Last updated at Dec. 8, 2016 by Teachoo

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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 with 2 all sides (90Β°)/2 < π₯/2 < (180Β°)/2 45Β° < π₯/2 < 90Β° So, π₯/2 lies in Ist quadrant In Ist quadrant, sin , cos & tan are positive β sin π₯/2 , cos π₯/2 and tan π₯/2 are positive Given tan x = β4/3 tan x = (2 tanβ‘(π₯/2))/(1 β π‘ππ2(π₯/2) ) β4/3 = (2 tanβ‘(π₯/2))/(1 β π‘ππ2(π₯/2) ) β4( 1 β tan2 (π₯/2)) = 3Γ 2 tan (π₯/2) β4 Γ 1 β (β4) Γ tan2 (π₯/2) = 6 tan (π₯/2) β4 + 4 tan2 (π₯/2) = 6 tan (π₯/2) β4 + 4 tan2 (π₯/2) β 6 tan (π₯/2) = 0 Replacing tan π₯/2 by a 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 = - 2 a = (β2)/( 4) a = β1/2 So, a = β1/2 or a = 2 Hence, tan π₯/2 = β1/2 or tan π₯/2 = 2 Since, π₯/2 lies in Ist quadrant tan π₯/2 is positive, β΄ tan π₯/2 = 2 Now, We know that 1 + tan2 x = sec2 x Replacing x with π₯/2 1 + tan2 π₯/2 = sec2 π₯/2 1 + (2)2 = sec2 π₯/2 1 + 4 = sec2 x/2 5 = sec2 π₯/2 sec2 π₯/2 = 5 sec π₯/2 = Β± β5 Since π₯/2 lie on the 1st Quadrant, sec π₯/2 is positive in the 1st Quadrant So sec π₯/2 = β5 sec π₯/2 = β5 β 1/cosβ‘γ π₯/2γ = β5 β 1/β5 = cos π₯/2 β cos π₯/2 = 1/β5 cos π₯/2 = 1/β5 Γ β5/β5 cos π₯/2 = β5/5 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 = β5/5 sin2 π₯/2 = 1 β (β5/5)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) = Β± β4/β5 = Β± 2/β5 = Β± 2/β5 Γ β5/β5 = Β± (2β5)/5 Since π₯/2 lie on the 1st Quadrant , sin π₯/2 is positive in the 1st Quadrant So sin π₯/2 = (2β5)/5 Hence, tan π₯/2 = 2 , cos π₯/2 = β5/5 & sin π₯/2 = (2β5)/5

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.