How do I get theta out of this equation?

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Discussion Overview

The discussion centers around solving for theta in the equation (4/5) = sin(theta) + cos(theta). Participants explore various methods to isolate theta, including algebraic manipulation and trigonometric identities. The context is primarily mathematical reasoning related to a statics problem.

Discussion Character

  • Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant suggests squaring both sides of the equation to derive a new equation involving sin(2theta) and proposes that theta can be expressed as theta = 1/2·arcsin(-16/25).
  • Another participant introduces a trigonometric identity, stating that sin(x) + cos(x) can be rewritten as √2 cos(x - π/4), leading to a different approach to isolate theta.
  • A participant cautions that squaring both sides may introduce extraneous solutions and emphasizes the importance of verifying any solutions against the original equation.
  • It is noted that the equation sin(2theta) = -16/25 has multiple solutions, contradicting the earlier claim of a single solution.

Areas of Agreement / Disagreement

Participants express differing methods for solving the equation, and there is no consensus on a single approach or solution. The discussion remains unresolved regarding the best method to isolate theta.

Contextual Notes

Participants highlight the potential for extraneous solutions when squaring both sides of the equation and the existence of multiple solutions for the derived equation involving sin(2theta).

benji
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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 :(
 
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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:
Thanks Swapnil.
 
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.
 

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