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

3 Answers
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

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

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.