Problem Statement
(a) Show that:
- (i) $ \sin 2\theta = 2\sin\theta\cos\theta $
- (ii) $ \cos 2\theta = 2\cos^2\theta - 1 $
Given that $ \theta \neq (90^\circ + 180^\circ n) $, where $ n \in \mathbb{Z} $.
(b) Use the results from part (a) to show that $ \sin 2\theta - \tan\theta $ can be written as $ \tan\theta\cos 2\theta $.
(c) Solve for $ 0^\circ < x < 360^\circ $ the equation:
$$ \sin 2x^\circ - \tan x^\circ = 0 $$
Solution
Part (a)
(i) Proof of $ \sin 2\theta = 2\sin\theta\cos\theta $
Using the double-angle identity for sine:
$$ \sin 2\theta = \sin(\theta + \theta) = \sin\theta\cos\theta + \cos\theta\sin\theta = 2\sin\theta\cos\theta $$
(ii) Proof of $ \cos 2\theta = 2\cos^2\theta - 1 $
Using the double-angle identity for cosine:
$$ \cos 2\theta = \cos^2\theta - \sin^2\theta $$
Substitute $ \sin^2\theta = 1 - \cos^2\theta $:
$$ \cos 2\theta = \cos^2\theta - (1 - \cos^2\theta) = 2\cos^2\theta - 1 $$
Part (b)
Starting with:
$$ \sin 2\theta - \tan\theta $$
Substitute $ \sin 2\theta = 2\sin\theta\cos\theta $ and $ \tan\theta = \frac{\sin\theta}{\cos\theta} $:
$$ \sin 2\theta - \tan\theta = 2\sin\theta\cos\theta - \frac{\sin\theta}{\cos\theta} $$
Combine terms over a common denominator:
$$ \sin 2\theta - \tan\theta = \frac{2\sin\theta\cos^2\theta - \sin\theta}{\cos\theta} $$
Factor out $ \sin\theta $:
$$ \sin 2\theta - \tan\theta = \frac{\sin\theta(2\cos^2\theta - 1)}{\cos\theta} $$
Using $ \cos 2\theta = 2\cos^2\theta - 1 $, we have:
$$ \sin 2\theta - \tan\theta = \frac{\sin\theta\cos 2\theta}{\cos\theta} $$
Simplify:
$$ \sin 2\theta - \tan\theta = \tan\theta\cos 2\theta $$
Part (c)
Solve the equation:
$$ \sin 2x^\circ - \tan x^\circ = 0 $$
Using part (b):
$$ \tan x^\circ \cos 2x^\circ = 0 $$
This implies:
$$ \tan x^\circ = 0 \quad \text{or} \quad \cos 2x^\circ = 0 $$
Case 1: $ \tan x^\circ = 0 $
$$ \tan x^\circ = 0 \implies x = 0^\circ,\ 180^\circ $$
Since $ 0^\circ < x < 360^\circ $, the solution is:
$$ x = 180^\circ $$
Case 2: $ \cos 2x^\circ = 0 $
$$ \cos 2x^\circ = 0 \implies 2x = 90^\circ + 180^\circ n \quad (n \in \mathbb{Z}) $$
$$ 2x = 90^\circ,\ 270^\circ,\ 450^\circ,\ 630^\circ \implies x = 45^\circ,\ 135^\circ,\ 225^\circ,\ 315^\circ $$
Final Solution
The solutions for $ 0^\circ < x < 360^\circ $ are:
$$ \boxed{x = 45^\circ,\ 135^\circ,\ 180^\circ,\ 225^\circ,\ 315^\circ} $$
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