How do I prove ((1-tan^2 theta) / (1+tan^2 theta))=cos(2 theta)?

Is proving ((1-tan^2 theta) / (1+tan^2 theta))=cos 2 theta the same as prove ((1-tan^2 theta) / (1+tan^2 theta))=cos (2 theta)?

I can prove the first, but don't know if the parentheses cos(2 theta) in the second is the same.

Is proving ((1-tan^2 theta) / (1+tan^2 theta))=cos 2 theta the same as prove ((1-tan^2 theta) / (1+tan^2 theta))=cos (2 theta)?

I can prove the first, but don't know if the parentheses cos(2 theta) in the second is the same.

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Aviv S. Share
Feb 8, 2018

Here are the identities that I used:

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

#tan^2theta=sin^2theta/cos^2theta#

#sin^2theta+cos^2theta=1#

#cos(2theta)=cos^2theta-sin^2theta#

Now here's the actual proof:

#(1-tan^2 theta) / (1+tan^2 theta)=cos(2 theta)#

#(-sin^2theta/cos^2theta+1)/(sin^2theta/cos^2theta+1)=cos(2theta)#

#((-sin^2theta+cos^2theta)/cos^2theta)/((sin^2theta+cos^2theta)/cos^2theta)=cos(2theta)#

#(-sin^2theta+cos^2theta)/cancel(sin^2theta+cos^2theta)=cos(2theta)#

#-sin^2theta+cos^2theta=cos(2theta)#

#cos^2theta-sin^2theta=cos(2theta)#

#cos(2theta)=cos(2theta)#

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Feb 10, 2018

Answer:

As proved below

Explanation:

#1 + tan^2 theta = sec^2 theta = 1/cos^2 theta#

#1 - 2sin^2 theta = 2cos^2 theta - 1 = cos^2 theta - sin^2 theta = cos (2theta)#

#(1-tan^2theta) / (1+tan^2theta) = (1-(sin^2theta/cos^2theta))/(1/cos^2(theta)# as #1 + tan^2theta = sec^2 theta = 1/cos^2theta#

#((cos^2theta-sin^2theta)/cancel(cos^2(theta))/ 1/cancel(cos^2 (theta))) = cos^2(theta) - sin^2theta = cos(2theta)#

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Steve M Share
Feb 8, 2018

We seek to prove that:

# (1-tan^2 theta) / (1+tan^2 theta) -= cos(2 theta) #

We will require the following trigonometric definitions/identities:

# tan phi = sin phi / cos phi #
# sin^2 phi + cos^phi -=1 #
# cos 2phi -= cos^2 phi - sin^2 phi #

Consider the LHS of the given expression:

# LHS = (1-tan^2 theta) / (1+tan^2 theta) #

# \ \ \ \ \ \ \ \ = (1-(sin^2 theta)/(cos^2 theta)) / (1+(sin^2 theta)/(cos^2 theta)) #

# \ \ \ \ \ \ \ \ = ((cos^2 theta -sin^2 theta)/(cos^2 theta)) / ( (cos^2 theta+sin^2 theta)/(cos^2 theta)) #

# \ \ \ \ \ \ \ \ = (cos^2 theta -sin^2 theta)/(cos^2 theta+sin^2 theta) #

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

# \ \ \ \ \ \ \ \ = cos 2theta #

# \ \ \ \ \ \ \ \ = RHS \ \ \ \ # QED

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