Check sibling questions

Misc 8 - tan x = -4/3, find sin x/2 , cos x/2 and tan x/2

Misc 8 - Chapter 3 Class 11 Trigonometric Functions - Part 2
Misc 8 - Chapter 3 Class 11 Trigonometric Functions - Part 3 Misc 8 - Chapter 3 Class 11 Trigonometric Functions - Part 4 Misc 8 - Chapter 3 Class 11 Trigonometric Functions - Part 5 Misc 8 - Chapter 3 Class 11 Trigonometric Functions - Part 6 Misc 8 - Chapter 3 Class 11 Trigonometric Functions - Part 7 Misc 8 - Chapter 3 Class 11 Trigonometric Functions - Part 8 Misc 8 - Chapter 3 Class 11 Trigonometric Functions - Part 9 Misc 8 - Chapter 3 Class 11 Trigonometric Functions - Part 10

This video is only available for Teachoo black users

Get Real time Doubt solving from 8pm to 12 am!


Transcript

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° < π‘₯/2 < 90Β° So, π‘₯/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 π‘‘π‘Žπ‘›β‘π‘₯)/(1 βˆ’ π‘‘π‘Žπ‘›2π‘₯) 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⁑(π‘₯/2))/(1 βˆ’ π‘‘π‘Žπ‘›2(π‘₯/2) ) = βˆ’4/3 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 (π‘₯/2) – 6 tan (π‘₯/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’1)/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 π‘₯/2 is positive, ∴ tan 𝒙/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 1 + 4 = sec2 x/2 5 = sec2 π‘₯/2 sec2 π‘₯/2 = 5 sec π‘₯/2 = Β± √5 Since π‘₯/2 lie on the 1st Quadrant, sec π‘₯/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 π‘₯/2 + cos2 π‘₯/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 π‘₯/2 = Β± (2√5)/5 Since π‘₯/2 lies on the 1st Quadrant sin π‘₯/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“

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.