Half-Angle, Double Angle, and Product Trig Identities

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:

\(\sin{(\frac{\theta}{2})}=\pm \sqrt{\frac{1-\cos{\theta}}{2}}\)

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 \(\theta\).

\(\theta=30°\)
 
\((\frac{\sqrt{3}}{2},\frac{1}{2})\)

Then, we simply substitute all of the appropriate values into the formula.

\(\sin{(\frac{\theta}{2})}= \sqrt{\frac{1-\cos{(\theta)}}{2}}\)
 
\(\sin{(\frac{30°}{2})}=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}\)

Finally, we simplify.

\(\sin{(15°)}=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{\sqrt{2}}}\)

This time we have a radical in our denominator so we need to rationalize it. The process for doing so is a bit different though 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.

\(\frac{\sqrt{1-\frac{\sqrt{3}}{2}}}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}\sqrt{1-\frac{\sqrt{3}}{2}}}{2}=\frac{\sqrt{2-\sqrt{3}}}{2}\)

And just like that, we have our answer.

As it turns out, we have half-angle identities for cosine and tangent as well. In fact, tangent has two!

\(\cos{(\frac{\theta}{2})}=\pm \sqrt{\frac{1+\cos{\theta}}{2}}\)     \(\tan{(\frac{\theta}{2})}=\frac{\sin{\theta}}{1+\cos{\theta}}\)
 
  \(\tan{(\frac{\theta}{2})}=\frac{1-\cos{\theta}}{\sin{\theta}}\)

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.

\(\theta=45°\)
 
\((\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})\)
 
\(\tan{(\frac{\theta}{2})}=\frac{\sin{\theta}}{1+\cos{\theta}}\)
 
\(\tan{(\frac{45°}{2})}=\frac{\frac{\sqrt{2}}{2}}{1+\frac{\sqrt{2}}{2}}\)

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 \((\frac{2}{2})\) and then rewriting the whole denominator.

\(\frac{\frac{\sqrt{2}}{2}}{1+\frac{\sqrt{2}}{2}}=\frac{\frac{\sqrt{2}}{2}}{\frac{2+\sqrt{2}}{2}}=\frac{\sqrt{2}}{2}\div \frac{2+\sqrt{2}}{2}= \frac{\sqrt{2}}{2}\cdot \frac{2}{2+\sqrt{2}}=\frac{\sqrt{2}}{2+\sqrt{2}}\)

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.

\(\frac{\sqrt{2}}{2+}\cdot \frac{2-\sqrt{2}}{2-\sqrt{2}}=\frac{2\sqrt{2-}\sqrt{4}}{4-2\sqrt{2}+2\sqrt{2}-\sqrt{4}}\)

Then, we can simplify even further and see if we can make the answers look the same.

\(\frac{2\sqrt{2}-2}{2}=\sqrt{2}-1\)

Next up are the double angle identities.

\(\sin{(2\theta)}=2 \sin{\theta} \cos{\theta}\)
 
\(\cos{(2\theta)}=\cos^{2}{\theta}-\sin^{2}{\theta}\)
 
\(\tan{(2\theta)}=\frac{2 \tan{\theta}}{1-\tan^{2}{\theta}}\)

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 \(\theta\) 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.

\(\theta=60°\)
 
\((\frac{1}{2},\frac{\sqrt{3}}{2})\)
 
\(\cos{(2\theta)}=\cos^{2}{\theta}-\sin^{2}{\theta}\)

Next, we substitute in our values.

\(\cos{(2(60°))}=(\frac{1}{2})^{2}-(\frac{\sqrt{3}}{2})^{2}\)

With the cosine double-angle formula, our substituted values are squared, which very nicely removes our radicals.

\(\frac{1}{4}-\frac{3}{4}\)

Finally, we subtract our fractions and get our answer of \(-\frac{1}{2}\).

Checking the unit circle, we see that the cosine of 120 degrees is \(-\frac{1}{2}\), 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 II, 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 a cosine and a sine.

Our problem is looking for \(\sin{(45°)}\cos{(15°)}\), so we want to use the third one.

We start by establishing \(\alpha\) as 45 degrees and \(\beta\) as 15 degrees.

\(\alpha=45°\)
 
\(\beta=15°\)

Then we write down our formula, making sure we have all the signs right.

\(\sin{(\alpha)}\cos{(\beta)}=\frac{1}{2}[\sin{(\alpha + \beta)}+ \sin{(\alpha – \beta)}]\)

Then we substitute in our values for \(\alpha\) and \(\beta\).

\(\sin{(45°)}\cos{(15°)}=\frac{1}{2}[\sin{(45°+15°)}+\sin{(45°-15°)}]\)

Next, we add and subtract our angles in the parentheses to find what we’re going to need.

\(\sin{(45°)}\cos{(15°)}=\frac{1}{2}[\sin{(60°)}+\sin{(30°)}]\)

We will need to know the sine of 60 degrees and the sine of 30 degrees. Heading back to the unit circle, we find those two values and plug them into our formula.

\(\sin{(45°)}\cos{(15°)}=\frac{1}{2}[\frac{\sqrt{3}}{2}+\frac{1}{2}]\)

Finally, we simplify.

\(\frac{1}{2}[\frac{\sqrt{3}+1}{2}]=\frac{\sqrt{3}+1}{4}\)

We can check our answer on a calculator by evaluating our answer and then finding out the approximate value of \(\sin{(45°)}\cos{(15°)}\). In both cases, we get this:

\(\frac{\sqrt{3}+1}{4}=0.683012701892\)
 
\(\sin{(45°)}\cos{(15°)}=0.683012701892\)

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!