Skip to 0 minutes and 12 seconds Hello. It’s your turn on the function arctangent. OK. Before to start with this exercise, let us remind something about the function arctangent. OK. The function arctangent is defined on the whole set of real numbers. And its range is equal to the open interval minus pi over 2, pi over 2.

Skip to 0 minutes and 47 seconds More, this is a bijective function. That is, it is both injective and surjective. Good. Let us start with the first point of our exercise. OK, we want to compute the arctangent of minus the square root of 3. Good. Let us denote by theta this value. Then, what is theta? Theta is the unique value inside this open interval, where the tangent of theta is equal to minus the square root of 3. OK. But now, we remember that the tangent of theta is the sine of theta over the cosine of theta. And we are looking for what? For a value of theta inside this interval, such that this fraction is equal to minus the square root of 3.

Skip to 2 minutes and 4 seconds But now, because this fraction is negative, surely theta will be a value between minus pi over 2 and 0. And precisely, which is the correct value? Theta equal to minus pi over 3. Indeed, if you consider theta equal to minus pi over 3, then what we have that the sine of theta over the cosine of theta is equal to minus the square root of 3/2 over 1/2. That is, minus the square root of 3. Exactly what we wanted. Therefore, the arctangent of minus the square root of 3 is minus pi over 3. OK. And now, let us consider the second point of our exercise. We want to compute the arctangent of the sine of pi over 2.

Skip to 3 minutes and 24 seconds Good. Let us denote by theta again this value. But the sine of pi over 2 is equal to 1. Therefore, what is theta is the arctangent of 1. OK. Then by the definition of the function arctangent, what we have, that 1 has to be equal to the tangent of theta. But the tangent of theta is, by definition, the sine of theta over the cosine of theta. OK. Therefore, theta has to be a value inside this open interval, where the sine and the cosine coincide. Which is the unique value inside the interval where this happens. We immediately get that theta has to be equal to pi over 4.

Skip to 4 minutes and 21 seconds Indeed, in pi over 4, both the sine and the cosine functions are equal to the square root of 2 over 2. Good. And now, let us consider the third point of our exercise. We have to compute the sine of the arctangent of 2. OK. Let us denote by theta the arctangent of 2. OK. Therefore what we have? We have that the tangent of theta is equal to 2. OK. But, what is the tangent of theta? We know then that the 2 equal to the tangent of theta has to be equal to the sine of theta over the cosine of theta. And now, attention. Theta is the arctangent of something. Therefore, theta lives inside this open interval.

Skip to 5 minutes and 28 seconds And for each angle inside this open interval, the cosine is always a positive number. OK? Then, what we can write here? OK. We’ll leave the sine of theta on the top. And write the square root plus the square root of 1 minus the sine squared of theta. Not only– and also, you see from this equality, what do we get? You see, this is a positive number– the square root. This is a positive number. Then also, this has to be a positive number. Then, not only we have this relation, but also we get, from this relation, that the sine of theta has to be a positive number. Good. And now, from this equality, what do we get?

Skip to 6 minutes and 28 seconds We take the square of this and this. And then we get 4 equal to sine squared of theta over one minus sine squared of theta. That is, multiplying on both sides by 1 minus sine squared, we get 4 minus 4 times sine squared of theta equal to sine squared of theta. That is, 5 times sine squared of theta equal to 4. Which implies sine squared of theta equal to 4/5. And now, because we know that the sine of theta is greater than 0, we can conclude that the sine of theta is equal to 2 over the square root of 5. Good.

Skip to 7 minutes and 36 seconds Then, we have that the sine of theta, that is the sine of the arctangent of 2, that was exactly what we had to compute, is equal to 2 over the square root of 5. Thank you very much for your attention.

# It's your turn on the function arctan

Do your best in trying to **solve the following problems**. In any case some of them are solved in the video and all of them are solved in the pdf file below.

### Exercise 1.

Find the values of the following: \[1.\ \ \arctan(-\sqrt 3);\] \[2.\ \ \arctan(\sin\pi/2);\] \[3.\ \ \sin(\arctan 2).\]

### Exercise 2.

Find the values of the following: \[1.\ \ \arctan(\cos\pi);\] \[2.\ \ \cos(\arctan 2).\]

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