Trigonometry & Inverse Trigonometry
General
Grade 12
Question:
<p>If \(\sin(2\cos^{-1}\frac{1}{\sqrt{5}})+\cos(2\tan^{-1}\frac{1}{3})=\frac{p}{q}\) (coprime), units digit of \((p-q)^{2k+1}\), \(k\in\mathbb{N}\) can be:</p>
1
<strong>3</strong>
<strong>7</strong>
9
Step-by-Step Solution
<div class="solution"><p><strong>Step 1:</strong> $\cos^{-1}(1/\sqrt{5})=\theta\implies\sin\theta=2/\sqrt{5}$. $\sin 2\theta=2\cdot(2/\sqrt{5})\cdot(1/\sqrt{5})=4/5$.</p><p><strong>Step 2:</strong> $\tan^{-1}(1/3)=\alpha$. $\cos 2\alpha=\frac{1-1/9}{1+1/9}=4/5$.</p><p><strong>Step 3:</strong> Sum $=4/5+4/5=8/5$. So $p=8,q=5,p-q=3$.</p><p><strong>Step 4:</strong> Units digit of $3^{2k+1}$: odd powers of 3 cycle 3,7,3,7,... \to units digits 3 or 7.</p><p><strong>Answer: (B),(C)</strong></p><div class="trap-box"><strong>Trap:</strong> Odd powers of 3 alternate between units digits 3 and 7.<div class="key-concept"><strong>Key Concept:</strong> Double-angle formulas + cyclicity of units digits of powers
Correct Answer: B,C