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aussy123

  • 3 years ago

Given: costheta= -4/5, sin x =-12/13 ,theta is in the third quadrant, and x is in the fourth quadrant; evaluate cos 2x. -119/169 119/169 -11/13

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  1. anonymous
    • 3 years ago
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    you don't need the sine part

  2. aussy123
    • 3 years ago
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    then how do I start

  3. terenzreignz
    • 3 years ago
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    Two variables?

  4. anonymous
    • 3 years ago
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    \[\cos(2x)=2\cos^2(x)-1\]

  5. tomhoffhim
    • 3 years ago
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    |dw:1364529999084:dw|

  6. anonymous
    • 3 years ago
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    oh i see there is a \(\theta\) and an \(x\)

  7. aussy123
    • 3 years ago
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    costheta= -4/5 I need cosx

  8. anonymous
    • 3 years ago
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    you are asked for \(\cos(2x)\) where \(\sin(x)=-\frac{12}{13}\)

  9. anonymous
    • 3 years ago
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    \[\cos(2x)=1-2\sin^2(x)\]

  10. aussy123
    • 3 years ago
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    When I do it I get 169

  11. aussy123
    • 3 years ago
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    a denominator of 169

  12. anonymous
    • 3 years ago
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    there is no \(\theta \) in your question

  13. tomhoffhim
    • 3 years ago
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    ok, I'll do it for you, give me a second, this is an easy question

  14. aussy123
    • 3 years ago
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    yea I had to put theta for that

  15. tomhoffhim
    • 3 years ago
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    satellite, you don't know what you're doing..

  16. anonymous
    • 3 years ago
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    \[\cos(x)=1-2\sin^2(x)\] \[\cos(x)=1-(\frac{-12}{13})^2\]\

  17. anonymous
    • 3 years ago
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    there is your 169

  18. anonymous
    • 3 years ago
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    \[\cos(2x)=1-(\frac{12}{13})^2\] \[\cos(2x)=1-\frac{144}{169}\]

  19. anonymous
    • 3 years ago
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    \[\cos(2x)=\frac{25}{169}\]

  20. aussy123
    • 3 years ago
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    I know, thats what I get, but it isnt a choice

  21. anonymous
    • 3 years ago
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    your question asks for \(\cos(2x)\) and doesn't mention any \(\theta\)

  22. aussy123
    • 3 years ago
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    I know but that is what is already given

  23. anonymous
    • 3 years ago
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    oh damn i made a mistake!! sorry

  24. tomhoffhim
    • 3 years ago
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    I'm too lazy to do it but you need to draw the triangles and either use theta or x for the angle

  25. aussy123
    • 3 years ago
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    lol its okay, Ive made plenty

  26. anonymous
    • 3 years ago
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    \[\cos(2x)=1-2\sin^2(x)\] \[\cos(2x)=1-2\times (\frac{12}{13})^2\] \[cos(2x)=1-2\times \frac{144}{169}\]

  27. anonymous
    • 3 years ago
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    \[\cos(2x)=1-\frac{288}{169}\]\[\cos(2x)=-\frac{119}{169}\]

  28. aussy123
    • 3 years ago
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    Lol this is a lot of work, can you use this same formula for tan2Theta

  29. aussy123
    • 3 years ago
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    nvm. I found the formula, Thank you!

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