# taking trig class and forgot how to do this problem.

• Jun 23rd 2010, 07:33 PM
coryjannis
taking trig class and forgot how to do this problem.
haven't taken math in a while and I'm taking a trigonometry class. I believe i learned this in algebra but i can't remember how its done.

cos = +/-√1-225/289

(basically i don't understand how to do the square root part of this. its the square root of 1- 225 over 289.

The answer for this is = +/- 8/17

**no clue how they got 8/17.
thx and godbless if anybody can help me

http://i48.photobucket.com/albums/f2...799/sqroot.jpg
• Jun 23rd 2010, 08:04 PM
Think about sin²(θ) + cos²(θ) = 1 and tell us what you get from that.
• Jun 23rd 2010, 08:05 PM
bigwave
Quote:

Originally Posted by coryjannis
haven't taken math in a while and I'm taking a trigonometry class. I believe i learned this in algebra but i can't remember how its done.

cos = +/-√1-225/289

(basically i don't understand how to do the square root part of this. its the square root of 1- 225 over 289.

The answer for this is = +/- 8/17

**no clue how they got 8/17.
thx and godbless if anybody can help me

http://i48.photobucket.com/albums/f2...799/sqroot.jpg

$
\sqrt{1 - \frac{225}{289}}
\Rightarrow \sqrt{\frac{289}{289} - \frac{225}{289}}
\Rightarrow \sqrt{\frac{64}{289}}
\Rightarrow \pm\frac{8}{17}
$

$\cos{\theta} = \pm\sqrt{1-\frac{225}{289}}$
square both sides
$\cos^2{\theta} = 1 - \frac{225}{289}
\Rightarrow
\cos^2{\theta}-1 = \frac{225}{289}
\Rightarrow
\sin^2{\theta} = \sqrt{\frac{225}{298}}
\Rightarrow
\sin{\theta = \pm\frac{15}{17}$

do you need help with more of this