00:01
We want to express g of x in terms of sign only and graph the result.
00:06
So to start, let's rewrite g of x as root 3, sine 2x plus cosine 2x by the commutative property of addition.
00:22
And this will just make it easier for us to relate the expression to the sum of signs and cosine theorem, which says that a sine 2x plus b cosine 2x is equal to k sine 2x plus fie, where k is equal to the square root of a squared plus b squared, and fee satisfies both cosine fee equals a over k, as well as sine fee equals a over k, as well as sine fee equals b over k so we'll start by recognizing that we have a equals root 3 and b equals 1 and with that we can solve for k by the square root of root 3 squared plus 1 squared and that gives us k equals the square root of 3 plus 1 3 plus 1 is 4 and the square root of 4 is 2 so we get k equals 2 and then we'll use k to help us solve for fee in the cosine fee and sine fee equations so we get cosine fee is equal to a over k or root 3 over 2 and we get sine fee is equal to b over k or 1 over 2 and since cosine fee and sine fee are both positive we know that fee is in quadrant 1 and we will take the inverse sign of one -half to get our value of fee.
02:38
And the inverse sign of one -half is pi over six.
02:45
So now we have values for k and fee, and that means that k -sign 2x plus fee is going to become 2 -sign 2x plus pi over 6.
03:10
And this is equivalent to g of x.
03:18
Now we want to graph g of x.
03:22
So the first thing we're going to want to do is take a look at this inside of the parentheses and set it equal to zero.
03:33
And this will tell us where g of x equals sine of zero.
03:40
So we get 2x plus pi over 6 equals 0, or 2x equals negative pi over 6, and that will give us x equals negative pi over 12.
03:54
So when x is equal to negative pi over 12, we end up with g of x equals sign of zero, which gives us g of x equals zero.
04:13
So when we go up to our graph, at x equals negative pi over 12, we'll put a dot at y equals zero...