Well, How Do I Simplify It?

Geometry Level 4

If sin ⁡ θ = 3 sin ⁡ ( θ + 2 β ) \sin\theta =3\sin(\theta +2\beta) , then what is the value of tan ⁡ ( θ + β ) + 2 tan ⁡ β \tan(\theta +\beta)+2\tan\beta ?

0 -1 None of these 1 3 4 2

Sravanth C. by Sravanth C. , 21, India

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1 solution

Sravanth C.
Aug 14, 2016

sin ⁡ θ − sin ⁡ ( θ + 2 β ) = 2 sin ⁡ ( θ + 2 β ) 2 sin ⁡ ( θ − ( θ + 2 β ) 2 ) cos ⁡ ( θ + ( θ + 2 β ) 2 ) = 2 sin ⁡ ( θ + 2 β ) Transformation Formula − sin ⁡ β cos ⁡ ( θ + β ) = sin ⁡ ( θ + β + β ) − sin ⁡ β cos ⁡ ( θ + β ) = sin ⁡ ( θ + β ) cos ⁡ β + cos ⁡ ( θ + β ) sin ⁡ β − sin ⁡ β = tan ⁡ ( θ + β ) cos ⁡ β + sin ⁡ β Dividing by cos tan ⁡ ( θ + β ) + 2 tan ⁡ β = 0 Dividing by sin \begin{aligned} \sin\theta-\sin(\theta +2\beta)&=2\sin(\theta +2\beta)&\\ 2\sin\left(\dfrac{\theta -(\theta+2\beta)}{2}\right)\cos\left(\dfrac{\theta +(\theta+2\beta)}{2}\right)&=2\sin(\theta+2\beta)&\text{Transformation Formula}\\ -\sin\beta\cos(\theta +\beta)&=\sin(\theta +\beta +\beta)&\\ -\sin\beta\cos(\theta +\beta)&=\sin(\theta +\beta)\cos\beta+\cos(\theta +\beta)\sin\beta&\\ -\sin\beta&=\tan(\theta+\beta)\cos\beta +\sin\beta&\text{Dividing by cos}\\ \tan(\theta +\beta)+2\tan\beta&=0&\text{Dividing by sin}\\ \end{aligned}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Of course this isn't a solution, but anybody could just assume that the answer is the same for all θ , β \theta, \beta which satisfy the given equation. Therefore you could just have θ , β = 0 \theta, \beta=0 and get 0 as the answer.

Wen Z - 4 years, 9 months ago

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