It’s your turn on More inverse trig functions, and a few others

Welcome to “It’s your turn on More inverse trigonometric functions”. OK. This exercise is dedicated to the functions arcsin and arccosine. Let us start reminding where they are defined and which is their range. OK. First, the function arcsine is defined on the closed interval minus 1, 1, and its range is the closed interval minus pi over 2, pi over 2. And the arccosine function is defined, again, in the closed interval minus 1, 1, but its range is the closed interval 0, pi. OK? Good. Both of these functions are bijective. That is, they are both injective and surjective functions. OK? Good. With this in mind, let us start to solve the points of our exercise. Let us start with the point 1.

We have to compute the arcsine of 1.

OK. Let us denote by theta this value. What means that theta is the arcsine of 1? This means that the sine of theta is equal to 1 and theta belongs to this interval. OK? Then which is the unique angle inside this interval where the sine is equal to 1? Clearly it’s theta equal to pi over 2. Good. And now let us move to the second point of this exercise. We have to compute the arcsine of the sine of 3 pi over 4. Good. Again, let us denote by theta the arcsine of the sine of 3 pi over 4. What means that the theta is equal to this value?

This means, again, that the sine of theta is equal to the sine of 3 pi over 4, but also that theta is to belong to this interval. You see? For this reason, we cannot declare that theta is equal to 3 pi over 4, because 3 pi over 4 is not in this interval. OK. But which is the sine in 3 pi over 4? 3 pi over 4 is pi minus pi over 4. Then this sine is equal to the sine of pi over 4 and the pi over 4 belongs to this interval. Then we immediately get that theta, exactly what we were looking for, is equal to pi over 4. Good.

And now let us consider the third point of our exercise. We have to compute the cosine of the arcsine of 3 over 5. OK. Let us denote by theta the arcsine of 3 over 5. OK. What means? Means that the sine of theta is equal to 3 over 5 not only, and that theta belongs to this interval. And now look, the cosine of all the angles in this interval are greater or equal than 0. OK. Then usually the cosine of an angle is plus or minus the square root of 1 minus sine squared of the same angle. But because theta belongs to this interval, I repeat, the cosine is positive. And then what do we get?

That the cosine of theta is equal to plus the square root of 1 minus sine squared of theta. That is the square root of 1 minus 9 over 25, which is equal to the square root of 16 over 25. That is 4 over 5. Then the cosine of theta, which is the cosine of the arcsine of 3 over 5, exactly what we wanted to compute, is equal to 4 over 5. And now let us finish our exercise with this fourth point. What do we have? OK, we want to compute the tangent of the arccosine of minus 2 over 3. Good. Let us denote by theta now the arccosine of minus 2 over 3. OK. What means?

Means that the cosine of theta is equal to minus 2 over 3– not only– and that theta belong to this interval, 0 pi. OK. Now observe that the sine of all the angles inside this interval is always greater or equal than 0. Then what do we get? We get that the sine of theta is a number greater or equal than 0. And then it’s equal to plus the square root of 1 minus the square of the cosine of theta. That is the square root of 1 minus 4 over 9, which is equal to the square root of 5 over 3. OK. Now we have the cosine of theta. We have the sine of theta.

We want to compute the tangent of this, which is the tangent of theta. OK. We are very close to the end of this point. Indeed, the tangent of theta, which is, by definition, the sine of theta over the cosine of theta, is equal to the square root of 5 over 3 over minus 2 over 3. Therefore it is equal to minus the square root of 5 over 2. OK. [IN ITALIAN] Goodbye to everyone