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Math Help - Trigonometric identities

  1. #1
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    Trigonometric identities

    Hi,

    I'm working with a question that I think mainly involves trig identities, but I'm not sure if I have the right approach and can't figure out the later parts.

    If cos(x)=-5/7 where pi < x < 3pi/2, find the values of cos(2x), sin(2x), cos(x/2), sin(x/2)
    I'm also not sure how it being in the third quadrant plays a role...

    For the first two, I worked out that cos(2x)=1/49 and sin(2x)=4sqrt6/7 with double angle formulas, but I'm not sure if those are correct and I don't know where to go now.

    Thanks!
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  2. #2
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    Quote Originally Posted by metaname90 View Post
    Hi,

    I'm working with a question that I think mainly involves trig identities, but I'm not sure if I have the right approach and can't figure out the later parts.



    I'm also not sure how it being in the third quadrant plays a role...

    For the first two, I worked out that cos(2x)=1/49 and sin(2x)=4sqrt6/7 ? with double angle formulas, but I'm not sure if those are correct and I don't know where to go now.

    Thanks!
    Given that the angle is in the 3rd quadrant, you may express the angle as x=180^o+A

    Hence, since Cosx<-\frac{1}{\sqrt{2}}\Rightarrow\ x<225^o\Rightarrow\ 2x<450^o or 2x<2{\pi}+\frac{{\pi}}{2}

    2x=360^o+2A\Rightarrow\ Cos2x=Cos2A=2Cos^2A-1=2\frac{25}{49}-\frac{49}{49}=\frac{1}{49}

    Cos^2(2x)+Sin^2(2x)=1\Rightarrow\ Sin2x=\sqrt{\frac{49^2-1}{49^2}}

    \frac{x}{2} is in the 2nd quadrant.

    \frac{x}{2}=90^o+\frac{A}{2}

    If you draw a circle and sketch the angles, you will see we can calculate

    Cos\left(90+\frac{A}{2}\right)=-Sin\left(\frac{A}{2}\right)

    and to find this you can use Sin^2B=\frac{1}{2}(1-Cos2B)\Rightarrow\ Sin^2\left(\frac{A}{2}\right)=\frac{1}{2}(1-CosA) etc
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