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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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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 : tan⁑π‘₯ = – 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° < 𝒙/𝟐 < 90Β° So, 𝒙/𝟐 lies in Ist quadrant In 1st quadrant, sin , cos & tan are positive ∴ sin π‘₯/2 , cos π‘₯/2 and tan π‘₯/2 are positive Given tan x = (βˆ’4)/3 We know that tan 2x = (𝟐 𝒕𝒂𝒏⁑𝒙)/(𝟏 βˆ’ π’•π’‚π’πŸπ’™) Replacing x with π‘₯/2 tan (2π‘₯/2) = (2 π‘‘π‘Žπ‘›β‘(π‘₯/2))/(1 βˆ’ π‘‘π‘Žπ‘›2(π‘₯/2) ) tan x = (2 π‘‘π‘Žπ‘›β‘(π‘₯/2))/(1 βˆ’ π‘‘π‘Žπ‘›2(π‘₯/2) ) (𝟐 𝒕𝒂𝒏⁑(𝒙/𝟐))/(𝟏 βˆ’ π’•π’‚π’πŸ(𝒙/𝟐) ) = βˆ’πŸ’/πŸ‘ βˆ’4/3 = (2 tan⁑(π‘₯/2))/(1 βˆ’ π‘‘π‘Žπ‘›2(π‘₯/2) ) –4(2π‘₯/2) = 3Γ— 2 tan (π‘₯/2) –4 Γ— 1 – (–4) Γ— tan2 (π‘₯/2) = 6 tan (π‘₯/2) –4 Γ— 1 – (–4) Γ— tan2 (π‘₯/2) = 6 tan (π‘₯/2) –4 + 4 tan2 (π‘₯/2) = 6 tan (π‘₯/2) –4 + 4 tan2 (𝒙/𝟐) – 6 tan (𝒙/𝟐) = 0 Replacing tan 𝒙/𝟐 by a Our equation becomes –4 + 4a2 – 6a = 0 4a2 – 6a – 4 = 0 4a2 – 8a + 2a – 4 = 0 4a(a – 2) + 2 (a – 2) = 0 (4a + 2) (a – 2) = 0 Hence 4a + 2 = 0 4a = βˆ’2 a = (βˆ’2)/( 4) a = (βˆ’πŸ)/𝟐 a – 2 = 0 a = 2 So, a = (βˆ’1)/2 or a = 2 Hence, tan π‘₯/2 = (βˆ’1)/2 or tan π‘₯/2 = 2 Since, π‘₯/2 lies in Ist quadrant tan 𝒙/𝟐 is positive, ∴ tan 𝒙/𝟐 = 2 Now, We know that 1 + tan2 x = sec2 x Replacing x with π‘₯/2 1 + tan2 𝒙/𝟐 = sec2 𝒙/𝟐 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 𝒙/𝟐 = Β± βˆšπŸ“ Since π‘₯/2 lie on the 1st Quadrant, sec 𝒙/𝟐 is positive in the 1st Quadrant So, sec 𝒙/𝟐 = βˆšπŸ“ Therefore, cos 𝒙/𝟐 = 𝟏/βˆšπŸ“ Now, We know that sin2x + cos2x = 1 Replacing x with π‘₯/2 sin2 𝒙/𝟐 + cos2 𝒙/𝟐 = 1 sin2 π‘₯/2 = 1 – cos2 π‘₯/2 Putting cos π‘₯/2 = √5/5 sin2 π‘₯/2 = 1 – (√5/5)2 sin2 π‘₯/2 = 1 – 5/25 sin2 π‘₯/2 = 1 – 1/5 sin2 π‘₯/2 = (5 βˆ’ 1)/5 sin2 π‘₯/2 = 4/5 sin π‘₯/2 = Β± √(4/5) sin π‘₯/2 = Β± √4/√5 sin π‘₯/2 = Β± 2/√5 sin π‘₯/2 = Β± 2/√5 Γ— √5/√5 sin 𝒙/𝟐 = Β± (πŸβˆšπŸ“)/πŸ“ Since π‘₯/2 lies on the 1st Quadrant sin 𝒙/𝟐 is positive in the 1st Quadrant So, sin 𝒙/𝟐 = (πŸβˆšπŸ“)/πŸ“ Therefore, tan π‘₯/2 = 2 , cos 𝒙/𝟐 = βˆšπŸ“/πŸ“ & sin 𝒙/𝟐 = (πŸβˆšπŸ“)/πŸ“

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.