How do you find the exact value of #tan 0# using the half angle identity?

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Answer 1 · 2018-03-17T06:49:24

#tan0=0#

I don't know why you should use the half angle identity when dealing with zero, but I'll do it anyway.

Here are the sine and cosine half-angle identities:

#sin(theta)=+-sqrt((1-costheta)/2)#

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

Here's the problem:

#color(white)=tan(0)#

#=tan(0/2)#

#=sin(0/2)/cos(0/2)#

#=(+-sqrt((1-cos0)/2))/(+-sqrt((1+cos0)/2))#

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

#=(+-sqrt(0/2))/(+-sqrt(2/2))#

#=(+-sqrt0)/(+-sqrt1)#

#=(+-0)/(+-1)#

#=0/(+-1)#

#=0#

That's the result. Hope this helped!