Question

Help me understand this solution (picture included, trig related).
Calculus, finding absolute max & minimum points.

Answer 1

where does cost=\[\frac{ \sqrt{3} }{ 2 }\] come from?

Answer 2

this is Calculus on finding absolute maximum & minimum points that involves a "TRIG function"

Answer 3

meh

Answer 4

I understand everything, but I don't know where cos t = \[\frac{ \sqrt{3} }{ 2 }\] comes from at the end.

Answer 5

plug in

Answer 6

Here's how they're getting that

\[\Large \cos^2(t)+\sin^2(t) = 1\]
\[\Large \cos^2(t)+\left(\frac{1}{2}\right)^2 = 1\]
\[\Large \cos^2(t)+\frac{1^2}{2^2} = 1\]
\[\Large \cos^2(t)+\frac{1}{4} = 1\]
\[\Large \cos^2(t)+\frac{1}{4}-\frac{1}{4} = 1-\frac{1}{4}\]
\[\Large \cos^2(t) = \frac{4}{4}-\frac{1}{4}\]
\[\Large \cos^2(t) = \frac{4-1}{4}\]
\[\Large \cos^2(t) = \frac{3}{4}\]
\[\Large \sqrt{\cos^2(t)} = \sqrt{\frac{3}{4}}\]
\[\Large |\cos(t)| = \sqrt{\frac{3}{4}}\]
\[\Large \cos(t) = \pm\sqrt{\frac{3}{4}}\]
\[\Large \cos(t) = \pm\frac{\sqrt{3}}{\sqrt{4}}\]
\[\Large \cos(t) = \pm\frac{\sqrt{3}}{2}\]

Answer 7

Oh, I see, I got it now. Thanks! :)

Answer 8

sin^2 (t) + cos^2(t) =1 is an identity. that means you can plug in any value for t and it is a true statement

Answer 9

no problem