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Question 14
12. (a) Prove \[ \frac{\cos 3\theta}{\sin \theta} + \frac{\sin 3\theta}{\cos \theta} = 2 \cot 2\theta \] \( \theta \pm (90n)^\circ, n \in \mathbb{Z} \) (b) Henc... show full transcript
Step 1
Answer
To prove the equation, we start with the left-hand side (LHS):
[ \frac{\cos 3\theta}{\sin \theta} + \frac{\sin 3\theta}{\cos \theta} = \frac{\cos 3\theta \cos \theta + \sin 3\theta \sin \theta}{\sin \theta \cos \theta} ]
Using the angle addition formula, we simplify:
[ \text{LHS} = \frac{\cos(3\theta - \theta)}{\sin \theta \cos \theta} = \frac{\cos 2\theta}{\sin \theta \cos \theta} ]
We know that ( \cot 2\theta = \frac{\cos 2\theta}{\sin 2\theta} ), which leads us to:
[ \frac{\cos 2\theta}{\sin \theta \cos \theta} \cdot \frac{2}{2} = 2 \cot 2\theta ]
Thus, confirming the given identity.
Step 2
Answer
From the earlier derived formula:
[ \frac{\cos 3\theta}{\sin \theta} + \frac{\sin 3\theta}{\cos \theta} = 2 \cot 2\theta = 4 ]
We can simplify this to:
[ 2 \cot 2\theta = 4 \implies \cot 2\theta = 2 ]
Taking the cotangent inverse:
[ 2\theta = \arctan\left(\frac{1}{2}\right) ]
Thus:
[ \theta = \frac{1}{2} \arctan\left(\frac{1}{2}\right) + n(90)^\circ ]
Calculating the angle within ( 90^\circ < \theta < 180^\circ ):
The solution is: ( \theta \approx 103.3^\circ ).
Thus, the only solution to one decimal place is ( \theta = 103.3^\circ ).
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