# Math Help - Difficult Trigo Problems

1. ## Difficult Trigo Problems

Hello people,

1) If $Cos(a) + Cos(b) + Cos(c) = Sin(a) + Sin(b) + Sin(c) = 0$,
then the value of $Cos(3a) + Cos(3b) + Cos(3c)$ = ?

2) If [e^cos(x)] - [e^-cos(x)] = 4. Find the value of $Cos(x)$

3) If $2Sin^2 \theta$ = $3 Cos \theta$, find the positive angle of $\theta$

4) In $\triangle ABC$, $\theta$ is an acute angle & $tan\theta$ is equal to three times of $Cot\theta$. Find the value of $(sin^2\theta) + (Cosec^2\theta) - (1/2) (Cot^2\theta)$

2. Originally Posted by saberteeth
Hello people,

1) If $Cos(a) + Cos(b) + Cos(c) = Sin(a) + Sin(b) + Sin(c) = 0$,
then the value of $Cos(3a) + Cos(3b) + Cos(3c)$ = ?

2) If [e^cos(x)] - [e^-cos(x)] = 4. Find the value of $Cos(x)$

3) If $2Sin^2 \theta$ = $3 Cos \theta$, find the positive angle of $\theta$

4) In $\triangle ABC$, $\theta$ is an acute angle & $tan\theta$ is equal to three times of $Cot\theta$. Find the value of $(sin^2\theta) + (Cosec^2\theta) - (1/2) (Cot^2\theta)$

$2) e^{\cos x} - e^{-\cos x} = 4$

$e^{2\cos x } - 1 = 4 e^{\cos x}$

$e^{2\cos x } - 4e^{\cos x} - 1 =0$

let $u=e^{\cos x}$

$u^2 - 4u - 1=0$

$u=\frac{4 \mp \sqrt{16+4}}{2}$

$u=2+\sqrt{5}$ or $u =2-\sqrt{5}$

$e^{\cos x} = 2+\sqrt{5} \Rightarrow \cos x = \ln (2+\sqrt{5} )$

$e^{\cos x } = 2 -\sqrt{5} \Rightarrow \cos x = \ln (2-\sqrt{5})$

3) $2\sin ^2 \theta = 3\cos \theta$

$2(1-\cos ^2\theta) = 3\cos \theta$

$2\cos ^2\theta +3\cos \theta -2=0$

let $u=\cos \theta$

$2u^2+3u-2 =0$ you can solve it after you find u values find x

3. Originally Posted by saberteeth
Hello people,

1) If $Cos(a) + Cos(b) + Cos(c) = Sin(a) + Sin(b) + Sin(c) = 0$,
then the value of $Cos(3a) + Cos(3b) + Cos(3c)$ = ?

2) If [e^cos(x)] - [e^-cos(x)] = 4. Find the value of $Cos(x)$

3) If $2Sin^2 \theta$ = $3 Cos \theta$, find the positive angle of $\theta$

4) In $\triangle ABC$, $\theta$ is an acute angle & $tan\theta$ is equal to three times of $Cot\theta$. Find the value of $(sin^2\theta) + (Cosec^2\theta) - (1/2) (Cot^2\theta)$

$4) \tan \theta = 3 \cos \theta$

$\tan \theta = \frac{3}{\tan \theta }$

$\tan ^2\theta = 3$

$\tan \theta = \sqrt{3}$ $\Rightarrow \theta =\frac{\pi}{3}$

the rest of question just you should know

$\sin \frac{\pi}{3} = ??$

$\sec \frac{\pi}{3}= ??$

$\cos \frac{\pi}{3} =??$
the rest for you