## LOST: cos(3theta)= -1/2 Find all values of theta

Trigonometric ratios and functions, the unit circle, inverse trig functions, identities, trig graphs, etc.

### LOST: cos(3theta)= -1/2 Find all values of theta

LOST: cos(3theta)= -1/2 Find all values of theta

Haven't done Trig in years. Can anybody walk me through this? Is it done using identies?
hmerr10

Posts: 3
Joined: Wed Mar 18, 2009 12:23 am

hmerr10 wrote:Haven't done Trig in years.

Aren't they covering the material again in the course you're currently taking...?

hmerr10 wrote:LOST: cos(3theta)= -1/2 Find all values of theta

Use the basic reference angles (from the table of values they gave you to memorize), along with your knowledge of the behavior of the cosine curve (for instance, being negative in the second and third quadrants), to solve for the values of $3\theta$. Then divide through by 3.

For instance, given:

. . . . .$\sin{(2\theta)}\, =\, \frac{\sqrt{3}}{2}$

...you would note that:

. . . . .$\sin{(\alpha)}\, =\, \frac{\sqrt{3}}{2}$ for $\alpha\, =\, \frac{\pi}{3},\, \frac{2\pi}{3},\, \frac{7\pi}{3},\, \frac{8\pi}{3},...$

Then it must be that:

. . . . .$\theta\, =\, \frac{\pi}{6},\, \frac{\pi}{3},\, \frac{7\pi}{6},\, \frac{4\pi}{3},...$.

Use the same reasoning for your exercise.

stapel_eliz

Posts: 1803
Joined: Mon Dec 08, 2008 4:22 pm

### Re: LOST: cos(3theta)= -1/2 Find all values of theta

Thank you so much. This makes perfect sense; a great explanation. It's remembering the relationships between sin, cos and tan that I'm having difficulty with........... need to re-orient my thinking.
hmerr10

Posts: 3
Joined: Wed Mar 18, 2009 12:23 am

### Re: LOST: cos(3theta)= -1/2 Find all values of theta

stapel_eliz:

for:

sin(alpha) = sqrt3/2

alpha = pi/3; 2pi/3; 7pi/3; 8pi/3

where did you get these values using the sin curve? Just to be super clear here......... I jumped the gun on seeing the full picture; didn't see how you got these values using sine.

hmerr10

Posts: 3
Joined: Wed Mar 18, 2009 12:23 am

Draw the sine wave, at least two cycles-worth, so you can see what sin(t) looks like.

Mark where t = pi/3, and draw the vertical line up to the sine wave.

Now move along to the right, marking every spot where the sine wave is at the same height. This will occur, on the way back down, in the first half of the first cycle. By symmetry, this must be at pi - pi/3 = (3/3)pi - (1/3)pi = (2/3)pi.

The sine wave will be negative in the second half of the first cycle, so...

Oh.

Okay, so the third and fourth values I listed previously were wrong. Argh!

The sine wave will be positive again in the first half of the next cycle, so add pi/3 and 2pi/3 to 2pi to get their values, etc, etc. Then do the division to solve the original equation.

Sorry!

stapel_eliz

Posts: 1803
Joined: Mon Dec 08, 2008 4:22 pm