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Quick Trick — Evaluate ∑θ=0811+tan⁡3(10θ)∘\displaystyle\sum_{\theta=0}^{8}\frac{1}{1+\tan^3(10\theta)^\circ}θ=0∑81+tan3(10θ)∘1 (MCQ)
Description:
Can you evaluate this sum fast using a trig identity?
We’re asked to find
∑θ=0811+tan⁡3(10θ)∘,\sum_{\theta=0}^{8}\frac{1}{1+\tan^{3}(10\theta)^\circ},θ=0∑81+tan3(10θ)∘1,
where θ=0,1,2,…,8\theta=0,1,2,\dots,8θ=0,1,2,…,8 (i.e. angles 0∘,10∘,…,80∘0^\circ,10^\circ,\dots,80^\circ0∘,10∘,…,80∘). Multiple choice options: (a) 555, (b) 214\tfrac{21}{4}421, (c) 143\tfrac{14}{3}314, (d) 92\tfrac{9}{2}29.
Quick hint / idea:
Pair each term at angle xxx with the term at its complementary angle 90∘−x90^\circ-x90∘−x. Let t=tan⁡xt=\tan xt=tanx. Since tan⁡(90∘−x)=cot⁡x=1t\tan(90^\circ-x)=\cot x=\tfrac{1}{t}tan(90∘−x)=cotx=t1, check
11+t3+11+(1/t)3=11+t3+t31+t3=1.\frac{1}{1+t^3}+\frac{1}{1+(1/t)^3}=\frac{1}{1+t^3}+\frac{t^3}{1+t^3}=1.1+t31+1+(1/t)31=1+t31+1+t3t3=1.
So each complementary pair (10°+80°, 20°+70°, 30°+60°, 40°+50°) contributes 1. Plus the θ=0\theta=0θ=0 term 11+tan⁡30∘=1\frac{1}{1+\tan^3 0^\circ}=11+tan30∘1=1. Total =4+1=5=4+1=5=4+1=5.
Answer: (a) 555.
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