Author Topic: Trig question (Read 1402 times) Tweet Share

fredrick · « **on:** April 28, 2008, 04:23:07 pm »

Solve $\frac{1}{5}(5cos\theta-3sin\theta)=sin\theta$ for $\theta$ , $0\le \theta\le 2\pi$ , giving answer in exact form.

Heres my attempt:
$(cos\theat-\frac{3}{5}sin\theta)^2=(sin\theta)^2$
$cos^2\theta-\frac{6}{5}cos\theta\sin\theta+\frac{9}{25}sin^2\theta=sin^2\theta$
$cos^2\theta-\frac{6}{5}cos\theta\sin\theta=\frac{16}{25}sin^2\theta$
$(1-sin^2\theta)-\frac{6}{5}cos\theta\sin\theta=\frac{16}{25}sin^2\theta$ ------>Replaced $cos^2\theta\ with\ (1-sin^2\theta)$
$1-\frac{6}{5}cos\theta\sin\theta=\frac{41}{25}sin^2\theta$
Now im stuck

dcc · « **Reply #1 on:** April 28, 2008, 05:40:30 pm »

Given:

$\cos \theta - \dfrac{3}{5}\sin \theta = \sin \theta$

$\cos \theta - \dfrac{8}{5}\sin \theta = 0$

$5\cos \theta = 8\sin \theta$ (dividing both sides by $\cos \theta$ )

$\tan \theta = \dfrac{5}{8}$

I dont think this has any exact values, so you'll have to use your knowledge of inverse trig to get exact answers:

First Angle:
$\theta_{1} = \mbox{ArcTan } \dfrac{5}{8}$

Second Angle (because there are two solutions in this interval, so finding the same value is just remembering the symmetry of the unit circle)

$\theta_{2} = \pi + \mbox{ArcTan } \dfrac{5}{8}$

which should give you your correct answers

As a general rule, dont square a trig equation unless you absolutely have to, because it can lead to solutions which dont exist, etc etc

fredrick · « **Reply #2 on:** April 28, 2008, 07:18:46 pm »

The answer in the back off the book is $\frac{\pi}{4} and \frac{5\pi}{4}$

Mao · « **Reply #3 on:** April 28, 2008, 07:23:08 pm »

what =\

$\frac{1}{5}\left(5\cos\frac{\pi}{4}-3\sin\frac{\pi}{4}\right)$

$=\frac{1}{5}\cdot \sqrt{2}$

and $\frac{\sqrt{2}}{5}\neq \frac{\sqrt{2}}{2}$ !!!

fredrick · « **Reply #4 on:** April 28, 2008, 07:27:17 pm »

funny ey? but thats what the back says, although their probably wrong anyway

Mao · « **Reply #5 on:** April 28, 2008, 07:31:23 pm »

or the question is wrong....?

maybe it meant $\frac{1}{2}$ instead of $\frac{1}{5}$ ?

fredrick · « **Reply #6 on:** April 28, 2008, 09:29:45 pm »

nop the question is right

dcc · « **Reply #7 on:** April 28, 2008, 09:37:44 pm »

Then your book is wrong

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