Solving x/2 in 3rd Quadrant: Cosx=-7/9

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SUMMARY

The discussion focuses on solving for cos(x/2) and sin(x/2) given that cos(x) = -7/9 in the third quadrant. The correct approach involves using the half-angle identities, specifically sin²(x/2) = (1 - cos(x))/2. Participants confirm that this method is valid and provide guidance on simplifying the results. The final expressions for sin(x/2) and cos(x/2) are derived from these identities, ensuring the correct signs are applied based on the quadrant.

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  • Study the derivation and application of half-angle identities in trigonometry.
  • Practice solving trigonometric equations in different quadrants.
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Students studying trigonometry, educators teaching trigonometric identities, and anyone needing to solve trigonometric equations involving half-angle formulas.

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Homework Statement



We have cosx = -7/9 in the third quadrant, and my question is how to find cos(x/2) and sin(x/2).

I've tried using the sin^2x=1-cos2x/2 formula and its adjacent cosine formula. Is this the correct formula I should be using to get the answer for this question?
 
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Yes. You say you tried, what happened? Try rewriting the equality as ## \sin^2\frac{x}{2} = \frac{1 - \cos x}{2} ##; make sure your answer has the correct sign.
 
MrAnchovy said:
Yes. You say you tried, what happened? Try rewriting the equality as ## \sin^2\frac{x}{2} = \frac{1 - \cos x}{2} ##; make sure your answer has the correct sign.

It worked for the sin(x/2), but when I tried for the cos(x/2)... I got 1+(-7/9) / 2 which becomes sqrt(2/18). Is this what you get ??
 
You can simplify \sqrt{\frac{2}{18}}, you know. :-p Also, there's one little thing you're forgetting. MrAnchovy mentions it in his post.
 

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