Prove that Cos α cos (β-α) - sin α sin (β-α) = cos β?

3 Answers
Narad T.
Jun 4, 2018

Please see the explanation below.

Explanation:

We need

#cos(a-b)=cosacosb+sinasinb#

#sin(a-b)=sinacosb-sinbcoas#

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

Therefore,

#LHS=cosacos(a-b)-sinasin(b-a)#

#=cosa(cosacosb+sinasinb)-sina(sinbcosa-sinacosb)#

#=cos^2acosb+cancel(cosasinasinb)-cancel(cosasinasinb)+sin^2acosb#

#=(cos^2a+sin^2a)cosb#

#=cosb#

#=RHS#

#QED#

Dean R.
Jun 4, 2018

That follows pretty quickly from the sum angle formula for cosine:

# cos(a + b) = cos a cos b - sin a sin b #

Let #a=alpha and b=beta-alpha#

#cos beta = cos(alpha + (beta-alpha) ) #

#cos beta = cos alpha cos (beta-alpha) - sin alpha sin(beta-alpha) quad sqrt#

That was too easy. How about a quick proof of the sum angle formulas:

# cos(a+b) + i sin (a+b) = e^{i(a+b)}=e^{ia}e^{ib}#

#= (cos a + i sin a)(cos b + i sin b) #

#= (cos a cos b - sin a sin b) + i(sin a cosb + cos a sin b)#

Equating respective real and imaginary parts,

#cos(a+b)=cos a cos b - sin a sin b quad sqrt#

#sin(a+b) = sin a cosb + cos a sin b quad sqrt#

Douglas K.
Jun 4, 2018

I shall prove by using axioms and identities to change only one side until it is identical to the other side.

Explanation:

Given:

#cos(alpha) cos(beta-alpha) - sin(alpha) sin(beta-alpha) = cos(beta)#

Substitute #cos(beta-alpha) = cos(beta)cos(alpha)+sin(beta)sin(alpha)#:

#cos(alpha)(cos(beta)cos(alpha)+sin(beta)sin(alpha)) - sin(alpha) sin(beta-alpha) = cos(beta)#

Distribute #cos(alpha)# across the terms within the parentheses:

#cos(beta)cos^2(alpha)+sin(beta)sin(alpha)cos(alpha) - sin(alpha) sin(beta-alpha) = cos(beta)#

Substitute #sin(beta-alpha) = sin(beta)cos(alpha)-cos(beta)sin(alpha)#:

#cos(beta)cos^2(alpha)+sin(beta)sin(alpha)cos(alpha) - sin(alpha)(sin(beta)cos(alpha)-cos(beta)sin(alpha)) = cos(beta)#

Distribute #-sin(alpha)# across the terms within the parentheses:

#cos(beta)cos^2(alpha)+sin(beta)sin(alpha)cos(alpha) - sin(beta)sin(alpha)cos(alpha)+cos(beta)sin^2(alpha) = cos(beta)#

Please observe that the two terms in #color(red)("red")# sum to 0:

#cos(beta)cos^2(alpha)+color(red)(sin(beta)sin(alpha)cos(alpha) - sin(beta)sin(alpha)cos(alpha))+cos(beta)sin^2(alpha) = cos(beta)#

Rewriting without the two terms:

#cos(beta)cos^2(alpha)+cos(beta)sin^2(alpha) = cos(beta)#

Factor out #cos(beta)#:

#cos(beta)(cos^2(alpha)+sin^2(alpha)) = cos(beta)#

We know that #(cos^2(alpha)+sin^2(alpha)) = 1#, therefore, we need not write the factor:

#cos(beta) = cos(beta)# Q.E.D.

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