Hello, and welcome to this video over some of the common trigonometric identities.
In this video, we will go over the half-angle identities, double angle identities, and the product properties.
Let’s jump right in by finding the sine of 15 degrees using the half-angle identities.
Here’s the half-angle identity for sine:
This is used to find the sine of half a known angle.
For our problem, we can use it to find half of 30 degrees, which is 15 degrees.
We could also use it to find half of 45 degrees, which is 22.5 degrees.
To find our sine of 15 degrees, we’ll use 30 degrees as theta, since half of that will give us 15.
As always, we start by gathering the information that we need and defining theta.
So theta is equal to 30 degrees.
And then we'll calculate the sine and the cosine of 30 degrees.
So the sine of 30 degrees is one-half and the cosine of 30 degrees is the square root of 3 over 2.
Then, we simply substitute all of the appropriate values into the formula.
So first just so that we remember it a little bit better we're going to write out our formula.
So we have the sine of theta over 2, remember this is our half angle identity which is why we're diving by 2, is equal to the square root of 1 minus cosine of theta over 2.
Now we plug in all the values that we know.
So we know that theta is equal to 30 degrees, so we have sine of 30 degrees over 2 is equal to the square root of 1 minus cosine of 30 degrees over 2.
So 30 degrees divided by 2 is 15 degrees so we'll get the sine of 15 degrees which is what we're looking for,
is equal to the square root of 1 minus the cosine of 30 degrees is the square root of 3 over 2 all over 2.
Finally, we simplify.
So because we have the square root of a fraction, we're going to take the numerator of our fraction and take the square root of that,
so the square root of 1 minus the square root of 3 over 2 over the square root of our denominator, so over the square root of 2.
This time we have a radical in our denominator so we need to rationalize it.
Since the radical is a monomial, we multiply the top and bottom of our answer by the denominator in order to make our new denominator an integer.
So we're going to multiply this by the square root of 2 over the square root of 2.
So to do that our numerator will be the square root of 2 times the square root of 1 minus the square root of 3 over 2
and then our denominator will be the square root of 2 times the square root of 2 which is 2.
From here we're going to distribute this to into our other radical.
Since they're both radicals, we can simply multiply this part by 2.
So we'll get 2 times 1 is 2, minus the square root of 3 over 2 times 2 is the square root of 3 and we're still taking the square root of this and that'll be over 2.
So this is the simplified form of sine 15 degrees.
As it turns out, we have half-angle identities for cosine and tangent as well.
In fact, tangent has two!
Let’s go ahead and use one of the tangent half-angle identities to find the tangent of 22.5 degrees.
Since 22.5 is half of 45, we can use 45 for theta.
It doesn’t matter which of the tangent half-angle identities we choose; we’ll get the same answer either way.
Let’s go with the first one.
Once again, our process is the same.
Write down the information we’re going to need and then plug that information into our formula.
So we said that theta is equal to 45 degrees and then we'll need to find the sine of theta so the sine of 45 degrees is the square root of 2 over 2
and we also need the cosine of 45 degrees which is also the square root of 2 over 2.
Now we're going to write out our tangent half-angle identity.
So we have the tangent of theta over 2.
Now remember you can use either one of the tangent identities but we're going to use the first one.
So we have the sine of theta over 1 plus the cosine of theta.
From here we plug in all of the values that we have so we know the tangent of theta which is 45 degrees over 2 is equal to the sine of 45 degrees
over 1 plus the cosine of 45 degrees.
Now we simplify.
So 45 degrees divided by 2 is 22.5 degrees and we're looking for the tangent of 22.5 degrees.
The sine of 45 degrees is the square root of 2 over 2, over 1 plus cosine of 45 degrees is also the square root of 2 over 2.
It gets a little tricky since we have fractions inside of fractions, but writing it out as fraction division makes that easier.
In order to do this, we need to write the denominator as one fraction,
which we accomplish by turning the one into a fraction (2 over 2) and then rewriting the whole denominator.
So if we go down here and change 1 into 2 over 2, we'll have the square root of 2 over 2, over 2 over 2, plus the square root of 2 over 2.
And now we're going to simplify the denominator.
So if we do that we'll get 2 plus the square root of 2 over 2.
Now we can turn this into fraction division.
So this is equal to the square root of 2 over 2 divided by 2 plus the square root of 2 over 2.
Remember when you divide fractions it's the same as flipping the second fraction and multiplying them.
So this is equal to the square root of 2 over 2 times 2 over 2 plus the square root of 2.
If we multiply this across, our twos will cancel out and we'll be left with the square root of 2 over 2 plus the square root of 2.
Finally, we have to rationalize our denominators by multiplying the top and bottom of our answers by the conjugate of the denominator,
which looks just like the denominator but has the opposite sign between the two terms.
Now we're going to multiply by the conjugate of the denominator.
So our denominator is 2 plus the square root of 2.
So if we multiply by our conjugate we'll have the square root of 2 over 2 plus the square root of 2,
times 2 minus the square root of 2 over 2 minus the square root of 2.
If we multiply 2 times the square root of 2 we'll get 2 square root of 2
and if we multiply negative square root of 2 by the square root of 2 we'll get the negative square root of 4 and the square root of 4 is 2 so we'll have minus 2.
Down here we're going to FOIL.
