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September 21, 2014

September 21, 2014

Posted by **mathstudent** on Tuesday, December 19, 2006 at 11:33pm.

(a * cos t + b * sin t)^2 <= a^2 + b^2

I will be happy to critique your work. Start on the left, square it,

(a * cos t + b * sin t)^2 =

a^2 (1 - sin^2t) + 2ab sin t cost+ b^2 (1 - cos^2 t)=

a^2 + b^2 - (a sin t - b cos t)^2

<= (a^2 + b^2)

because (a sin t - b cos t)^2 mus be positive or zero.

I don't know how you figured that out, but thanks. That works perfectly!

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