Complete the following exercises before our next class (after the break).
Exercise Set 13A.3 questions 6bdij and 12
Also, have a look at the question below (which was discussed in class). It was mentioned in class that a solution can be found (to both parts!) without a calculator. Can you figure it out over the break?
\arctan\left(\frac{1}{2}\right)-\arctan\left(\frac{1}{3}\right) = \arctan(a), a \in \mathbb{Q}^+
- Find the value of a.
- Hence, or otherwise, solve the equation \arcsin (x)=\arctan(a).
I’ll post a hint in the comments below [update: one error in that comment has been corrected, and the hint has been extended]—you may find it useful.
Let x=\arctan\left(\frac{1}{2}\right). Then \tan(x)=\frac{1}{2}. Since the range of the arctangent function is ]-\frac{\pi}{2}, \frac{\pi}{2}[, and the input is positive, we also know that 0 < x < \frac{\pi}{2}. So, it's appropriate to consider a right triangle with angle x. Such a triangle can be constructed with an opposite side length of 1 and an adjacent side length of 2; similarly for y=\arctan\left(\frac{1}{3}\right).
Also, Sam suggested an even more direct solution that makes use of the compound angle formula for the tangent function. Can you also find a solution that uses that formula?