Prove cos(a+b)cos(a-b)=cos^2b-sin^2a ?

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Answer 1 · 2017-04-22T09:30:17

See proof below

We need

$(x+y)(x-y)=x^2-y^2$

$cos(a+b)=cosacosb-sina sinb$

$cos(a-b)=cosacosb+sina sinb$

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

$cos^2b+sin^2b=1$

Therefore,

$LHS=cos(a+b)cos(a-b)$

$=(cosacosb-sina sinb)(cosacosb+sina sinb)$

$=cos^2acos^2b-sin^2a sin^2b$

$=cos^2b(1-sin^2a)-sin^2a(1-cos^2b)$

$=cos^2b-cancel(cos^2bsin^2a)-sin^2a+cancel(cos^2bsin^2a)$

$=cos^2b-sin^2a$

$=RHS$

$QED$