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

[hmerr10]

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?

[stapel_eliz]

Haven't done Trig in years.

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

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 . Then divide through by 3.

For instance, given:

. . . . .

...you would note that:

. . . . . for

Then it must be that:

. . . . ..

Use the same reasoning for your exercise.

[hmerr10]

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]

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.

Thank you in advance!!

[stapel_eliz]

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!