How to solve $(sin^2(alpha+beta)-cos^2alpha-cos^2beta)/(sin(alpha+beta)-sin^2alpha-sin^2beta)$ if $alpha=pi/3$ and $beta=2pi/3$ ?

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Answer 1 · 2018-06-12T07:41:24

Given $alpha=pi/3$ and $beta=(2pi)/3$

We have $(beta+alpha)=piand (beta-alpha)=pi/3$

To evaluate

$(sin^2(alpha+beta)-cos^2alpha-cos^2beta)/(sin(alpha+beta)-sin^2alpha-sin^2beta)$

$=(sin^2(alpha+beta)-(cos^2alpha+cos^2beta))/(sin(alpha+beta)-(sin^2alpha+sin^2beta))$

$=(sin^2(alpha+beta)-(1-sin^2alpha+cos^2beta))/(sin(alpha+beta)-(sin^2alpha+1-cos^2beta))$

$=(sin^2(alpha+beta)-(1+cos^2beta-sin^2alpha))/(sin(alpha+beta)-(1-(cos^2beta-sin^2alpha)))$

$=(sin^2(alpha+beta)-(1+cos(beta+alpha)cos(beta-alpha)))/(sin(alpha+beta)-(1-cos(beta+alpha)cos(beta-alpha))$

$=(sin^2(pi)-(1+cos(pi)cos(pi/3)))/(sin(pi)-(1-cos(pi)cos(pi/3))$

$=(0-(1-1/2))/(0-(1+1/2))=1/3$

Formula used

$cos(beta+alpha)cos(beta-alpha)$

$=cos^2betacos^2alpha-sin^2betasin^2alpha$

$=cos^2beta(1-sin^2alpha)-(1-cos^2beta)sin^2alpha$

$=cos^2beta-sin^2alpha$

Alternative

$(sin^2(alpha+beta)-cos^2alpha-cos^2beta)/(sin(alpha+beta)-sin^2alpha-sin^2beta)$

$=(sin^2(pi)-cos^2(pi/3)-cos^2((2pi)/3))/(sin(pi)-sin^2(pi/3)-sin^2((2pi)/3))$

$=(0-1/4-1/4)/(0-3/4-3/4)=(1/2)/(3/2)=1/3$

How to solve (sin^2(alpha+beta)-cos^2alpha-cos^2beta)/(sin(alpha+beta)-sin^2alpha-sin^2beta)(sin^2(alpha+beta)-cos^2alpha-cos^2beta)/(sin(alpha+beta)-sin^2alpha-sin^2beta) if alpha=pi/3alpha=pi/3 and beta=2pi/3beta=2pi/3?