Double Angle Identities - Online Tutor, Practice Problems & Exam Prep

Double angle identities are derived from sum formulas for the same angle, enhancing the ability to simplify trigonometric expressions. Key identities include: sin(2θ)=2sin(θ)cos(θ), cos(2θ)=cos(θ)^2-sin(θ)^2, and tan(2θ)=2tan(θ)/(1-tan(θ)^2). Recognizing parts of these identities aids in simplification, allowing for efficient problem-solving in trigonometry.

Double Angle Identities

Hey, everyone. We just learned a bunch of different trig identities. And now that we know so many trig identities, we can begin to use those that we already know in order to come up with even more trig identities. Like, if we take our sum formula but for the same two angles, we end up with what's called our double angle identities. Now your first question here might be why. Why do we need even more trig identities? And the answer is that they're just going to keep allowing us to work through more and more trig problems with ease. So here I'm going to walk you through exactly what our double angle identities are and how to use them in order to continue simplifying trig expressions. So let's go ahead and get started.

Now I mentioned that these double angle identities come from using our sum formulas but for the same two angles. So if I take the sine of some angle theta plus that same angle theta, I end up with this expression here. Now these two terms are actually exactly the same now, so this simplifies to 2 times the sine of theta times the cosine of theta, giving me my double angle identity for sine. That sine of 2 theta is equal to 2sinθcosθ. Now we can do the same exact thing for cosine and tangent, and this ends up giving me that the cosine of 2 theta is equal to cosθ2-sinθ2, and the tangent of 2 theta is equal to 2tanθ1-tanθ2.

Now looking at our expression here, the cosine squared of pi over 12 minus the sine squared of pi over 12, I don't know the cosine or sine of pi over 12 off the top of my head, but I do recognize this as one of my double angle identities, and specifically my cosine double angle identity, the cosine squared of some angle minus the sine squared of some angle. So this is really just the cosine of 2 times that angle. So I can rewrite this cosine squared of pi over 12 minus the sine squared of pi over 12 as the cosine of 2 times my angle here, pi over 12. Now 2 times pi over 12 is just pi over 6, so this is really the cosine of pi over 6, a value that I know really well from my unit circle. So I can come to an answer rather easily, that this is just equal to the square root of 3 over 2 having used my double angle identity.

Now something that you may have noticed here is that we have 2 other forms of our cosine double angle formula. And these are just alternate forms that come from taking this first identity here and rewriting it using our Pythagorean identities. So these alternate forms are just going to help us in different scenarios to simplify different trig expressions. Now that we've seen all of these double angle identities, how do we know when to use them? Well, something that might be rather obvious is that whenever our argument contains 2 times some angle, we should use our double angle identities because that's the exact argument of all of these identities. Now something that might be less obvious is that whenever we recognize a part of the identity, we should go ahead and use these identities.

Now this is kind of what we did in our example here. We recognize that this cosine squared minus sine squared was this particular part of my cosine double angle formula. Now here, this was a bit more simple, but it's not always going to be exactly like that. Because remember that for our other identities, we had to recognize a bunch of different forms and rearranged versions of our identities in order to successfully simplify expressions and verify identities, and it's going to be no different with our double angle identities. So let's go ahead and work through another example here.

Here, we're asked to simplify the expression but not to evaluate. The sine of 15 degrees times the cosine of 15 degrees. Now remember that when simplifying an expression, we want to be constantly scanning for identities. And here, even though I don't know the sine or cosine of 15 degrees, I do recognize that this looks like part of my double angle formula for sine. Specifically here, I see the sine of theta times the cosine of theta. I'm just missing that 2. So let's see what we can do here. I'm going to go ahead and rewrite my sine double angle identity here that the sine of 2 theta is equal to 2 times the sine of theta times the cosine of theta. Now I only want this part of this identity. So what can I do here? Well, if I go ahead and divide both sides by 2, that 2 will cancel on that right side, leaving me with the sine of 2 theta over 2 is equal to the sine of theta times the cosine of theta, which is the exact expression that I'm working with here, just with theta being equal to 15 degrees. So with this in mind, I can use that double angle identity in order to rewrite this as the sine of 2 times that angle 15 degrees over 2. Now what is 2 times 15 degrees? It's just 30 degrees. So this is really a sine of 30 degrees over 2, and we have successfully simplified this expression, taking it from 2 different trig functions to just 1 trig function. Now we're not asked to evaluate here, but if we were, we easily could because the sine of 30 degrees is a value that I know really well from the unit circle.

Now as we continue to learn more and more identities, it can become overwhelming to figure out how to simplify a trig expression. But with each new identity that we learn, we have a greater ability to work through the problems that are asked of us. So now that we know our double angle identities, let's continue practicing with them. Thanks for watching, and let me know if you have questions.

