Trig help

**coremaths** · November 2nd 2008, 09:34 AM

hey can someone show me how to solve these questions.....thanks

(1) Prove 1 + sin theta - cos theta = 2 sin (1/2 theta) [ cos (1/2 theta) + sin (1/2 theta) ]

(2) hence or otherwise, solve this equation

1 + sin theta - cos theta = 0.

thanks

**Soroban** · November 2nd 2008, 10:30 AM

Hello, coremaths!

(1) Prove: . $1 + \sin\theta - \cos\theta \;=\; 2\sin\tfrac{\theta}{2}\left(\cos\tfrac{\theta}{2} + \sin\tfrac{\theta}{2}\right)$

On the right: . $\underbrace{2\sin\tfrac{\theta}{2}\cos\tfrac{\thet a}{2}} + \underbrace{2\sin^2\!\tfrac{\theta}{2}}$

. . . . . . . . $= \quad\;\sin\theta \quad+\quad 1 - \cos\theta$

2) Hence or otherwise, solve this equation: . $1 + \sin\theta - \cos\theta \;=\; 0$

We have: . $2\sin\tfrac{\theta}{2}\left(\cos\tfrac{\theta}{a} - \sin\tfrac{\theta}{2}\right) \;=\;0$

Then: . $\sin\tfrac{\theta}{2} = 0 \quad\Rightarrow\quad \tfrac{\theta}{2} = \pi n \quad\Rightarrow\quad \boxed{\theta \:=\:2\pi n}$

And: . $\cos\tfrac{\theta}{2} - \sin\tfrac{\theta}{2} \:=\:0\quad\Rightarrow\quad \sin\tfrac{\theta}{2} \:=\:\cos\tfrac{\theta}{2} \quad\Rightarrow\quad \frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}} \:=\:1$

. . $\tan\tfrac{\theta}{2} \:=\:1\quad\Rightarrow\quad \tfrac{\theta}{2} \:=\:\tfrac{\pi}{4} + \pi n \quad\Rightarrow\quad\boxed{\theta \:=\:\tfrac{\pi}{2} + 2\pi n}$

**coremaths** · November 2nd 2008, 10:37 AM

hi there

thanks for the reply, would you mind just explaining question 1, in a bit more depth as i don't understand how u got the sin theta and 1 - cos theta.

thanks

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