How do I get theta out of this equation?

  • Level: High School 
  • Thread starter Thread starter benji
  • Start date Start date
  • Tags Tags
    Theta
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
3 replies · 5K views
benji
Messages
48
Reaction score
0
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 :(
 
Mathematics news on Phys.org
You can square both sides and after simplification you should get

[tex]9/25 = 1 + 2\sin(\theta)\cos(\theta) = 1 + \sin(2\theta)[/tex]

[tex]\Rightarrow -16/25 = \sin(2\theta) \Rightarrow \theta = 1/2\cdot\arcsin(-16/25)[/tex]
 
Last edited:
benji said:
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:
[tex]\sin x + \cos x = \sqrt{2} \cos \left( x - \frac{\pi}{4} \right)[/tex]
So:
[tex]\frac{4}{5} = \sin x + \cos x = \sqrt{2} \cos \left( x - \frac{\pi}{4} \right)[/tex]
[tex]\Leftrigharrow \frac{\sqrt{2 ^ 3}}{5} = \cos \left( x - \frac{\pi}{4} \right)[/tex]
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:
[tex]\frac{-16}{25} = \sin (2 \theta)[/tex], not just one as Swapnil mentioned.