Double Angle Identities Example 1

Hey everyone, in this problem we're asked to use our double angle formulas in order to verify the identity. The sine of 2θ over the tangent of θ is equal to the cosine of 2θ plus 1. Now you should go ahead and try this problem on your own before checking back in with me, but let's go ahead and get started. Now remember when verifying an identity, you're going to want to start with your more complicated side and it's not always going to be really obvious which side is more complicated. So remember that if you really don't know, starting with either side is really fine. Now here, I'm going to start with this left side because it has a fraction in it. So let's go ahead and get started here using our simplifying strategies. Now remember that we want to be constantly scanning for identities and the first thing I notice here is that I'm taking the sine of 2θ, which tells me that I can use my double angle formula for sine to rewrite this as 2 sine θ cosine θ. Now I'm going to keep my denominator the same for now the tangent of θ. Now I can't really do anything yet here but remember one of our strategies is to break down everything in terms of sine and cosine. So let's go ahead and break down this tangent in terms of sine and cosine because everything else already is. I'm gonna keep my numerator the same for now again, 2 sine θ cosine θ, and then I'm going to rewrite that tangent as the sine of θ over the cosine of θ. Now, where do we go from here? Well, remember when we're dividing by a fraction, we're really just multiplying by the reciprocal. So I can think of this as 2 sine θ cosine θ times the reciprocal of that fraction on the bottom, cosine θ over sine θ. Now from here those sine θs will cancel out and I'm left with 2 times the cosine squared of θ since those cosines are multiplying each other. Now where do I go from here? Because remember our goal is to make this side equal to the cosine of 2θ plus 1. Now remember we're scanning for identities and here I see that I have the 2 times the cosine squared of θ, which I recognize as a part of one of my cosine double angle formulas. So let's go ahead and use that here. Now the cosine of 2θ, if I go ahead and add 1 to both sides here, I'm left with exactly the expression that I have over here. So I can replace this with the cosine of 2θ plus 1. Now let's remember what that right side of our equation is. It is the cosine of 2θ plus 1. So from here I'm done and I have fully verified this identity. Cosine of 2θ plus 1. Now we're done here. Please let me know if you have any questions. Thanks for watching.

Double Angle Identities Example 2

Hey everyone. In this problem, we're asked to use our double-angle formulas in order to verify the identity. The cotangent of 2θ is equal to the cotangent of θ minus 1 over 2 cosine θ sine θ. Now you should go ahead and try this on your own before checking back in with me. And remember that there are always multiple ways to verify an identity. So, if you did this slightly differently than me but you still ended up with both sides being equal to each other, that's totally fine. Now here, let's go ahead and jump in.

Now remember, when verifying an identity, we want to start with our more complicated side first. And here, this right side is clearly more complicated, so let's start there. Now using our simplifying strategies, one thing that I notice is that one of my terms is in terms of cosine and sine and the other one is not. So let's go ahead and break down everything in terms of that sine and cosine. Now the cotangent of θ is equal to the cosine of θ over the sine of θ. So I'm going to go ahead and replace that cotangent using that identity. Now I'm still subtracting 1 over 2 times the cosine of θ times the sine of θ.

Now, where do we go from here? We want to go ahead and combine any fractions using a common denominator since we have 2 separate fractions here. Now, here, my common denominator is going to be 2 times the cosine θ times the sine θ by multiplying this fraction by 2 cosine θ on the top and at the bottom. Now what happens when I do that? Well, I have 2 cosine squared θ minus 1 over that common denominator, 2 cosine θ sine θ. Now let's go ahead and do some scanning for identities. In that numerator, I see 2 cosine squared θ minus 1, which I recognize from my double-angle identity for cosine here.

cos ( 2 θ ) sin ( 2 θ )

Now I can replace that with the cosine of 2θ. But what about my denominator? Well, in my denominator, I see 2 cosine θ sine θ, which is exactly what the sine of 2θ is. So that becomes my denominator here, sine of 2θ. Now let's remember what our goal is, that left side of our equation, the cotangent of 2θ. Our goal is that this right side is equal to this left side and look at where we are. We have the cosine of 2θ over the sine of 2θ. And what is that equal to? It's equal to the cotangent of 2θ, which is exactly what that left side already is. So we have successfully verified this identity having our right and left side of our equation being equal to each other. Thanks for watching, and let me know if you have questions.

