Find cos ( a + b ) \cos\left(a+b\right) cos ( a + b ) given cos a = 1 2 \cos a=\frac12 cos a = 2 1 , sin b = 1 2 \sin b=\frac12 sin b = 2 1 , & a a a is in Q IV and b b b is in Q II.

In this problem we're asked to find the cosine of a plus b given that the cosine of a is equal to 1 half and the sine of b is also equal to 1 half with angle a in quadrant 4 and angle b in quadrant 2. Now what are we going to do with all that information? Well, let's go ahead and get started with our steps. Now in step 1, we're told to go ahead and expand our identity and then identify any unknown angles. Now here we're asked to find the cosine of a plus b. So we wanna go ahead and expand that using our sum identity for cosine. So the cosine of a plus b is equal to the cosine of a times the cosine of b minus the sine of a times the sine of b. So what of that do we already know? Well, in my problem, I'm told that the cosine of a is equal to 1 half and the sine of b is equal to 1 half. So those two trig values I already know, but I don't know the cosine of b and the sine of a. So we need to find that information. So let's continue on in our steps. Step 1 is done. We want to move on to step number 2, where from our given info, we want to sketch and label our right triangles in their proper quadrant, remembering to pay attention to the sine. Now we're told that the cosine of a is equal to 1 half and that angle a is in quadrant 4. So let's go ahead and sketch that triangle in quadrant 4 and label that angle angle a. Now since the cosine of a is equal to 1 half, that tells me that my adjacent side is equal to 1 and my hypotenuse is equal to 2, both of these being positive values. Now for angle b, we're told that angle b is in quadrant 2. So we wanna go ahead and sketch that right triangle over there and label our angle angle b. Now we're told that the sine of b is equal to 1 half. So here, our opposite side is going to be equal to 1, and our hypotenuse is equal to 2. Both of these are again positive values. Now we've completed step number 2. Step number 3 tells us to find any of our missing side lengths. Now here in this triangle right here, we're missing this side length. So we have 1 and a 2 for each of our side lengths that we know already, and we can use our pythagorean theorem to find that missing third side length. But seeing as I have a side length of 1 and a side length of 2, you may already notice that this side length is going to end up being the square root of 3. Feel free to work this out using the pythagorean theorem on your own. Now let's look at our other triangle. We also have a side length of 1 and a side length of 2. So again, our missing side length is going to be the square root of 3. But wait, remember that we want to pay close attention to the sign and this is a negative square root of 3. And on my other triangle, this is also a negative square root of 3 because of where they're located on my graph. Here we have a negative y values, and here we have negative x values. So remember to pay close attention to that sign. Now we've completed step number 3. Moving on to step number 4, we want to solve for unknown trig values that we found in step number 1. Now our unknown trig values, remember, are the sine of a and the cosine of b. So from this triangle here, I wanna find the sine. Now the sine of this angle a is going to be equal to that opposite side, negative square root of 3, over that hypotenuse 2. Then for this triangle, we wanna find the cosine of b. So the cosine of b is equal to that adjacent side, the square root of 3, negative square root of 3, over the hypotenuse of 2. So now we have all the information that we need to actually evaluate this sum. We've completed step 4. Moving on to step 5, we're actually going to plug in all of those values. Now remember that we already know the cosine of a and the sine of b from our problem. So the cosine of a, based on that problem statement, is equal to onetwo. And I'm multiplying that by the cosine of b, which we found as the negative square root of 3 over 2. Now here I'm going to subtract the sine of a, which I also just found in that last step, as being the negative square root of 3 over 2. And then I'm multiplying that by the sine of b, which is in my problem statement, 1 half. Now let's see what happens when I multiply this across because all that's left to do is algebra. Now multiplying these first two fractions across, I end up with the negative root 3 over 4. Then I'm subtracting these 2 multiplied together, which is also the negative square root of 3 over 4. But what happens when I subtract a negative? I'm really just adding. So here I have negative square root of 3 over 4 plus square root of 3 over 4. So what happens here? Well, these end up just canceling out and all that I'm left with is 0. And that's my final answer here. The cosine of a plus b is just equal to 0. Feel free to let me know if you have any questions. Thanks for watching.

Find cos ⁡ ( a + b ) \cos\left(a+b\right) cos ( a + b ) given cos ⁡ a = 1 2 \cos a=\frac12 cos a = 2 1 , sin ⁡ b = 1 2 \sin b=\frac12 sin b = 2 1 , & a a a is in Q IV and b b b is in Q II.

3 4 \frac{\sqrt3}{4} 4 3 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z">

− 3 2 -\frac{\sqrt3}{2} − 2 3 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z">

6. Trigonometric Identities and More Equations

Sum and Difference Identities

Trigonometry

6. Trigonometric Identities and More Equations

Sum and Difference Identities

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Multiple Choice

Find $\cos\left(a+b\right)$ given $\cos a=\frac12$ , $\sin b=\frac12$ , & $a$ is in Q IV and $b$ is in Q II.

$\frac{\sqrt3}{4}$

$-\frac{\sqrt3}{2}$

Verified step by step guidance

First, recall the angle addition formula for cosine: $ \cos(a+b) = \cos a \cos b - \sin a \sin b $. We will use this formula to find $ \cos(a+b) $.

Given $ \cos a = \frac{1}{2} $ and $ a $ is in the fourth quadrant, we need to find $ \sin a $. In the fourth quadrant, sine is negative. Use the Pythagorean identity: $ \sin^2 a + \cos^2 a = 1 $. Substitute $ \cos a = \frac{1}{2} $ to find $ \sin a $.

Calculate $ \sin a $ using the identity: $ \sin^2 a = 1 - \cos^2 a = 1 - \left(\frac{1}{2}\right)^2 = \frac{3}{4} $. Therefore, $ \sin a = -\frac{\sqrt{3}}{2} $ because $ a $ is in the fourth quadrant.

Next, find $ \cos b $ using the given $ \sin b = \frac{1}{2} $ and $ b $ is in the second quadrant. In the second quadrant, cosine is negative. Use the identity: $ \cos^2 b = 1 - \sin^2 b = 1 - \left(\frac{1}{2}\right)^2 = \frac{3}{4} $. Therefore, $ \cos b = -\frac{\sqrt{3}}{2} $.

Substitute $ \cos a = \frac{1}{2} $, $ \sin a = -\frac{\sqrt{3}}{2} $, $ \cos b = -\frac{\sqrt{3}}{2} $, and $ \sin b = \frac{1}{2} $ into the angle addition formula: $ \cos(a+b) = \left(\frac{1}{2}\right)\left(-\frac{\sqrt{3}}{2}\right) - \left(-\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) $. Simplify this expression to find $ \cos(a+b) $.