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Misc 9 - cos x = -1/3, find sin x/2 , cos x/2 and tan x/2

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

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Misc 9 Find sin /2, cos /2 and tan /2 for cos = 1/3 , in quadrant III Since x is in quadrant III 180 < x < 270 Dividing by 2 all sides (180 )/2 < /2 < (270 )/2 90 < /2 < 135 So, /2 lies in IInd quadrant In IInd quadrant, sin is positive, cos & tan are negative sin /2 Positive and cos /2 and tan /2 negative Given, cos x = 1/3 cos x = 2 cos2 /2 1 1/3 = 2cos2 /2 1 1 1/3 = 2cos2 /2 (3 1)/3 = 2cos2 /2 2/3 = 2cos2 /2 2cos2 /2 = 2/3 cos2 /2 = 2/3 1/2 cos2 /2 = 1/3 cos /2 = (1/3) = 1/ 3 Since /2 lie is llnd Quadrant , cos /2 is negative So cos /2 = 1/ 3 = 1/ 3 3/ 3 = 3/3 We know that sin2x + cos2x = 1 Replacing x with /2 sin2 /2 + cos2 /2 = 1 sin2 /2 = 1 cos2 /2 Putting cos /2 = 1/ 3 sin2 /2 = 1 (" " 1/ 3)2 sin2 /2 = 1 1/3 sin2 /2 = (3 1)/3 sin2 /2 = 2/3 sin /2 = (2/3) = 2/ 3 = 2/ 3 3/ 3 = (2 3)/3 = 6/3 Since /2 lie on the llnd Quadrant , sin /2 is positive in the llnd Quadrant So sin /2 = 6/3 We know that tan x = sin / Replacing x with /2 tan /2 = /2 / /2 = ( 6/3)/( 3/3) = 6/3 ( 3)/ 3 = 6/ 3 = (6/3) = 2 Hence, tan /2 = 2 Hence, tan /2 = 2 , cos /2 = 3/3 & sin /2 = 6/3

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.