Misc 9 - cos x = -1/3, find sin x/2 , cos x/2 and tan x/2 - Miscellaneous

  1. Chapter 3 Class 11 Trigonometric Functions
  2. Serial order wise
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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° < π‘₯/2 < 135Β° So, π‘₯/2 lies in IInd quadrant In IInd quadrant, sin is positive, cos & tan are negative β‡’ sin π‘₯/2 Positive and cos π‘₯/2 and tan π‘₯/2 negative Given, cos x = βˆ’ 1/3 cos x = 2 cos2 π‘₯/2 – 1 – 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 cos2 π‘₯/2 = 2/3 Γ— 1/2 cos2 π‘₯/2 = 1/3 cos π‘₯/2 = ±√(1/3) = Β± 1/√3 Since π‘₯/2 lie is llnd Quadrant , cos π‘₯/2 is negative So cos π‘₯/2 = βˆ’1/√3 =βˆ’1/√3 Γ— √3/√3 = βˆ’βˆš3/3 We know that sin2x + cos2x = 1 Replacing x with π‘₯/2 sin2 π‘₯/2 + cos2 π‘₯/2 = 1 sin2 π‘₯/2 = 1 – cos2 π‘₯/2 Putting cos π‘₯/2 = βˆ’1/√3 sin2 π‘₯/2 = 1 – ("–" 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) = Β± √2/√3 = Β± √2/√3 Γ— √3/√3 = Β± √(2 Γ— 3)/3 = Β± √6/3 Since π‘₯/2 lie on the llnd Quadrant , sin π‘₯/2 is positive in the llnd Quadrant So sin π‘₯/2 = √6/3 We know that tan x = sin⁑π‘₯/π‘π‘œπ‘ β‘π‘₯ Replacing x with π‘₯/2 tan π‘₯/2 = 𝑠𝑖𝑛⁑〖 π‘₯/2γ€—/γ€–π‘π‘œπ‘  〗⁑〖π‘₯/2γ€— = (√6/3)/(βˆ’ √3/3) = √6/3 Γ— (βˆ’ 3)/√3 = – √6/√3 = – √(6/3) = – √2 Hence, tan π‘₯/2 = – √2 Hence, tan π‘₯/2 = – √2 , cos π‘₯/2 = βˆ’βˆš3/3 & sin π‘₯/2 = √6/3

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