Miscellaneous

Chapter 3 Class 11 Trigonometric Functions
Serial order wise

### Transcript

Misc 8 Find the value of sin π₯/2 , cos π₯/2 and tan π₯/2 in each of the following : tanβ‘π₯ = β 4/3 , π₯ in quadrant II Given that x is in Quadrant II So, 90Β° < x < 180Β° Dividing by 2 all sides (90Β°)/2 < π₯/2 < (180Β°)/2 45Β° < π/π < 90Β° So, π/π lies in Ist quadrant In 1st quadrant, sin , cos & tan are positive β΄ sin π₯/2 , cos π₯/2 and tan π₯/2 are positive Given tan x = (β4)/3 We know that tan 2x = (π πππβ‘π)/(π β πππππ) Replacing x with π₯/2 tan (2π₯/2) = (2 π‘ππβ‘(π₯/2))/(1 β π‘ππ2(π₯/2) ) tan x = (2 π‘ππβ‘(π₯/2))/(1 β π‘ππ2(π₯/2) ) (π πππβ‘(π/π))/(π β ππππ(π/π) ) = βπ/π β4/3 = (2 tanβ‘(π₯/2))/(1 β π‘ππ2(π₯/2) ) β4(2π₯/2) = 3Γ 2 tan (π₯/2) β4 Γ 1 β (β4) Γ tan2 (π₯/2) = 6 tan (π₯/2) β4 Γ 1 β (β4) Γ tan2 (π₯/2) = 6 tan (π₯/2) β4 + 4 tan2 (π₯/2) = 6 tan (π₯/2) β4 + 4 tan2 (π/π) β 6 tan (π/π) = 0 Replacing tan π/π by a Our equation becomes β4 + 4a2 β 6a = 0 4a2 β 6a β 4 = 0 4a2 β 8a + 2a β 4 = 0 4a(a β 2) + 2 (a β 2) = 0 (4a + 2) (a β 2) = 0 Hence 4a + 2 = 0 4a = β2 a = (β2)/( 4) a = (βπ)/π a β 2 = 0 a = 2 So, a = (β1)/2 or a = 2 Hence, tan π₯/2 = (β1)/2 or tan π₯/2 = 2 Since, π₯/2 lies in Ist quadrant tan π/π is positive, β΄ tan π/π = 2 Now, We know that 1 + tan2 x = sec2 x Replacing x with π₯/2 1 + tan2 π/π = sec2 π/π 1 + (2)2 = sec2 π₯/2 1 + 4 = sec2 x/2 1 + 4 = sec2 x/2 5 = sec2 π₯/2 sec2 π₯/2 = 5 sec π/π = Β± βπ Since π₯/2 lie on the 1st Quadrant, sec π/π is positive in the 1st Quadrant So, sec π/π = βπ Therefore, cos π/π = π/βπ Now, We know that sin2x + cos2x = 1 Replacing x with π₯/2 sin2 π/π + cos2 π/π = 1 sin2 π₯/2 = 1 β cos2 π₯/2 Putting cos π₯/2 = β5/5 sin2 π₯/2 = 1 β (β5/5)2 sin2 π₯/2 = 1 β 5/25 sin2 π₯/2 = 1 β 1/5 sin2 π₯/2 = (5 β 1)/5 sin2 π₯/2 = 4/5 sin π₯/2 = Β± β(4/5) sin π₯/2 = Β± β4/β5 sin π₯/2 = Β± 2/β5 sin π₯/2 = Β± 2/β5 Γ β5/β5 sin π/π = Β± (πβπ)/π Since π₯/2 lies on the 1st Quadrant sin π/π is positive in the 1st Quadrant So, sin π/π = (πβπ)/π Therefore, tan π₯/2 = 2 , cos π/π = βπ/π & sin π/π = (πβπ)/π