# proving that $\sin\beta \cos(\beta+\theta)=-\sin\theta$ implies $\tan\theta=-\tan\beta$

Show that $\sin\beta \cos(\beta+\theta)=-\sin\theta$ implies $\tan\theta=-\tan\beta$

I expand the cosinus: $$\cos(\beta+\theta)=\left(1-\frac{\theta^2}{2}\right)\left(1-\frac{\beta^2}{2}\right)-\beta\theta$$ but I can't get any tan.

• I get $\cos(\beta + \theta) = \cos\beta\cos\theta - \sin\beta\sin\theta$. Your expansion is new to me, and frankly doesn't look right. Is it just the first terms of the Taylor expansion or something? Commented Jan 23, 2018 at 8:48
• It's false that $\cos(\beta+\theta)$ equals the expression on the right hand side. You can use Taylor expansion up to some degree for computing limits and other tasks, but generally not for showing equalities. Commented Jan 23, 2018 at 9:00
• Write RHS as $\cos (\beta + \theta - \beta)$ Commented Jan 23, 2018 at 10:00
• Where did you get that expansion formula for cosine I never see that in m life Commented Jan 23, 2018 at 13:25
• Please, if you are ok, you can accept the answer and set it as solved. Thanks!
– user
Commented Jan 28, 2018 at 7:48

Separating out $\sin\theta,\cos\theta$ after using $$\cos(A+B)$$ formula

$$\sin\theta(\sin^2\beta-1)=\sin\beta\cos\beta\cos\theta$$

$$\iff\dfrac{\sin\theta}{\cos\theta}=\dfrac{\sin\beta\cos\beta}{\sin^2\beta-1}=?$$

Indeed, since $\cos(\beta + \theta) = \cos\beta\cos\theta - \sin\beta\sin\theta$ your equation becomes $$\sin\beta( \cos\beta\cos\theta - \sin\beta\sin\theta) =\sin\theta$$ Dividing by $\cos\theta$ $$\sin\beta( \cos\beta - \sin\beta\tan\theta) =-\tan\theta \implies \sin\beta \cos\beta - \sin^2\beta\tan\theta =-\tan\theta$$ Dividing by $\cos^2\beta = \frac{1}{1+\tan^2\beta}$ $$\tan\beta - \tan^2\beta\tan\theta =-\tan\theta(1+\tan^2\beta) \implies \tan\beta =-\tan\theta$$

Note that

$$\cos(\beta+\theta)=\cos\beta \cos\theta-\sin\beta \sin\theta$$

thus step by step

$$\sin\beta \cos(\beta+\theta)=\sin\beta \cos\beta \cos\theta-\sin^2\beta \sin\theta=-\sin \theta$$

$$\sin\beta \cos\beta \cos\theta-\sin^2\beta \sin\theta=-\sin \theta$$

$$\sin\beta \cos\beta \cos\theta=-\sin \theta+\sin^2\beta \sin\theta$$

$$\sin\beta \cos\beta \cos\theta=-\sin \theta(1-\sin^2\beta)$$

$$\sin\beta \cos\beta \cos\theta=-\sin \theta \cos^2\beta$$

$$\sin\beta \cos\theta=-\sin \theta \cos\beta$$

$$\frac{\sin\beta} {\cos\beta}=-\frac{\sin \theta} {\cos\theta}\iff\tan\theta=-\tan \beta$$