Tuesday

November 25, 2014

November 25, 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!

**Answer this Question**

**Related Questions**

algebra - Can someone please help me do this problem? That would be great! ...

Mathematics - Trigonometric Identities - Let y represent theta Prove: 1 + 1/tan^...

Calculus - I wanted to confirm that I solved these problems correctly (we had to...

calculus - conic sections prove that the line x-2y+10=0 touches the ellipse 9x^2...

Precalculus with Trigonometry - Prove or disprove the following Identities: cos...

Math- Trig - 1. Determine the exact value of cos^-1 (pi/2). Give number and ...

Math - Calculus - The identity below is significant because it relates 3 ...

Trigonometry desperate help, clueless girl here - 2. solve cos 2x-3sin x cos 2x...

math - Prove the identity cos theta - sin theta / cos theta + sin theta = sec 2 ...

math(Trigonometry) - sin 2 theta + cos theta = 0 so, we use sin 2 theta = 2 sin ...