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Trig ratio solve for theta

  • Thread starter thejosh
  • Start date

thejosh

1. Homework Statement
Sino - ucos o =x

2. Homework Equations

N.A.
3. The Attempt at a Solution
I am asking if it's possible to make theta(o) the subject of formula in relation to some physics homework which I narrowed down to this but I don't know how to solve for theta, x is a known value but it's long so to make it easier I used a symbol and u is also a known value , in my attempt to make o the subject of formula I factorised out o only to find out cos and sin don"t work without a definite value, any suggestions.If it's not possible I may then post the full problem for guidance.Your help is much appreciated.
 

cnh1995

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Sino - ucos o =x
sinθ-ucosθ=x.
You can add greek letters in the post. Click on the '∑' symbol in the options above the typing space.

What are the values of u and x? It is possible to solve for θ with some trig and algebraic manipulations. Please post the full problem.
 

MarkFL

If I understand correctly, you have:

##\sin(\theta)-u\cos(\theta)=x##

Using a linear combination identity, you may write:

##\sqrt{u^2+1}\sin\left(\theta-\arctan(u)\right)=x##

Can you proceed?
 

thejosh

That will take some lessons, it's way over my head to be honest but if you could direct me to a plausible learning site I would be happy to learn that , is it calculus?
 

thejosh

@cnh1995 I think I was ok until I got to the final equation which is what I posted, the whole question requires pictures which may not be very clear because of their size but besides that I just wanted to knowhow to do the last equation and solve for Θ.Nevertheless I will try post the whole question hopefully within 2 days (school is a savage time consumer) oh and u is 0.4 whilst x is 0.127420999
 

cnh1995

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oh and u is 0.4 whilst x is
Ok. If you only want to know how to find θ, you don't need to post the whole question.

So,
sinθ-0.4cosθ=0.1274.
∴sinθ+0.1274=0.4cosθ.

Use cosθ=√(1-sin2θ) and square both the sides. You'll get a quadratic equation in sinθ. Solve for sinθ and get θ.
 

haruspex

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Another method.
Given Acos(θ)+Bsin(θ)=C, divide through by D=√(A2+B2). The coefficients of the trig functions, A/D and B/D, are then the sine and cosine of some angle φ. Can you solve sin(φ)cos(θ)+cos(φ)sin(θ)=C/D?
 

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