# How do I get theta out of this equation?

1. Sep 4, 2006

### benji

I've been working on a statics problem and I need to solve for theta, the equation I have right now is (4/5) = sin(theta) + cos(theta).

I can't remember how to get theta out of this, my brain isn't functioning very well tonight... Been up too long :( :yuck:

2. Sep 4, 2006

### Swapnil

You can square both sides and after simplification you should get

$$9/25 = 1 + 2\sin(\theta)\cos(\theta) = 1 + \sin(2\theta)$$

$$\Rightarrow -16/25 = \sin(2\theta) \Rightarrow \theta = 1/2\cdot\arcsin(-16/25)$$

Last edited: Sep 4, 2006
3. Sep 4, 2006

### benji

Thanks Swapnil.

4. Sep 4, 2006

### VietDao29

One more way is to use the identity:
$$\sin x + \cos x = \sqrt{2} \cos \left( x - \frac{\pi}{4} \right)$$
So:
$$\frac{4}{5} = \sin x + \cos x = \sqrt{2} \cos \left( x - \frac{\pi}{4} \right)$$
$$\Leftrigharrow \frac{\sqrt{2 ^ 3}}{5} = \cos \left( x - \frac{\pi}{4} \right)$$
Can you go from here? :)
By the way, when using the first mentioned method, you should check if the solution gives out really satisfies the problem, since if you square both sides, like you have (-2)2 = (2)2, but -2 is not equal to 2. Do you follow me? :)
And another point is that, there are infinite numbers of solutions to the equation:
$$\frac{-16}{25} = \sin (2 \theta)$$, not just one as Swapnil mentioned.