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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Misc 9 Find sin π‘₯/2, cos π‘₯/2 and tan π‘₯/2 for cos π‘₯ = βˆ’ 1/3 , π‘₯ in quadrant III Since x is in quadrant III 180Β° < x < 270Β° Dividing by 2 all sides (180Β°)/2 < π‘₯/2 < (270Β°)/2 90Β° < 𝒙/𝟐 < 135Β° So, π‘₯/2 lies in IInd quadrant In IInd quadrant, sin is positive, cos & tan are negative sin π‘₯/2 Positive and cos π‘₯/2 and tan π‘₯/2 negative Given, cos x = βˆ’1/3 2 cos2 𝒙/𝟐 – 1 = βˆ’πŸ/πŸ‘ In IInd quadrant, sin is positive, cos & tan are negative sin π‘₯/2 Positive and cos π‘₯/2 and tan π‘₯/2 negative Given, cos x = βˆ’1/3 2 cos2 𝒙/𝟐 – 1 = βˆ’πŸ/πŸ‘ – 1/3 = 2cos2 π‘₯/2 – 1 1 – 1/3 = 2cos2 π‘₯/2 (3 βˆ’ 1)/3 = 2cos2 π‘₯/2 2/3 = 2cos2 π‘₯/2 2cos2 π‘₯/2 = 2/3 2cos2 π‘₯/2 = 2/3 cos2 π‘₯/2 = 2/3 Γ— 1/2 cos2 π‘₯/2 = 1/3 cos π‘₯/2 = ±√(1/3) cos π‘₯/2 = Β± 1/√3 cos π‘₯/2 = Β± 1/√3 Γ— √3/√3 cos 𝒙/𝟐 = Β± βˆšπŸ‘/πŸ‘ Since π‘₯/2 lies is llnd Quadrant , cos 𝒙/𝟐 is negative So, cos 𝒙/𝟐 = (βˆ’βˆšπŸ‘)/πŸ‘ We know that sin2x + cos2x = 1 Replacing x with π‘₯/2 sin2 𝒙/𝟐 + cos2 𝒙/𝟐 = 1 sin2 π‘₯/2 = 1 – cos2 π‘₯/2 Putting cos π‘₯/2 = (βˆ’1)/√3 sin2 π‘₯/2 = 1 – ((βˆ’1)/√3)2 sin2 π‘₯/2 = 1 – 1/3 sin2 π‘₯/2 = (3 βˆ’ 1)/3 sin2 π‘₯/2 = 2/3 sin π‘₯/2 = Β± √(2/3) sin π‘₯/2 = Β± √2/√3 Γ— √3/√3 sin π‘₯/2 = Β± √(2 Γ— 3)/3 sin 𝒙/𝟐 = Β± βˆšπŸ”/πŸ‘ Since π‘₯/2 lie on the llnd Quadrant, sin 𝒙/𝟐 is positive in the llnd Quadrant So, sin 𝒙/𝟐 = βˆšπŸ”/πŸ‘ We know that tan x = sin⁑π‘₯/π‘π‘œπ‘ β‘π‘₯ Replacing x with π‘₯/2 tan 𝒙/𝟐 = π’”π’Šπ’β‘γ€– 𝒙/πŸγ€—/〖𝒄𝒐𝒔 〗⁑〖𝒙/πŸγ€— tan π‘₯/2 = (√6/3)/(βˆ’ √3/3) = √6/3 Γ— (βˆ’ 3)/√3 = – √6/√3 = – √(6/3) = – √2 Hence, tan 𝒙/𝟐 = – √𝟐 Therefore, tan π‘₯/2 = – √2 , cos π‘₯/2 = βˆ’βˆš3/3 & sin π‘₯/2 = √6/3

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.