# Help Me with trigs please

• Dec 10th 2008, 02:39 PM
I have a test tomrow and i cant do some of the problems on the topic please help. i want to know how to do these before my test tomrow

express sin 320degrees as a sin function of a positive acute angle

find the value of cos^2 x - 2cos x= 0

thx
• Dec 10th 2008, 04:14 PM
skeeter
Quote:

I have a test tomrow and i cant do some of the problems on the topic please help. i want to know how to do these before my test tomrow

express sin 320degrees as a sin function of a positive acute angle

find the value of cos^2 x - 2cos x= 0

thx

$\sin(320) = -\sin(40)$

320 degrees is 40 degrees below the (+) x-axis

$\cos^2{x} - 2\cos{x}= 0$

factor out $\cos{x}$ ...

$\cos{x}(\cos{x} - 2) = 0$

$\cos{x} = 0$ at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$

$\cos{x} = 2$ has no solution.
• Dec 10th 2008, 05:08 PM
thank you very much
can you also help me with this
sinx = - 1/2 solve for x in neareast degree of 0 equal or greater than x equal or greater 360

also for the second question you asnwered for me i made a mistake
it is deta not x
• Dec 10th 2008, 05:29 PM
skeeter
Quote:

thank you very much
can you also help me with this
sinx = - 1/2 solve for x in neareast degree of 0 equal or greater than x equal or greater 360

also for the second question you asnwered for me i made a mistake
it is deta not x

theta , not deta ... so, change all the x's to theta's

$\sin{\theta} = -\frac{1}{2}$

this value is on your unit circle in quad III and IV ...

$\theta = 210^{\circ}$ and $\theta = 330^{\circ}$

next time start a new thread for a new question.