Miscellaneous

Chapter 3 Class 11 Trigonometric Functions
Serial order wise

### Transcript

Misc 10 Find sin π₯/2, cos π₯/2 and tan π₯/2 for sinβ‘π₯ = 1/4 , π₯ in quadrant II Given that x is in Quadrant II So, 90Β° < x < 180Β° Replacing x with π₯/2 (90Β°)/2 < π₯/2 < (180Β°)/2 45Β° < π/π < 90Β° So, π/π lies in Ist quadrant In Ist quadrant, sin , cos & tan are positive sin π₯/2 , cos π₯/2 and tan π₯/2 are positive Given sin x = 1/4 2 sin π/π cos π/π = π/π sin π₯/2 cos π₯/2 = 1/4 Γ 1/2 sin π/π cos π/π = π/π We are given sin x , lets first calculate cos x We know that sin2 x + cos2 x = 1 (1/4)^2 + cos2 x = 1 1/16 + cos2 x = 1 cos2 x = 1 β 1/16 cos2 x = (16 β 1)/16 cos2 x = 15/16 cos x = Β±β(15/16) cos x = Β± βππ/π Given that x is in llnd Quadrant So cos x is negative β΄ cos x = (ββππ)/π We know that cos 2x = 2cos2 x β 1 Replacing x by π₯/2 cos x = 2 cos2 π₯/2 β 1 2cos2 π₯/2 β 1 = ββ15/4 2cos2 π₯/2 = (4 β β15)/4 cos2 π₯/2 = (4 β β15)/4 Γ 1/2 cos2 π₯/2 = (4 β β15)/(4 Γ 2) cos2 π₯/2 = (4 β β15)/8 cos π₯/2 = Β± β((4 β β15)/8) = Β± β((4 β β15)/8Γ 2/2) = Β± β((2(4 β β15))/16) = Β± β((8 β2β15))/4 Since π₯/2 lie on the lst Quadrant, cos π/π is positive in the lst Quadrant So, cos π₯/2 = β((8 β2β15))/4 So, cos π/π = β((π βπβππ))/π Also, from (1) sin π₯/2 cos π₯/2 = 1/8 Putting value of cos π₯/2 "sin " π₯/2 Γ β((8 β2β15))/4 = 1/8 "sin " π₯/2 = 1/8 Γ 4/β((8 β2β15)) "sin " π₯/2 = 1/2 Γ 1/β((8 β2β15)) "sin " π₯/2 = 1/2 Γ 1/β((8 β2β15) Γ ((8 + 2β15))/((8 + 2β15) )) "sin " π₯/2 = 1/2 Γ 1/β( ((8 β2β15) Γ (8 + 2β15))/((8 + 2β15) )) "sin " π₯/2 Γ β((8 β2β15))/4 = 1/8 "sin " π₯/2 = 1/8 Γ 4/β((8 β2β15)) "sin " π/π = 1/2 Γ 1/β((8 β2β15)) = 1/2 Γ β(((8 + 2β15))/((64 β (2)2 (β15)2 )) = 1/2 Γ β(((8 + 2β15))/((64 β 4 Γ 15) )) = 1/2 Γ β(((8 + 2β15))/((64 β 4 Γ 15) )) = 1/2 Γ β(((8 + 2β15))/((64 β 60) )) = 1/2 Γ β(((8 + 2β15))/4) = 1/2 Γ β((8 + 2β15) )/β4 = 1/2 Γ β((8 + 2β15) )/2 = β((8 + 2β15) )/4 β΄ sin π/π = β((π + πβππ) )/π We know that tan x = sinβ‘π₯/πππ β‘π₯ Replacing x with π₯/2 tan π/π = πππβ‘γ π/πγ/γπππ γβ‘γπ/πγ = (β((8 + 2β15) )/4)/(β((8 β2β15))/4) = β((8 + 2β15) )/β((8 β 2β15) ) = β(((8 + 2β15))/((8 β 2β15) )Γ((8 + 2β15))/((8 + 2β15) )) = β((8 + 2β15)2/(8 β 2β15)(8 + 2β15) ) = β(((8 + 2β15))/((8 β 2β15) )Γ((8 + 2β15))/((8 + 2β15) )) = β((8 + 2β15)2/(8 β 2β15)(8 + 2β15) ) = β((8 + 2β15)2/((82 β(2β15)2) )) = β((8 + 2β15)2/((64 β (4 Γ 15)) )) = β((8 + 2β15)2/((64 β 60) )) = β((8 + 2β15)2/4) = β((8 + 2β15)2/22) = β((((8 + 2β15))/2)^2 ) = β((((8 + 2β15))/2)^2 ) = ((π + πβππ))/π = 2(4 + β15)/2 \ Hence, tan π/π = (π + βππ)