Complement, Supplement, Coterminal and principle angles?

• July 21st 2009, 09:30 PM
crosser43
Complement, Supplement, Coterminal and principle angles?
Complement, Supplement, Coterminal and principle angles?
I don't really know what these terms mean so i need help on their meanings and how to find them.

Example
find the Complement, Supplement, Coterminal and principle angles for each $\theta$

a) $\theta=92^o, for -720^o\leq\theta\leq720^o$

and

b) $\theta=(11\pi)/6, for -4\pi\leq\theta\leq4\pi$

Thanks.
• July 21st 2009, 10:12 PM
yeongil
Quote:

Originally Posted by crosser43
Complement, Supplement, Coterminal and principle angles?
I don't really know what these terms mean so i need help on their meanings and how to find them.

You should know what complementary and supplementary angles are, at least! (Worried) I'll forgive you for not knowing coterminal, if you're learning trig for the first time. I actually don't know what you mean by "principle angle" -- probably we call it differently in the US. I'm going to guess that "principle angle" means reference angle.

Quote:

a) $\theta=92^o, for -720^o\leq\theta\leq720^o$
Two angles are complementary if they add up to 90°. Two angles are supplementary if they add up to 180°. I'll let you do those. (Although, I've never had to find the complement of an obtuse angle before. I have seen in one book that it's possible if the other angle is negative.)

Two angles are coterminal if their difference is a multiple of 360°. So take 92° and add and subtract 360° repeatedly until you find all angles in the range you want.

For any given angle in standard position, a reference angle is the smallest angle between the terminal side and the x-axis. Knowing these is helpful because the values of trig functions of angle $\theta$ is the same as the trig values of the reference angle of $\theta$, though many will have a minus sign.

To find the reference angle for $\theta$ depends on the quadrant. Obviously, there is no need to find the reference angle of an angle in Quadrant I -- it's just $\theta$.

For an angle in Quadrant II, the reference angle is $180^{\circ} - \theta$.
For an angle in Quadrant III, the reference angle is $\theta - 180^{\circ}$.
For an angle in Quadrant IV, the reference angle is $360^{\circ} - \theta$.

Now find the reference angle for $92^{\circ}$.

Quote:

b) $\theta=(11\pi)/6, for -4\pi\leq\theta\leq4\pi$
I'll let you try this yourself. Just remember that the numbers I used above need to be converted to radians.

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