Trigonometry & Inverse Trigonometry Question - None

Question

Ex. 57: Which of the following statements are always correct? (where Q denotes the set of rationals)

(a) $\cos 2\theta \in \mathbb{Q}$ and $\sin 2\theta \in \mathbb{Q}$ $\Rightarrow$ $\tan\theta \in \mathbb{Q}$ (if defined)

(b) $\tan\theta \in \mathbb{Q}$ $\Rightarrow$ $\sin 2\theta, \cos 2\theta$ and $\tan 2\theta \in \mathbb{Q}$ (if defined)

(c) If $\sin\theta \in \mathbb{Q}$ and $\cos\theta \in \mathbb{Q}$ $\Rightarrow$ $\tan 3\theta \in \mathbb{Q}$ (if defined)

(d) If $\sin\theta \in \mathbb{Q}$ $\Rightarrow$ $\cos 3\theta \in \mathbb{Q}$

Accepted Answer

Step 1: Check (a): $ an heta = \frac{1-\cos 2 heta}{\sin 2 heta}$. If both $\cos 2 heta$ and $\sin 2 heta$ are rational, then $ an heta$ is rational. ✓

Step 2: Check (b): $\sin 2 heta = \frac{2 an heta}{1+ an^2 heta}$ and $\cos 2 heta = \frac{1- an^2 heta}{1+ an^2 heta}$ and $ an 2 heta = \frac{2 an heta}{1- an^2 heta}$. If $ an heta \in \mathbb{Q}$, all these are rational. ✓

Step 3: Check (c): $ an 3 heta = \frac{\sin 3 heta}{\cos 3 heta}$. Since both $\sin heta$ and $\cos heta$ are rational, $ an 3 heta$ is rational. ✓

Step 4: Check (d): $\cos 3 heta = \cos heta(4\cos^2 heta - 3)$. Even if $\sin heta$ is rational, $\cos heta$ may be irrational, making $\cos 3 heta$ irrational. ✗

∴ Answer is (a, b, c).

Question:

Step-by-Step Solution

Master Trigonometry & Inverse Trigonometry with Mathbee