csc a=2, pi/2<a<pi find sin a, cos a, tan a

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gsenviro's profile pic

gsenviro | College Teacher | (Level 1) Educator Emeritus

Posted on

given: cosec a = 2 and  `pi/2 < a < pi`

cosec is inverse of sine, thus

csc a = 1/sin a = 2

or, sin a = 1/2

cos a = `sqrt (1-sin^2 a) = sqrt(1-(1/2)^2) = sqrt (1-1/4) = sqrt (3/4) = - sqrt 3/2`

and tan a = `sina /cosa = (1/2)/(-sqrt3/2) = -1/sqrt3`  

we chose negative signs for cos and tan, since a is between pi/2 and pi, and between these ranges, cos a and tan a will be negative.

Hope this helps.

kspcr111's profile picture

kspcr111 | In Training Educator

Posted on

From the below tables givne in the attachments we can easily find the values of sin a, cos a, tan a

Given ,

csc a= 2, `pi/2<a<pi` ie the angle a is in the second quadrant .

From the table I , we get the values and from the table II , we assign the signs that is the final answer 

Now , Finding 

1) 

sin a = `1/(csc a )` = 1/2

2) cos a = `(+- sqrt(csc ^2 a -1)/ (csc a))`

            = `(+- sqrt(2^2 -1)/ (2))`

             = `(+- sqrt(4 -1)/ (2))`

             = `(+- sqrt(3)/ (2))`

As the angle is in the Second quadrent  cos is negative (see table II )

so `cos a = (- sqrt(3)/ (2))`

3)

`Tan a = (+- 1/(sqrt(csc^2 a -1)))`

         = `(+- 1/(sqrt(2^2 -1)))`

         = `(+- 1/(sqrt(4 -1)))`

         =`(+- 1/(sqrt(3)))`

As the angle is in the Second quadrent  tan is negative (see table II )

so ,

`tan a =(- 1/(sqrt(3)))`

simple :)

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