Double Angle Identities Example 3

Hey, everyone. You may come across this specific type of problem in which you're given certain information about theta and asked to find a function of 2 theta. Now this is really similar to other problem types that we've seen before, but now just for our double angle identities. So let's go ahead and jump in here. We're gonna use the same exact steps that we saw with our sum and difference identities when doing a really similar problem type. So let's look at this problem. We are given the cosine of theta is equal to 4/5 and that theta is in between 0 and π/2, and we want to find the sine of 2θ. Now, remembering our steps here in evaluating trig functions when we're given different conditions, our step number 1 is to expand our identity out and identify any of our unknown trig values. Now here we're asked to find the sine of 2θ and when we find the sine of 2θ we're going to look at this identity here. The sine of 2θ is equal to 2 times the sine of theta times the cosine of theta. Now we want to identify our unknown trig values, and in our problem statement, we're told that the cosine of theta is equal to 4/5. So we already know that and all that's left to find is the sine of theta. So let's move on to our next step and figure out how to get there. Step 1 is done. In step 2, we're told that from our given info, we want to sketch and label our triangles in the proper quadrant, remembering to pay close attention to the sign. Here we're told that the cosine of theta is equal to 4/5 and that theta is in that first quadrant. So I'm going to go ahead and draw my right triangle here and label my angle theta on that inside. Now since the cosine of this angle theta is equal to 4/5 that tells me that this adjacent side is 4 and my hypotenuse is 5 and we have completed step number 2. Now, from step number 3, we want to find any missing side lengths using the Pythagorean theorem. Now here, I don't even need to use the Pythagorean theorem because I see that this is a 3, 4, 5 triangle. So my missing side length is simply 3 and step number 3 is done. Now moving on to step number 4, we want to solve for any unknown trig values that we identified in step number 1. Now in step 1, we needed to find the sine of beta. So let's go ahead and do that here. Now the sine of beta is gonna be equal to that opposite side that we just found, 3 over that hypotenuse 5, and that's my only unknown trig value. So we've completed step number 4. Now with that unknown trig value, we can move on to step number 5 and plug in all of our values and simplify. So here I'm taking 2, I'm multiplying it by that sine of theta that I just found in step 4, 3/5, and then multiplying it by that cosine that I was given in my problem, 4/5. Now all that’s left to do is simplify doing some algebra here. Now multiplying 2 times 3 times 4 will give me 24. Then in my denominator, I just have 5 times 5, which is 25. So that gives me my final answer here. Step 5 is done and I have successfully solved this problem with my final answer 24/25. Thanks for watching, and I'll see you in the next one.

Given $\tan\theta=\frac{5}{12}$ and 0 < $\theta$ < $\frac{\pi}{2}$, find $\cos\left(2\theta\right)$.

• A. 0
• B. $-\frac{199}{169}$
• C. $\frac{119}{169}$
• D. $\frac{144}{169}$

Here’s what students ask on this topic:

What are the double angle identities in trigonometry?

Double angle identities are trigonometric formulas that express trigonometric functions of double angles (2θ) in terms of single angles (θ). The key double angle identities are:

$\sin \left(2\theta \right)=2\sin \left(\theta \right)\cos \left(\theta \right)$

$\cos \left(2\theta \right)=\cos \left(\theta \right)^2-\sin \left(\theta \right)^2$

$\tan \left(2\theta \right)=\frac{2\tan \left(\theta \right)}{1-\tan \left(\theta \right)^2}$

How do you derive the double angle identity for sine?

The double angle identity for sine is derived from the sum formula for sine. The sum formula states:

$\sin (A+B)=\sin \left(A\right)\cos \left(B\right)+\cos \left(A\right)\sin \left(B\right)$

For the same angle θ, we have:

$\sin (\theta +\theta )=\sin \left(\theta \right)\cos \left(\theta \right)+\cos \left(\theta \right)\sin \left(\theta \right)$

Since $\sin \left(\theta \right)\cos \left(\theta \right)=\cos \left(\theta \right)\sin \left(\theta \right)$, this simplifies to:

$\sin \left(2\theta \right)=2\sin \left(\theta \right)\cos \left(\theta \right)$

When should you use double angle identities in trigonometry?

Double angle identities should be used whenever the argument of a trigonometric function is twice an angle, i.e., 2θ. They are particularly useful for simplifying expressions and solving equations involving trigonometric functions. Additionally, if you recognize parts of the double angle identities within an expression, you can use these identities to simplify the expression further. For example, if you see $2\sin \left(\theta \right)\cos \left(\theta \right)$, you can replace it with $\sin \left(2\theta \right)$. Recognizing these patterns helps in solving trigonometric problems more efficiently.

How do you simplify trigonometric expressions using double angle identities?

To simplify trigonometric expressions using double angle identities, follow these steps:

1. Identify parts of the expression that match double angle identities. For example, look for terms like $2\sin \left(\theta \right)\cos \left(\theta \right)$ or $\cos \left(\theta \right)^2-\sin \left(\theta \right)^2$.
2. Replace these parts with their corresponding double angle identities. For example, $2\sin \left(\theta \right)\cos \left(\theta \right)$ becomes $\sin \left(2\theta \right)$.
3. Simplify the resulting expression further if possible.

For example, to simplify $\sin (15°)\cos (15°)$, recognize it as part of the sine double angle identity and rewrite it as $\frac{\sin }{(}$.

What are the alternate forms of the cosine double angle identity?

The cosine double angle identity has three forms, derived from the Pythagorean identities. The primary form is:

$\cos \left(2\theta \right)=\cos \left(\theta \right)^2-\sin \left(\theta \right)^2$

Using the Pythagorean identity $\sin \left(\theta \right)^2=1-\cos \left(\theta \right)^2$, we get:

$\cos \left(2\theta \right)=2\cos \left(\theta \right)^2-1$

Similarly, using $\cos \left(\theta \right)^2=1-\sin \left(\theta \right)^2$, we get:

$\cos \left(2\theta \right)=1-2\sin \left(\theta \right)^2$

These alternate forms are useful in different scenarios for simplifying trigonometric expressions.