I need help trying to determine what number is suppose to be in place foralpha? I do not understand how they came up with Cos = 1/4 in question (d).

Sin(90 deg - alpha) = Cos alpha = 1/4

http://i333.photobucket.com/albums/m...g?t=1221333118

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- Sep 13th 2008, 11:18 AMrobascTrigonometric Identities
I need help trying to determine what number is suppose to be in place for

**alpha**? I do not understand how they came up with Cos = 1/4 in question (d).

Sin(90 deg - alpha) = Cos alpha = 1/4

http://i333.photobucket.com/albums/m...g?t=1221333118 - Sep 13th 2008, 11:32 AMMoo
Hello,

$\displaystyle \sin(90°-\alpha)=\cos(\alpha)$ can be checked on a unit circle.

Otherwise, use this formula : $\displaystyle \sin(a-b)=\sin(a)\cos(b)-\sin(b)\cos(a)$

So here $\displaystyle \sin(90°-\alpha)={\color{red}\sin(90°)}\cos(\alpha)-\sin(\alpha) {\color{red}\cos(90°)}$

The red parts are supposed to be known (Wink)