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he66666

  • 5 years ago

Trig Identity question? The angle x lies in the interval pi/2≤x≤pi, and sin²x=4/9. Determine cos(x/2). answer: √[(3-√5)/6] How do you solve this question (using trig identities)? It gets confusing because it's x/2, not just x. I keep getting 7/9 as my answer..

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  1. anonymous
    • 5 years ago
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    \[cos(\frac{x}{2})=\pm \sqrt{\frac{1+cos(x)}{2}}\]

  2. anonymous
    • 5 years ago
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    square root should be over whole thing.

  3. he66666
    • 5 years ago
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    How does it have a square root? :S Is it derived from a trig identity?

  4. anonymous
    • 5 years ago
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    \[sin^2(x)=\frac{4}{9}\] \[sin(x)=\pm \frac{3}{4}\]

  5. anonymous
    • 5 years ago
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    derived from "double angle" formula \[cos(2x)=2cos^2(x)-1\] replace 2x by x t so x gets replaced by \[\frac{x}{2}\] and then solve for \[cos(\frac{x}{2})\]

  6. anonymous
    • 5 years ago
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    \[cos(x)=2cos^2(\frac{x}{2})-1\] \[cos(x)+1=2cos^2(\frac{x}{2})\] \[\frac{cos(x)-1}{2}=cos^2(\frac{x}{2})\]

  7. anonymous
    • 5 years ago
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    typo should be \[\frac{cos(x)+1}{2}=cos^2(\frac{x}{2})\]

  8. anonymous
    • 5 years ago
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    then take the square root of the whole thing. plus or minus of course. i cannot seem to type set it.

  9. anonymous
    • 5 years ago
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    in case \[sin(x)=\frac{3}{4}\] do you know how to find \[cos(x)\]?

  10. he66666
    • 5 years ago
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    yes, pythagorean theorem?

  11. anonymous
    • 5 years ago
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    whoa hold the phone i am tired. it is \[sin(x)=\frac{4}{9}\]

  12. anonymous
    • 5 years ago
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    damn \[sin^2(x)=\frac{4}{9}\] \[sin(x)=\frac{2}{3}\]

  13. anonymous
    • 5 years ago
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    draw a triangle opposite side 2 and hypotenuse 3 my mistake sorry

  14. anonymous
    • 5 years ago
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    adjacent side is \[\sqrt{3^2-2^2}=\sqrt{9-4}=\sqrt{5}\]

  15. anonymous
    • 5 years ago
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    so \[cos(x)=\frac{\sqrt{5}}{3}\]

  16. anonymous
    • 5 years ago
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    now plug in \[\frac{\sqrt{5}}{3}\] in the formula up top to get your answer. i will write it if you like.

  17. he66666
    • 5 years ago
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    no it's alright, I got it. Thanks so much for the help satellite! :)

  18. anonymous
    • 5 years ago
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    oh another mistake i forgot x was in quad II so cosine is negative. it is \[-\frac{\sqrt{5}}{3}\]

  19. anonymous
    • 5 years ago
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    make sure you use \[-\frac{\sqrt{5}}{3}\]

  20. he66666
    • 5 years ago
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    alright, I will :) thanks

  21. anonymous
    • 5 years ago
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    welcome

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