Trigonometric Equations Using Compound Angle Formulae

**Lee66** · August 11th 2008, 09:22 AM

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.

**Simplicity** · August 11th 2008, 09:38 AM

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?

**skeeter** · August 11th 2008, 09:52 AM

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

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

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

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

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

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

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

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

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

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

**Simplicity** · August 11th 2008, 10:10 AM

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 $\theta$ .

**skeeter** · August 11th 2008, 11:03 AM

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

**Lee66** · August 12th 2008, 03:34 AM

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 $\pi$ ?

**Simplicity** · August 12th 2008, 04:21 AM

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: $\theta = 1.18, (1.18+\pi)$

An accurate answer would be:
$\mathrm{arctan} \left(\frac{21}{5\sqrt{3}}\right), \pi + \mathrm{arctan} \left(\frac{21}{5\sqrt{3}}\right)$

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