this expression: cos(2x-3pi/4)

    so i did this way

    arcsin(-2/(5^0.5) = x = -63.4

    cos(2*(-63.4)-3pi/4)

    are i doing right ?
    Not quite!

    You are being asked to calculate an exact value, so you need to proceed
    similarly to Prove It's method.

    \displaystyle\ Sinx=-\frac{2}{\sqrt{5}}

    You could reference the Unit Circle also and write

    Cos({\pi}-A)=-CosA,\;\;\;Sin({\pi}-A)=SinA

    Also, the angle x is in the 4th quadrant from the range of x given.

    You could sketch a right-angled triangle and use Pythagoras' theorem to find Cosx

    \displaystyle\ 2^2+1^2=5\Rightarrow\ Cosx=\frac{1}{\sqrt{5}}


    Therefore....

    \displaystyle\ Cos\left(2x-\frac{3{\pi}}{4}\right)=Cos(2x)Cos\left(\frac{3{\p  i}}{4}\right)+Sin(2x)Sin\left(\frac{3{\pi}}{4}\rig  ht)

    from Cos(A-B)=CosACosB+SinASinB

    and since

    \displaystyle\ Cos\left(\frac{3{\pi}}{4}\right)=-Cos\left(\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}}

    while

    \displaystyle\ Sin\left(\frac{3{\pi}}{4}\right)=Sin\left(\frac{\p  i}{4}\right)=\frac{1}{\sqrt{2}}

    then

    \displaystyle\ Cos\left(2x-\frac{3{\pi}}{4}\right)=-\frac{Cos(2x)}{\sqrt{2}}+\frac{Sin(2x)}{\sqrt{2}}

    Using

    \displaystyle\ Sin^2x=\frac{1}{2}\left[1-Cos(2x)\right]\Rightarrow\ Cos(2x)=1-2Sin^2x=1-2\left[\frac{4}{5}\right]=-\frac{3}{5}

    and

    \displaystyle\ Sin(2x)=2SinxCosx=-\frac{4}{\sqrt{5}}\;\left[\frac{1}{\sqrt{5}}\right]=-\frac{4}{5}

    \displaystyle\Rightarrow\ Cos\left(2x-\frac{3{\pi}}{4}\right)=\frac{3}{5\sqrt{2}}-\frac{4}{5\sqrt{2}}=-\frac{1}{5\sqrt{2}}
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  6. #6
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    Quote Originally Posted by Archie Meade View Post
    Not quite!

    You are being asked to calculate an exact value, so you need to proceed
    similarly to Prove It's method.

    \displaystyle\ Sinx=-\frac{2}{\sqrt{5}}

    You could reference the Unit Circle also and write

    Cos({\pi}-A)=-CosA,\;\;\;Sin({\pi}-A)=SinA

    Also, the angle x is in the 4th quadrant from the range of x given.

    You could sketch a right-angled triangle and use Pythagoras' theorem to find Cosx

    \displaystyle\ 2^2+1^2=5\Rightarrow\ Cosx=\frac{1}{\sqrt{5}}


    Therefore....

    \displaystyle\ Cos\left(2x-\frac{3{\pi}}{4}\right)=Cos(2x)Cos\left(\frac{3{\p  i}}{4}\right)+Sin(2x)Sin\left(\frac{3{\pi}}{4}\rig  ht)

    from Cos(A-B)=CosACosB+SinASinB

    and since

    \displaystyle\ Cos\left(\frac{3{\pi}}{4}\right)=-Cos\left(\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}}

    while

    \displaystyle\ Sin\left(\frac{3{\pi}}{4}\right)=Sin\left(\frac{\p  i}{4}\right)=\frac{1}{\sqrt{2}}

    then

    \displaystyle\ Cos\left(2x-\frac{3{\pi}}{4}\right)=-\frac{Cos(2x)}{\sqrt{2}}+\frac{Sin(2x)}{\sqrt{2}}

    Using

    \displaystyle\ Sin^2x=\frac{1}{2}\left[1-Cos(2x)\right]\Rightarrow\ Cos(2x)=1-2Sin^2x=1-2\left[\frac{4}{5}\right]=-\frac{3}{5}

    and

    \displaystyle\ Sin(2x)=2SinxCosx=-\frac{4}{\sqrt{5}}\;\left[\frac{1}{\sqrt{5}}\right]=-\frac{4}{5}

    \displaystyle\Rightarrow\ Cos\left(2x-\frac{3{\pi}}{4}\right)=\frac{3}{5\sqrt{2}}-\frac{4}{5\sqrt{2}}=-\frac{1}{5\sqrt{2}}
    \displaystyle = -\frac{\sqrt{2}}{10} :P
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