Thread: Trigonometric Equations Using Compound Angle Formulae

1. Trigonometric Equations Using Compound Angle Formulae

Hi all,

Can someone help with the following question please?

Solve 5 sin (θ – Π/6) = 8 cos θ for values 0 < θ < 2Π

To begin (using the compound angle formulae), I have:

5 sin (θ – Π/6) = 5 [sinθ cos Π/6 – cosθ sin Π/6]

The trouble is that when I expand the expression I start to get complicated decimal numbers, whereas I wish to maintain Π in the answers, which I want to give in radians not degrees.

Can anyone help please, or offer an explanatory website that will assist me.

Thanks.

2. Originally Posted by Lee66
Hi all,

Can someone help with the following question please?

Solve 5 sin (θ – Π/6) = 8 cos θ for values 0 < θ < 2Π

To begin (using the compound angle formulae), I have:

5 sin (θ – Π/6) = 5 [sinθ cos Π/6 – cosθ sin Π/6]

The trouble is that when I expand the expression I start to get complicated decimal numbers, whereas I wish to maintain Π in the answers, which I want to give in radians not degrees.

Can anyone help please, or offer an explanatory website that will assist me.

Thanks.
$\cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$

$\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$

So, you will obtain:

$5\sin \theta \cos \left(\frac{\pi}{6}\right) - 5\cos\theta\sin\left(\frac{\pi}{6}\right)=8\cos \theta$
$\frac{5\sqrt{3}}{2}\sin \theta -\frac52 \cos \theta = 8\cos \theta$
$\frac{5\sqrt{3}}{2}\sin \theta=\frac{21}{2} \cos \theta$

Will you be able to continue?

3. $5\sin \left(\theta - \frac{\pi}{6} \right) = 8\cos{\theta}$

$5 \left(\sin{\theta} \cos{\frac{\pi}{6}} - \cos{\theta}\sin{\frac{\pi}{6}}\right) = 8\cos{\theta}$

$5 \left(\sin{\theta} \cdot \frac{\sqrt{3}}{2} - \cos{\theta} \cdot \frac{1}{2}\right) = 8\cos{\theta}$

$\frac{5\sqrt{3}}{2} \sin{\theta} - \frac{5}{2} \cos{\theta} = 8\cos{\theta}$

$5\sqrt{3} \sin{\theta} = 21 \cos{\theta}$

$75 \sin^2{\theta} = 441 \cos^2{\theta}$

$75 \sin^2{\theta} = 441 (1 - \sin^2{\theta})$

$\sin^2{\theta} = \frac{441}{516}$

$\sin{\theta} = \pm \frac{21}{2\sqrt{129}}$

$\theta = \arcsin\left(\frac{21}{2\sqrt{129}}\right)$ ... solution checks

$\theta = \pi - \arcsin\left(\frac{21}{2\sqrt{129}}\right)$ ... solution is extraneous

$\theta = \arcsin\left(-\frac{21}{2\sqrt{129}}\right)$ ... solution is extraneous

$\theta = \pi + \arcsin\left(\frac{21}{2\sqrt{129}}\right)$ ... solution checks

4. Originally Posted by skeeter
...

$5\sqrt{3} \sin{\theta} = 21 \cos{\theta}$

$75 \sin^2{\theta} = 441 \cos^2{\theta}$

...
$5\sqrt{3}\sin \theta=21 \cos \theta$

You could have divided both sides by $\cos \theta$ to obtain:

$5\sqrt{3}\tan \theta=21$

$\tan \theta = \frac{21}{5\sqrt{3}}$

Then, solved for $\theta$.

5. could have ... but, it's not always wise to divide by a variable quantity when solving an equation. call it force of habit.

6. Thanks both for the replies.

I worked through the problem and also obtained:

$\tan \theta = \frac{21}{5\sqrt{3}}$

How do I solve this for $\theta$ with an answer containing $\pi$?

7. Originally Posted by Lee66
Thanks both for the replies.

I worked through the problem and also obtained:

$\tan \theta = \frac{21}{5\sqrt{3}}$

How do I solve this for $\theta$ with an answer containing $\pi$?

The range they have provided is $0 < \theta < 2\pi$

So, you can provide the answer in decimal places as $\tan ^{-1} \left(\frac{21}{5\sqrt{3}}\right)$ will not give a standard $\pi$ answers.

So, for the range provided, the $\theta$ values to $3$sf would be:

$\theta = 1.18, 4.32$
^ This is: $\theta = 1.18, (1.18+\pi)$

$\mathrm{arctan} \left(\frac{21}{5\sqrt{3}}\right), \pi + \mathrm{arctan} \left(\frac{21}{5\sqrt{3}}\right)$