How to Simplify Trigonometric Expressions?

[bomba923]

\tan 2\theta = \frac{{2r + k\cos \theta }}{{2h - k\sin \theta }}

How to (isolate) find \theta \; {?}
(h,k,r >0)

 

[bomba923]

!Anyone??
(It's just trigonometry!)

 

[HallsofIvy]

That's right, it's just trigonometry so how about you showing some idea of how you would at least try to solve the proglem!

 

[dimensionless]

There is a function that is the reverse of a tangent.

 

[malaygoel]

\tan 2\theta = \frac{{2r + k\cos \theta }}{{2h - k\sin \theta }}

How to (isolate) find \theta \; {?}
(h,k,r >0)

are h,k ,r independent of each other?

 

[HallsofIvy]

tan(2\theta)= \frac{2tan(\theta)}{1+ tan^2(\theta)}
= \frac{\frac{2sin(\theta)}{cos(\theta)}}{1+ \frac{sin^2(\theta)}{cos^2(\theta)}}
= \frac{2sin(\theta)cos(\theta)}{sin^2(\theta)+ cos^2(\theta)}= 2sin(\theta)cos(\theta)
So your equation is really
2 sin(\theta)cos(\theta)= \frac{2r+ kcos(\theta)}{2h-ksin(\theta)}

Multiply both sides by the denominator on the right and you will have a quadratic equation for sin(\theta).

 

[VietDao29]

tan(2\theta)= \frac{2tan(\theta)}{1+ tan^2(\theta)}
= \frac{\frac{2sin(\theta)}{cos(\theta)}}{1+ \frac{sin^2(\theta)}{cos^2(\theta)}}
= \frac{2sin(\theta)cos(\theta)}{sin^2(\theta)+ cos^2(\theta)}= 2sin(\theta)cos(\theta)

Err, so are you saying that: tan(2x) = 2sin(x) cos(x) = sin(2x)?
There's a slight error in the first step. It should be:
\tan (2 \theta) = \frac{2 \tan \theta}{1 - \tan ^ 2 \theta}. :)
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@ bomba923, have you done anything? I just wonder whether you are asking others to help you or you are just challenging people...