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?