+505 Here's the question:
There is an angle \(\theta \) such that \(\sin\left(\frac\theta2\right) + \cos\left(\frac\theta2\right) = -\frac65.\) what is the value of sin(\(\theta\)).
Here's what I tried:
So first I tried replacing the theta/2 with their corresponding trig identities. So for sin(theta/2) = \(\sqrt{\frac{1-cos(x)}{2}}\) and for cos(theta/2) = \(\sqrt{\frac{cos(x) + 1}{2}}\). Unfortunately for me, that didn't work. Then I tried the sin(x) + cos(x) = sin x + sin(pi/2 - x) thing. However, that didn't make any sense to me. I may be doing something wrong, so some hints to push me to the right answer would be appreciated :)
+506 Why not just solve for \(\sin(\frac{\theta}{2})\) and then use the double angle identity?
Anyway, let \(x=\frac{\theta}{2}\) for simplicity.
\(\sin(x)+\cos(x)=-\frac{6}{5}\\ \cos(x)=-\sin(x)-\frac{6}{5}\)
(by the way, it might be tempting to square both sides of \(\sin(x)+\cos(x)=-\frac{6}{5}\), and then use trig identities. Don't do that; I have made this mistake before. Unless the problem specifies that the whole thing is always nonnegative, it's better to do the way that I'm about to demonstrate.)
Substitute it into the Pythagorean identity \(\sin^2(x)+\cos^2(x)=1\):
\(\sin^2(x)+(-\sin(x)-\frac{6}{5})^2=1\\ \sin^2(x)+\sin^2(x) + \frac{12}{5}\sin(x)+\frac{36}{25}=1\\ 2\sin^2(x)+ \frac{12}{5}\sin(x)+\frac{36}{25}=1\\ 50\sin^2(x)+60\sin(x)+11=0\\\)
Quadratic formula gives:
\(\sin(x)=\frac{-6\pm\sqrt{14}}{10}\)
Now we can solve the final question using the double angle identity:
\(\sin(\theta)\\ =\sin(2x)\\ =2\cos(x)\sin(x)\\ =2(-\sin(x)-\frac{6}{5})(\sin(x))\\ =\)
Now just substitute the 2 solutions of the quadratic equation for sin(x), and you're done!