So we have 2 times 2 which is 4, 2 times the negative square root of 2 is minus 2 square root of 2, square root of 2 times 2 is positive 2 square root of 2
and then the square root of 2 times the negative square root of 2 is negative square root of 4, again the square root of 4 is 2 so minus 2.
And then here we can simplify our denominator a little bit.
Our negative 2 square root of 2 will cancel out with our positive 2 square root of 2 and we'll be left with 2 square root of 2 minus 2 over 4 minus 2 which is 2.
Before we're done we have one final step.
If you'll notice there's a 2 in each one of these terms so we can divide the entire thing by 2 so divide each term by 2.
If we do that, 2 square root of 2 divided by 2 is the square root of 2,
2 divided by 2 is 1 so minus 1 and then 2 divided by 2 is 1 so it'll be over 1 which we know is just the square root of 2 minus 1.
Next up are the double angle identities.
The double angle identities aren’t quite as useful when dealing with unit circle angles, since we already know the values for double the angles given on it.
But they could definitely be used for a problem where a different angle was given (say 10 degrees) and we needed to find the cosine of 20 degrees.
Or if we wanted to check to make sure the answer we found when using a half-angle formula was correct,
the double angle formula will “undo” what we did with the half-angle formula.
To see how these work, let’s find the cosine of 120 degrees by using the cosine double-angle identity.
That means that theta will be 60 degrees, which is on our unit circle.
120 degrees is also on our unit circle, so we will know quickly if our answer is correct!
As usual, we set up the information we need and then write down our formula.
So for this problem theta is equal to 60 degrees and the sine of 60 degrees is the square root of 3 over 2 and the cosine of 60 degrees is one-half.
Now we write down our cosine double-angle identity.
So the cosine of 2 theta is equal to cosine squared theta minus sine squared theta.
Next, we substitute in our values.
So we have cosine of 2 times theta is 60 degrees is equal to cosine squared of 60 degrees minus sine squared of 60 degrees.
Now we're going to simplify a little bit.
So the cosine of 2 times 60 degrees is the cosine of 120 degrees which is what we're asked to look for and then when we have a trig function squared,
we just take the value of the trig function of cosine of 60 degrees and then square that value.
So the cosine of 60 degrees is one-half and we'll square one-half minus the sine of 60 degrees is the square root of 3 over 2 and we square that as well.
With the cosine double-angle formula, our substituted values are squared, which very nicely removes our radicals.
So one-half squared is one-fourth minus the square root of 3 over 2 squared is 3 over 4.
Finally, we subtract our fractions and get our answer of negative one-half.
Checking the unit circle, we see that the cosine of 120 degrees is negative one-half, so we’re right!
This is actually one of those cases where the calculator will give us the exact value too, since it’s a rational answer.
Also, note that the cosine of 120 degrees is negative.
This is because 120 degrees is in quadrant two, where cosine is always negative.
Our last identities to cover are the formulas to find the product of the sines or cosines of two angles.
For instance, what if we wanted to know the exact value of the sine of 45 degrees times the cosine of 15 degrees?
We could do this by finding the cosine of 15 degrees using the half-angle cosine identity and then multiplying by the sine of 45 degrees from the unit circle.
But there is another way, using the product formulas, which look like this:
Notice that each one covers a different product.
The first one is when we’re finding the product of two cosines.
The second when we’re finding the product of two sines.
The third when the product is of a sine and a cosine, and the last when it’s a product of the cosine and a sine.
Our problem is looking for the sine of 45 degrees times the cosine of 15 degrees, so we want to use the third one.
We start by establishing alpha as 45 degrees and beta as 15 degrees.
So we'll write that information down.
Alpha equals 45 degrees and beta equals 15 degrees.
Then we write down our formula, making sure we have all the signs right.
So our formula is the sine of alpha times the cosine of beta is equal to one-half times the sine of alpha plus beta, plus the sine of alpha minus beta.
Then we substitute in our values for alpha and beta.
So we have the sine of alpha which is 45 degrees times the cosine of beta which is 15 degrees is equal to one-half times the sine of 45 degrees plus 15 degrees,
plus the sine of 45 degrees minus 15 degrees.
Next, we add and subtract our angles in the parentheses to find what we’re going to need.
So we have one-half times the sine of 45 plus 15 which is 60 so the sine of 60 degrees, plus the sine of 45 degrees minus 15 degrees which is 30 degrees.
We will need to know the sine of 60 degrees and the sine of 30 degrees.
Heading back to the unit circle, we find these two values and plug them into our formula.
So we have one-half times the sine of 60 degrees,
the sine of 60 degrees from our unit circle is the square root of 3 over 2 plus the sine of 30 degrees which is one-half.
Finally, we simplify.
So we have one-half times the square root of 3 over 2 plus 1 over 2 can be simplified to the square root of 3 plus 1 over 2
and then we multiply this by one-half to get the square root of 3 plus 1 over 4.
We can check our answer on a calculator by evaluating our answer
and then finding out the approximate value of the sine of 45 degrees times the cosine of 15 degrees.
In both cases, we get this:
These formulas can also be used to turn products into sums, or sums into products, especially when angles are given as variables.
Hopefully, this review was helpful and you’re well on your way to mastering these trig formulas.
Thanks for watching, and happy studying!