How do I get theta out of this equation?

[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 :(

 

[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)

 

[benji]

Thanks Swapnil.

 

[VietDao29]

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 :(

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.