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

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.