Double Angle Identities - Video Tutorials & Practice Problems

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Double Angle Identities

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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 some 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 trade problems with. So here I'm gonna 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 some formulas but for the same two angles. So if I take the sign of some angle, theta plus that same angle, the I end up with this expression here. Now these two terms are actually exactly the same now. So this simplifies to two times the sign of the times the code of the giving me my double angle identity for sine, that sine of two data is equal to two sine theta cosine theta. Now we can do the same exact thing for cosine and tangent. And this ends up giving me that the cosine of two the is equal to the cosine squared of data minus sine squared of data. And the tangent of two data is equal to two times the tangent of data over one minus the tangent squared of data. Now looking at our expression here, the cosine squared of pi over 12 of minus the science 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 two 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 two times my angle here pi over 12. Now two times pi over two 12 is just pi over six. So this is really the cosign of pi over six, 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 3/2. Having used my double angle identity. Now, something that you may have noticed here is that we have two 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 just are going to help us in different scenarios to simplify different trigger 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 whatever argument contains two 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, 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 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 gonna 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 sign 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 sign or cosine of 15 degrees, I do recognize that this looks like part of my double angle formula for sign specifically here, I see the sign of the times the cosine of data. I'm just missing that too. So let's see what we can do here. I'm gonna go ahead and rewrite my sine double angle identity here that the sine of two theta is equal to two times the sine of the times the cosine of data. 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 two, that two will cancel on that right side leaving me with the sine of two theta over two is equal to the sign of the times the cosine of data, which is the exact expression that I'm working with here just with the being equal to 15 degrees. So with this in mind, I can use that double angle identity in order to rewrite this as the sign of two times that angle, 15 degrees over two. Now what is two times 15 degrees? It's just 30 degrees. So this is really a sign of 30 degrees over two. And we have successfully simplified this expression taking it from two different trig functions to just one function. Now, we're not asked to evaluate here. But if we were, we easily could, because the sign 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.

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example

Double Angle Identities Example 1

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Hey, everyone in this problem, we're asked to use our double angle formulas in order to verify the identity, the sign of two data over the tangent of data is equal to the cosine of two data plus one. 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 gonna wanna 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 sign of two, the which tells me that I can use my double angle formula for sign to rewrite this as to sign the cosign the. Now I'm gonna keep my denominator the same for now the tangent of data 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 sign and cosine because everything else already is, I'm gonna keep my numerator the same for now again to sin the cosine theta and then I'm gonna rewrite that tangent as the sign of data over the cosine of data. Now, where do we go from here? Or remember when we're dividing by a fraction, we're really just multiplying by the reciprocal. So I can think of this as two sine theta cosine theta times the reciprocal of that fraction on the bottom cosine theta over sine data. Now, from here, those sine theta will cancel out and I'm left with two times the cosine squared of theta 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 two theta plus one. Now remember we're scanning for identities and here I see that I have two times the cosine squared of theta, 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 two theta, if I go ahead and add one 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 two theta plus one. Now, let's remember what that right side of our equation is. It is the cosine of two theta plus one. So from here, I'm done and I have fully verified this identity. The cosine of two, theta plus one is equal to the cosine of two theta plus one. Now we're done here. Please let me know if you have any questions. Thanks for watching.

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Double Angle Identities Example 2

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Hey, everyone in this problem, we're asked to use our double angle formulas in order to verify the identity, the code tangent of two, the is equal to the code tangent of the minus 1/2 cosine theta sine theta. 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 noticed 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 sign and cosine. Now the code tangent of data is equal to the cosine of data over the sine of data. So I'm gonna go ahead and replace that cotangent using that identity. Now, I'm still subtracting 1/2 times the cosine of data times the sine of data. Now, where do we go from here? We want to go ahead and combine any fractions using a common denominator since we have two separate fractions here. Now here my common denominator is gonna be two times the cosine theta times the sine of data by multiplying this fraction by two cosine data on the top and the bottom. Now what happens when I do that? Well, I have two cosine squared of theta minus one over that common denominator two cosine theta sine theta. Now let's go ahead and do some scanning for identities in that numerator. I see two cosine squared theta minus one which I recognize from my double angle identity for cosine here. Now I can replace with the cosine of two theta. But what about my denominator? Well, in my denominator, I see two cosine theta sine theta, which is exactly what the sine of two theta is. So that becomes my denominator here sine of two theta. Now, let's remember what our goal is that left side of our equation, the cotangent of two theta. Our goal is that this right side is equal to this left side and look at where we are, we have the cosine of two data over the sign of tooth data. And what is that equal to, it's equal to the cotangent of two data 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.

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Double Angle Identities Example 3

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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 two 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 going to use the same exact steps that we saw with our sum in different identities when doing a really similar problem type. So let's look at this problem. We are given that the cosign of the is equal to 4/5 and that theta is in between zero and pi over two. And we want to find the sign of two theta. Now remembering our steps here in evaluating trig functions when we're given different conditions, our step number one is to expand our identity out and identify any of our unknown trig values. Now, here, we're asked to find the sign of two, the and when we find the sign of two theta, we're going to look at this identity here. The sign of two, the is equal to two times the sine of data times the cosine of data Now we want to identify our unknown trig values. And in our problem statement, we're told that the cosine of data is equal to 4/5. So we already know that and all that's left to find is the sine of data. So let's move on to our next step and figure out how to get there. Step one is done in step two. 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 gonna 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 four and my hypotenuse is five and we have completed step number two. Now, from step number three, we want to find any missing side links using the Pythagorean theorem. Now, here I don't even need to use the Pythagorean theorem because I see that this is a 345 triangle. So my missing side link is simply three and step number three is done. Now, moving on to step number four, we wanna solve for any unknown trig values that we identified in step number one. Now, in step one, we needed to find the sign of beta. So let's go ahead and do that here. Now, the sign of data is going to be equal to that opposite side that we just found three over that hypotenuse five. And that's my only unknown trig value. So we've completed step number four. Now, with that unknown trig value, we can move on to step number five and plug in all of our values and simplify. So here I'm taking two, I'm multiplying it by that sign of data that I just found in step 4 3/5 and then multiplying it by that cosign that I was given in my problem forfeits. Now, all that's left to do is simplify doing some algebra here. Now, multiplying two times three times four will give me 24. Then in my denominator, I just have five times five, which is 25. So that gives me my final answer here. Step five 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.

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Problem

Problem

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