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

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.

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