Answer to Question #249964 in Trigonometry for aisme

Question

If t = tan($\theta$ /2), show that sin($\theta$ )=2t/(1-t²) and cos ($\theta$ )=(1-t²)/(1+t²). hence solve the equation cos($\theta$ )-2sin($\theta$ )=2

Answer 1

Let $t = \tan(\frac{\theta}2).$ Then

$\sin(\theta )=2\sin\frac{\theta}2\cos\frac{\theta}2=2\tan\frac{\theta}2\cos^2\frac{\theta}2 =\frac{2\tan\frac{\theta}2}{\frac{1}{\cos^2\frac{\theta}2}} =\frac{2\tan\frac{\theta}2}{1+\tan^2\frac{\theta}2}=\frac{2t}{1+t^2},$ and

$\cos (\theta )=\cos^2\frac{\theta}2-\sin^2\frac{\theta}2 =\frac{\cos^2\frac{\theta}2-\sin^2\frac{\theta}2}{\cos^2\frac{\theta}2\frac{1}{\cos^2\frac{\theta}2}} =\frac{1-\tan^2\frac{\theta}2}{1+\tan^2\frac{\theta}2} =\frac{1-t^2}{1+t^2}.$

Let us solve the equation $\cos(\theta )-2\sin(\theta )=2.$ Let $t = \tan(\frac{\theta}2).$ Then we get the equation

$\frac{1-t^2}{1+t^2}-2\frac{2t}{t^2+1}=2,$ and hence $\frac{1-t^2-4t}{1+t^2}=2.$ It follows that $1-t^2-4t=2+2t^2,$ and thus $3t^2+4t+1=0.$ The last equation is equivalent to$(3t+1)(t+1)=0,$ and hence has the roots $t_1=-1,\ t_2=-\frac{1}3.$ Then $\tan(\frac{\theta}2)=-1$ or $\tan(\frac{\theta}2)=-\frac{1}3.$ It follows that $\frac{\theta}2=-\frac{\pi}4+\pi n$ or $\frac{\theta}2=-\arctan\frac{1}3+\pi n,\ n\in\mathbb N.$

We conclude that the solutions of the equation $\cos(\theta )-2\sin(\theta )=2$ are of the form:

$\theta=-\frac{\pi}2+2\pi n$ or $\theta=-2\arctan\frac{1}3+2\pi n,$ where $n\in\mathbb N.$