How do you find the exact value of # tan ( (3pi)/4)#?

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Answer 1 · 2016-07-09T06:56:00

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

#tan((3pi)/(4))=sin((3pi)/(4)) / cos((3pi)/(4))#

Knowing the unit circle, we can see that

#sin((3pi)/(4)) = (sqrt(2))/(2)#

and

#cos((3pi)/(4)) = -(sqrt(2))/(2)#

so

#tan((3pi)/(4))=(((sqrt(2))/(2)) * (-2/sqrt(2))) / cancel(((-(sqrt(2))/(2))* (-2/sqrt(2))))#

#tan((3pi)/(4))=(-2*cancel(sqrt(2)))/(2*cancel(sqrt(2))) =-2/2=-1#