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₯ = β 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 = β 1/3 cos x = 2 cos2 π₯/2 β 1 β 1/3 = 2cos2 π₯/2 β 1 1 β 1/3 = 2cos2 π₯/2 (3 β 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 = Β±β(1/3) = Β± 1/β3 Since π₯/2 lie is llnd Quadrant , cos π₯/2 is negative So cos π₯/2 = β1/β3 =β1/β3 Γ β3/β3 = ββ3/3 We know that sin2x + cos2x = 1 Replacing x with π₯/2 sin2 π₯/2 + cos2 π₯/2 = 1 sin2 π₯/2 = 1 β cos2 π₯/2 Putting cos π₯/2 = β1/β3 sin2 π₯/2 = 1 β ("β" 1/β3)2 sin2 π₯/2 = 1 β 1/3 sin2 π₯/2 = (3 β 1)/3 sin2 π₯/2 = 2/3 sin π₯/2 = Β± β(2/3) = Β± β2/β3 = Β± β2/β3 Γ β3/β3 = Β± β(2 Γ 3)/3 = Β± β6/3 Since π₯/2 lie on the llnd Quadrant , sin π₯/2 is positive in the llnd Quadrant So sin π₯/2 = β6/3 We know that tan x = sinβ‘π₯/πππ β‘π₯ Replacing x with π₯/2 tan π₯/2 = π ππβ‘γ π₯/2γ/γπππ  γβ‘γπ₯/2γ = (β6/3)/(β β3/3) = β6/3 Γ (β 3)/β3 = β β6/β3 = β β(6/3) = β β2 Hence, tan π₯/2 = β β2 Hence, tan π₯/2 = β β2 , cos π₯/2 = ββ3/3 & sin π₯/2 = β6/3