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sunni0$U

  • 2 years ago

Find sin x/2, cos x/2, and tan x/2, if cos x = -12/13, 180 degrees is less than x which is less than 270 degrees

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  1. myininaya
    • 2 years ago
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    |dw:1352334915182:dw| But anyways, mainly I drew this to show what quadrant we are in. sine and cosine are both negative in this quadrant Do you know the half angle identities for sine and cosine?

  2. sunni0$U
    • 2 years ago
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    I know them but I don't know how to use them.

  3. myininaya
    • 2 years ago
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    \[\sin(\frac{x}{2})=\pm \sqrt{\frac{1}{2}(1-\cos(x))}\] \[\cos(\frac{x}{2})=\pm \sqrt{\frac{1}{2}(1+\cos(x))}\] \[\tan(\frac{x}{2})=\frac{\sin(\frac{x}{2})}{\cos(\frac{x}{2})}\]

  4. sunni0$U
    • 2 years ago
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    I know how to get the decimal answers, but I need exact answers.

  5. myininaya
    • 2 years ago
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    Since we are looking at the third quadrant and I said sine and cosine are both negative there, then we are actually looking at: \[\sin(\frac{x}{2})=-\sqrt{\frac{1}{2}(1-\cos(x))}\] \[\cos(\frac{x}{2})=-\sqrt{\frac{1}{2}(1+\cos(x))}\] Replace cos(x) with what it equals which is -12/13 then simplify

  6. sunni0$U
    • 2 years ago
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    For sin x/2 I got -sqrt of 25/26, which is wrong apparently, but I got cos x/2=-sqrt of 1/26 and it was right. Also, how do I find tan x/2 from all of this?

  7. myininaya
    • 2 years ago
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    \[\sin(\frac{x}{2})=-\sqrt{\frac{1}{2}(1-\frac{-12}{13})} =- \sqrt{\frac{1}{2}(1+\frac{12}{13})}=-\sqrt{\frac{1}{2}(\frac{13}{13}+\frac{12}{13})}\] \[=-\sqrt{\frac{1}{2}(\frac{25}{13})}=-\sqrt{\frac{25}{26}}=-\frac{\sqrt{25}}{\sqrt{26}}=-\frac{5}{\sqrt{26}}\] This answer can be simplified more by choosing to rationalize the denominator

  8. sunni0$U
    • 2 years ago
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    Ok thanks. I figured out the tan x/2 part

  9. myininaya
    • 2 years ago
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    np.

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