Given #cos(pi/2-x)=3/5, cosx=4/5# to find the remaining trigonometric function?

1 Answer
Jan 5, 2017

Answer:

To find the value of the sine, use the identity for the cosine of the difference of two angles:
#cos(A - B) = cos(A)cos(B) + sin(A)sin(B)#. Please see below.

Explanation:

The identity for the cosine of the difference of two angles is:

#cos(A - B) = cos(A)cos(B) + sin(A)sin(B)#

Given: #A = pi/2 and B = x#

#cos(pi/2 - x) = cos(pi/2)cos(x) + sin(pi/2)sin(x)#

The first term on the right disappears, because #cos(pi/2) = 0#, and the second term on the right becomes #sin(x)#, because #sin(pi/2) = 1#:

#cos(pi/2 - x) = sin(x) = 3/5#

#sin(x) = 3/5#

#tan(x) = sin(x)/cos(x) = (3/5)/(4/5)#

#tan(x) = 3/4#

#cot(x) = 1/tan(x)#

#cot(x) = 4/3#

#csc(x) = 1/sin(x)#

#csc(x) = 5/3#

#sec(x) = 1/cos(x)#

#sec(x) = 5/4#