# mth, trig

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Given cos a=-7/25 in quadrant II , and sin b=-12/13, in quadrant IV , find cos(a+b) and sin2a

• mth, trig -

sin a = 24/25
cos b = 5/13

• mth, trig -

Thanks but how did you find that? My numbers came out completely different -_-

• mth, trig -

If a is in QII, sin is positive, cos is negative.

7^2 + 24^2 = 25^2, so it's a 7-24-25 triangle

If b is in QIV, sin is negative, cos is positive.

5^2 + 12^2 = 13^2, so it's a 5-12-13 triangle.

what did you do to find the missing sin/cos values?

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