Need help solving this trig equation

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To solve the equation 3cos(θ) + 1.595*sin(θ) = 3.114, the identity (sinθ)^2 + (cosθ)^2 = 1 is applied, leading to an incorrect result of cosθ = +/- 1.0526, which is outside the valid range for cosine. A suggested alternative method involves rewriting the equation in the form a cos(θ) + b sin(θ) = R cos(θ - φ), where R = √(a² + b²) and φ = arctan(b/a). This transformation simplifies the problem and helps find the correct angle θ. The discussion emphasizes the importance of ensuring results stay within the defined ranges for trigonometric functions. Understanding these identities is crucial for solving trigonometric equations effectively.
flgdx
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Homework Statement


what's the best way to solve this equation: 3cos(θ) + 1.595*sin(θ) = 3.114

Homework Equations


(sinθ)^2 + (cosθ)^2 = 1

The Attempt at a Solution


I tried using the identity above to solve this equation and ended up with cosθ = +/- 1.0526.
 
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Could you show your steps for your solution?
 
Comeback City said:
Could you show your steps for your solution?

so since (sinθ)^2 + (cosθ)^2 = 1, I solved for sinθ and got sqrt(1 - (cosθ)^2). Then I subtituted it into the equation 3cos(θ) + 1.595*sin(θ) = 3.114 and solved for θ.
 
flgdx said:

Homework Statement


what's the best way to solve this equation: 3cos(θ) + 1.595*sin(θ) = 3.114

Homework Equations


(sinθ)^2 + (cosθ)^2 = 1

The Attempt at a Solution


I tried using the identity above to solve this equation and ended up with cosθ = +/- 1.0526.
Well, that's obviously wrong because -1 ≤ cos θ ≤ 1, and 1.0526 is outside that range. Your approach should work though.

Another approach is to use
$$a \cos\theta + b \sin\theta = \sqrt{a^2+b^2} \cos(\theta-\phi)$$ where ##\tan\phi = b/a##.
 
could you pls show me how you got that equation?
 
Consider a right triangle below and use the identity ##\cos(\theta-\phi) = \cos\theta\,\cos\phi + \sin\theta\,\sin\phi##.

triangle.png
 

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