Trigonometry elementary problem

[sambarbarian]

Homework Statement

If 3 sin theta + 5cos theta = 5 ... prove that 5sin theta - 3 cos theta = +- 3

Homework Equations

The Attempt at a Solution

i tried many things here , mostly those including squares , because i need +-3 , but this one has me stumped :/

 

[Curious3141]

Homework Statement

If 3 sin theta + 5cos theta = 5 ... prove that 5sin theta - 3 cos theta = +- 3

Homework Equations

The Attempt at a Solution

i tried many things here , mostly those including squares , because i need +-3 , but this one has me stumped :/

Are you familiar with this "trick": a\sin \theta + b\cos \theta = \sqrt{a^2 + b^2}\sin(\theta + \arctan{\frac{b}{a}})?

 

[sambarbarian]

never heard of it

 

[Curious3141]

never heard of it

It's basically the angle sum formula for sine. Try expanding R\sin(\theta + \alpha) and compare coefficients to a \sin \theta + b \cos \theta.

 

[sambarbarian]

can this question be solved without it ?>

 

[Curious3141]

can this question be solved without it ?>

Not easily, I think. But why don't you want to try this? Haven't you covered the angle sum formula at all?

 

[Saitama]

You can try writing 5cosθ=5\sqrt{1-sin^2θ}. Taking this term to RHS,
you will get an equation
3sinθ-5=-5\sqrt{1-sin^2θ}
Square both the sides, the equation will be easy to solve and you will get two values for θ.

 

[SammyS]

You can try writing 5cosθ=5\sqrt{1-sin^2θ}. Taking this term to RHS,
you will get an equation
3sinθ-5=-5\sqrt{1-sin^2θ}
Square both the sides, the equation will be easy to solve and you will get two values for θ.

Very good !

 

[sambarbarian]

You can try writing 5cosθ=5\sqrt{1-sin^2θ}. Taking this term to RHS,
you will get an equation
3sinθ-5=-5\sqrt{1-sin^2θ}
Square both the sides, the equation will be easy to solve and you will get two values for θ.

thank you! that did the trick

 

[Saitama]

Very good !

Thank you SammyS!