How do you find the exact values of #costheta# and #sintheta# when #tantheta=1#?

2 Answers
Douglas K.
Nov 15, 2016

#sin(theta) = cos(theta) = sqrt(2)/2#

OR

#sin(theta) = cos(theta) = -sqrt(2)/2#

Explanation:

Given: #tan(theta) = 1#

Use the identity:

#tan^2(theta) + 1 = sec^2(theta)#

Substitute #1^2# for #tan^2(theta)#

#1^2 + 1 = sec^2(theta)#

#2 = sec^2(theta)#

Because #sec(theta) = 1/cos(theta)# we can change the above equation to:

#cos^2(theta) = 1/2#

#cos(theta) = +-1/sqrt(2)#

#cos(theta) = +-sqrt(2)/2#

#tan(theta) = sin(theta)/cos(theta) = 1#

#sin(theta) = cos(theta)#

#sin(theta) = +-sqrt(2)/2#

Because we are given nothing to determine whether #theta# is in the first of the third quadrant:

#sin(theta) = cos(theta) = sqrt(2)/2#

OR

#sin(theta) = cos(theta) = -sqrt(2)/2#

Annie
Nov 15, 2016

# cosθ=sinθ=+-sqrt2/2#

Explanation:

Draw a right isosceles triangle. The angles are 45, 45,90
Tan 45 =1,
Using Pythagoras the sides are in the ratio #1:1:sqrt2#
cos 45=sin45=#1/sqrt2#=#sqrt2/2#
But then we need to look at the angle in the third quadrant to get the negative result.

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