2
$\begingroup$

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.

$\endgroup$
5
  • 1
    $\begingroup$ 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? $\endgroup$
    – Arthur
    Jan 23, 2018 at 8:48
  • 1
    $\begingroup$ 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. $\endgroup$
    – egreg
    Jan 23, 2018 at 9:00
  • $\begingroup$ Write RHS as $\cos (\beta + \theta - \beta)$ $\endgroup$ Jan 23, 2018 at 10:00
  • $\begingroup$ Where did you get that expansion formula for cosine I never see that in m life $\endgroup$
    – Guy Fsone
    Jan 23, 2018 at 13:25
  • $\begingroup$ Please, if you are ok, you can accept the answer and set it as solved. Thanks! $\endgroup$
    – user
    Jan 28, 2018 at 7:48

3 Answers 3

1
$\begingroup$

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

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

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

$\endgroup$
0
$\begingroup$

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 $\cos^2\beta = \frac{1}{1+\tan^2\beta}$ $$ \tan\beta - \tan^2\beta\tan\theta =-\tan\theta(1+\tan^2\beta) \implies \tan\beta =-\tan\theta $$

$\endgroup$
0
$\begingroup$

Note that

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

thus step by step

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

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

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

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

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

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

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .