A Nice Trigonometric Problem | Math Olympiad
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- Опубликовано: 15 окт 2024
- A Nice Trigonometric Problem | Math Olympiad
Let's explore the world of trigonometry with this nice trigonometric problem from Math Olympiad! In this video, evaluating a challenging trigonometric expression, offering detailed explanations and strategies to simplify it. Perfect for students gearing up for math competitions or anyone who loves a challenging math puzzle. Don't miss out-like, subscribe, and hit the bell icon for more Math Olympiad problems and solutions!
Topics Covered:
Math Olympiad
Trigonometry
Expressions
Trigonometric identities
Math Olympiad
Simplification
Math Olympiad Preparation
Math Tutorial
Trigonometric expression
How to Evaluate Trig Expression
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Let the given expression =E and x = pi/16. Then 5pi/16 = pi/2-3x and 7pi/x = pi/2-x. So, E = tan^x + cot^x + tan^2(3x) + cot^2(3x) = sec^2x + cosec^2x + sec^2(3x) + cosec^2(3x) -4 = 4[cosec^2(2x) + cosec^2(6x) -1] . But cosec^2(6x) = sec^(pi/8) > E = 4[sec^(2x)+cosec^(2x) + 1] = 4[4cosec^2(4x)-1] = 4[4/(1/2) - 1] = 28. So, E=28.
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{tan^2+tan^2 ➖ }=tan^4 pi+pi ➖/16+16 ➖}=tan^4pi^2/32 {tan^2+tan^2 ➖ }=tan^2 {3pi+3pi ➖ /16+16 ➖ }= {tan^4+9pi^2/32 }=tan^49pi^2/32 {tan^4pi^2/32+tan^49pi^2/32}=:tan^89pi^4/64 {tan^2+tan^2 ➖ }=tan^4{ 5pi+5pi ➖/16+16 ➖}= 10pi^2/32.{tan^4+10pi/32}=tan^410pi^2/32 {tan^89pi^4/64+tan^410pi^2/32}= tan^12 19pi^6/96 { tan^2+tan^2 ➖ }=tan^4{ 7pi+7pi ➖ }/{16+16 ➖ }= {tan^4+14pi^2/32}= tan^414pi^2/32 tan^1219pi^6/96+tan^414pi^2/32}= tan^1633pi^8/128 tan^4^43^11pi^2^3/10^10^7^4 tan^2^2^2^2 3^1^1p^2^3/2^5^2^57^12^2,1^11^13^1^1pi1^11^1^11^11^2 32 ( tanPi/ ➖ 3tanpi/+2)
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