00:01
Hello everybody.
00:02
In this video, i'm going to be showing you how to solve exercise 23 in chapter 9, section 1 of cohen's precalculus 7th edition.
00:11
Now, in this problem, they want us to use the addition formulas for sine and cosine to simplify the expression, cosine of pi over 3 minus theta, minus cosine of pi over 3 plus theta, where theta is some angle.
00:26
Now, to do this, what we first want to do is recall the addition formulas for cosine, which we can express in this condensed form.
00:33
For two angles, s and t, the cosine of s plus or minus t, is equal to the cosine of s times the cosine of t, minus plus the sign of s times the sign of t.
01:00
And this really represents both equations for cosine, but this plus or minus and minus plus just indicate that a different version of this equation must be used, depending on whether or not the terms and cosine's argument are subjection.
01:12
From each other were added together.
01:16
And so with this formula, all we need to do is plug both of these terms into this expression and expand, using the substitutions s equals pi over 3 as well as t equals theta.
01:32
So let's do this down here.
01:35
For the first term, cosine of pi over 3 minus theta, we want to use the minus version of this equation, which means that we use a plus over here.
01:43
We have that this is equal to the cosine of s, which is the one, is pi over 3 times the cosine of t, which is theta, plus the sign of s, which is pi over 3, times the sign of t, which is theta.
02:05
Now subtracted from this, we have the expansion for the second term...