# trig identity

• Feb 26th 2008, 04:58 PM
boonies
trig identity
http://img.photobucket.com/albums/v6...igidentity.jpg

i need help with this identity, could someone tell me where i went wrong & help me?
• Feb 26th 2008, 06:44 PM
Soroban
Hello, boonies!

Your simplifying was off . . .

Quote:

$({\color{blue}\tan\theta\cos\theta} + 2\cos\theta)^2 + \left(2\sin\theta - {\color{red}\frac{\sin\theta}{\tan\theta}}\right)^ 2 \;=\;5$
. . ${\color{blue}\tan\theta\cos\theta \:=\:\frac{\sin\theta}{\cos\theta}\!\cdot\!\cos\th eta \:=\:\sin\theta}$

. . ${\color{red}\frac{\sin\theta}{\tan\theta} \:=\:\frac{\sin\theta}{\frac{\sin\theta}{\cos\thet a}} \:=\:\cos\theta}$

The problem becomes: . $({\color{blue}\sin\theta} + 2\cos\theta)^2 + (2\sin\theta - {\color{red}\cos\theta})^2$

. . $= \;\sin^2\!\theta + 4\sin\theta\cos\theta + 4\cos^2\!\theta + 4\sin^2\!\theta - 4\sin\theta\cos\theta + \cos^2\!\theta$

. . $= \;5\sin^2\!\theta + 5\cos^2\!\theta \;=\;5\underbrace{(\sin^2\!\theta + \cos^2\!\theta)}_{\text{This is 1}} \;=\;5(1) \;=\;5$