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If $A$ and $B$ are the solutions to ${\displaystyle 7\cos\theta+4\sin\theta+5=0\mbox{ where }A>0,0<B<2\pi;}$

Without finding the solutions to the trig equation, show that;

$$\cot\left(\frac{A}{2}\right)+\cot\left(\frac{B}{2}\right)=-\frac{2}{3}$$

This is my effort so far:

Since A and B are solutions then:

\begin{align*} 7\cos A+4\sin A+5 & =0\\ 7\left( 2\cos^{2}\frac{A}{2}-1\right) +4\times2\sin\frac{A}{2}\cos\frac{A}{2}+5 & =0\\ 14\cos^{2}\frac{A}{2}+8\sin\frac{A}{2}\cos\frac{A}{2}-2 & =0\\ \mbox{Now divide by }\sin\frac{A}{2}\cos\frac{A}{2} & \mbox{ gives:}\\ 14\cot\frac{A}{2}+8-\frac{2}{\sin\frac{A}{2}\cos\frac{A}{2}} & =0\\ \mbox{similarly for B;}\\ 14\cot\frac{B}{2}+8-\frac{2}{\sin\frac{B}{2}\cos\frac{B}{2}} & =0 \end{align*}

When I add these two equations I have $\cot\left(\frac{A}{2}\right)+\cot\left(\frac{B}{2}\right)$, but am unsure how to progress.

miracle173
  • 11,049
Anthony Macks
  • 229

2 Answers2

8

$\displaystyle 7\cos\theta+4\sin\theta+5=0$

$14\cos^{2}\frac{\theta}{2}+8\sin\frac{\theta}{2}\cos\frac{\theta}{2}-2 =0$

Dividing by $\sin ^2 \frac{\theta}{2} \ $,

$14 \cot^2\frac{\theta}{2} + 8 \cot \frac{\theta}{2} - 2 \csc^2\frac{\theta}{2} = 0$

Using $\csc^2\frac{\theta}{2} = 1 + \cot^2\frac{\theta}{2}$,

$6 \cot^2\frac{\theta}{2} + 4 \cot \frac{\theta}{2} - 1 = 0$

As we know, sum of roots of quadratic $ax^2 + bx + c = 0$ is $ - \frac{b}{a}$.

So, $\cot \frac{A}{2} + \cot \frac{B}{2} = - \frac{2}{3}$

Math Lover
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0

We use following relations:

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

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

plugging these in equation we get:

$6cot^2(\frac{(\theta)}2)+4cot(\frac{(\theta)}2)-1=0$

Comparing with $ax^2+bx+c=0$ we have:

$x_1+x_2=-\frac ba=-\frac 23$

which gives what is required.

sirous
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