suppose a right angle triangle with angle theta. Given cos(90-theta) = 5/[3(11)^0.5], find the exact value of cos(theta).

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- Feb 20th 2010, 11:28 AMmephisto50Pretty simple trigonometry but need help
suppose a right angle triangle with angle theta. Given cos(90-theta) = 5/[3(11)^0.5], find the exact value of cos(theta).

- Feb 20th 2010, 11:30 AMicemanfan
$\displaystyle \cos (a - b) = \cos a \cos b + \sin a \sin b$

$\displaystyle \cos (90 - \theta) = \cos 90 \cos \theta + \sin 90 \sin \theta =$

$\displaystyle \sin \theta$

Use this and the fact that

$\displaystyle \sin^2 \theta + \cos^2 \theta = 1$ - Feb 20th 2010, 11:45 AMmephisto50
thanks iceman i finally see the light haha

- Feb 20th 2010, 12:25 PMDiagonal
I don't think you need to get that complicated. Just use the definitions of sine and cosine within the right triangle with sides x, y, r, and the fact that if one angle is A, the other is 90 - A, and vice versa. So cos(90 - A) = x/r = sin(A) directly.