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Given that sin(θ)=1/2 and that θ is in the second quadrant, Sal finds tan(θ) using the Pythagorean identity. Created by Sal Khan.
Video transcript
Let's say that we're told that some angle theta, which is going to be expressed in radians, is between negative 3 pi over 2 and negative pi. It's greater than negative 3 pi over 2 and it's less than negative pi. And we're also told that sine of theta is equal to 1/2. Just from this information can we figure out what the tangent of theta is going to be equal to? And I encourage you to pause the video and try this on your own. In case you're stumped, I will give you a hint. You should use the Pythagorean identity, the fact that sine squared theta, plus cosine squared theta is equal to 1. So let's do it. So we know the Pythagorean identity, sine squared theta, plus cosine squared theta is equal to 1. We know what sine squared theta is. Sine theta is 1/2. So this could be rewritten as 1/2 squared, plus cosine squared theta, is equal to 1. Or we could write this as 1/4 plus cosine squared theta is equal to 1. Or we could subtract 1/4 from both sides, and we get cosine squared theta is equal to-- let's see. You subtract 1/4 from the left hand side, then this 1/4 goes away. That was the whole point. 1 minus 1/4 is 3/4. So what could cosine of theta be? Well, when I square it, I get positive 3/4. So it could be the positive or negative square root of 3/4. So cosine of theta could be equal to the positive or negative square root of 3 over 4, which is the same thing as the positive or negative square root of 3, over the square root of 4, which is 2. So it's the positive or negative square root of 3 over 2. But how do we know which one of these it actually is? Well, that's where this information becomes useful. Let's draw our unit circle. If you're saying, well, why am I even worried about cosine of theta? Well, if you know sine of theta you know cosine of theta. Tangent of theta is just sine of theta over cosine theta. So then you will know the tangent of theta. But let's look at the unit circle to figure out which value of cosine we should use. So let me draw it, the unit circle. That's my y-axis. That is my x-axis. And I will draw the unit circle in pink. So that's my best attempt at drawing a circle. Please forgive me for its lack of perfect roundness. And it says theta is greater than negative 3 pi over 2. So where is negative 3 pi over 2? So let's see. This is negative pi over 2. So this is one side of the angle. Let me do this in a color. So this one side of the angle is going to be along the positive x-axis. And we want to figure out where the other side is. So this right over here that's negative pi over 2. This is negative pi. So it's between negative pi, which is right over here. So let me make that clear. Negative pi is right over here. It's between negative pi and negative 3 pi over 2. Negative 3 pi over 2 is right over here. So our angle theta is going to put us someplace over here. And the whole reason I did this-- so this whole arc right here-- you could think of this as the measure of angle theta right over there. And the whole reason I did that is to think about whether the cosine of theta is going to be positive or negative. We clearly see it's in the second quadrant. The cosine of theta is the x-coordinate of this point where our angle intersects the unit circle. So this point right over here-- actually let me do it in that orange color again-- this right over here, that is the cosine of theta. Now is that a positive or negative value? Well it's clearly a negative value. So for the sake of this example, our cosine theta is not a positive 1. It is a negative 1 So we could write the cosine theta is equal to the negative square root of 3 over 2. So we figured out cosine theta, but we still have to figure out tangent of theta. And we just have to remind ourselves that the tangent of theta is going to be equal to the sine of theta over the cosine of theta. Well they told us the sine of theta is 1/2. So it's going to be 1/2 over cosine of theta, which is negative square root of 3 over 2. And what does that equal? Well that's the same thing as 1/2 times the reciprocal of this. So times negative 2 over the square root of 3. These twos will cancel out and we are left with negative 1 over the square root of 3. Now some people don't like a radical in the denominator, like this. They don't like an irrational denominator. So we could rationalize the denominator here by multiplying by square root of 3 over square root of 3. And so this will be equal to negative square root of 3 over 3 is the tangent of this angle right over here. And that actually makes sense, because the tangent of the angle is the slope of this line. And we see that it is indeed a negative slope.