# Thread: 1 Last question Guys

1. Yes. That method looks good.

You could just draw out the triangle I believe and not have to mess with a calculator, which may be what your teacher is getting at.

2. Originally Posted by topsquark
The problem with the double angle formulas is that the triangle where you have an angle of 2x doesn't have a simple relation to the triangle where you have an angle x. There is possibly a way to do it this way, but it is beyond my ken. Sometimes you just have to go algebraic instead of visual, I'm afraid.

-Dan
$sinx=\sqrt{\frac{1-cos2x}{2}}$
$cosx=\sqrt{\frac{1+cos2x}{2}}$
$tanx=\sqrt{\frac{1-cos2x}{1+cos2x}}$

Keep Smiling
Malay

3. Originally Posted by malaygoel
$sinx=\sqrt{\frac{1-cos2x}{2}}$
$cosx=\sqrt{\frac{1+cos2x}{2}}$
$tanx=\sqrt{\frac{1-cos2x}{1+cos2x}}$

Keep Smiling
Malay
In reality it should be (I mentioned this before a lot).
For example for sine function,
$\sin x=\mbox{sgn } x \sqrt{\frac{1-\cos 2x}{2}}$
Where,
$\mbox{sgn } x=\left\{ \begin{array}{c} 1\mbox{ if }0+2\pi k\leq x\leq \pi+2\pi k\\ -1\mbox{ otherwise } \end{array} \right\}$

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