How to prove that ? #cos x = 1-tan^2(x/2)⁄1+tan^2(x/2)#

1 Answer
Steve M
Jun 26, 2017

We will need the following identities:

# sin^2A+cos^2A -= 1 #
# tan^2A+1-=sec^2A #

Recall the cosine sum formula:

# cos(A+B) -= cosAcosB - sinAsinB #

From which we get the cosine double angle formula:

# cos(2A) -= cos^2A - sin^2A #
# " " -= (1-sin^2A) - sin^2A #
# " " -= 1-2sin^2A #

Now if we put #A=x/2#, then we get:

# cosx -= 1-2sin^2(x/2) #
# " " -= {1-2sin^2(x/2)} * sec^2(x/2)/sec^2(x/2)#
# " " -= {sec^2(x/2)-2sin^2(x/2)sec^2(x/2)}/sec^2(x/2)#
# " " -= {1+tan^2(x/2)-2sin^2(x/2)/cos^2(x/2)}/(1+tan^2(x/2))#
# " " -= {1+tan^2(x/2)-2tan^2(x/2)}/(1+tan^2(x/2))#
# " " -= {1-tan^2(x/2)}/(1+tan^2(x/2)) # QED

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