How to solve $\frac{{sin}^{2} (α + β) - {cos}^{2} α - {cos}^{2} β}{sin (α + β) - {sin}^{2} α - {sin}^{2} β}$ $(sin^2(alpha+beta)-cos^2alpha-cos^2beta)/(sin(alpha+beta)-sin^2alpha-sin^2beta)$ if $α = \frac{π}{3}$ $alpha=pi/3$ and $β = 2 \frac{π}{3}$ $beta=2pi/3$ ?

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How to solve $\frac{{sin}^{2} (α + β) - {cos}^{2} α - {cos}^{2} β}{sin (α + β) - {sin}^{2} α - {sin}^{2} β}$ $(sin^2(alpha+beta)-cos^2alpha-cos^2beta)/(sin(alpha+beta)-sin^2alpha-sin^2beta)$ if $α = \frac{π}{3}$ $alpha=pi/3$ and $β = 2 \frac{π}{3}$ $beta=2pi/3$ ?

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Answer 1 · 2018-06-12T07:41:24

Given $α = \frac{π}{3}$ $alpha=pi/3$ and $β = \frac{2 π}{3}$ $beta=(2pi)/3$

We have $(β + α) = π and (β - α) = \frac{π}{3}$ $(beta+alpha)=piand (beta-alpha)=pi/3$

To evaluate

$\frac{{sin}^{2} (α + β) - {cos}^{2} α - {cos}^{2} β}{sin (α + β) - {sin}^{2} α - {sin}^{2} β}$ $(sin^2(alpha+beta)-cos^2alpha-cos^2beta)/(sin(alpha+beta)-sin^2alpha-sin^2beta)$

$= \frac{{sin}^{2} (α + β) - ({cos}^{2} α + {cos}^{2} β)}{sin (α + β) - ({sin}^{2} α + {sin}^{2} β)}$ $=(sin^2(alpha+beta)-(cos^2alpha+cos^2beta))/(sin(alpha+beta)-(sin^2alpha+sin^2beta))$

$= \frac{{sin}^{2} (α + β) - (1 - {sin}^{2} α + {cos}^{2} β)}{sin (α + β) - ({sin}^{2} α + 1 - {cos}^{2} β)}$ $=(sin^2(alpha+beta)-(1-sin^2alpha+cos^2beta))/(sin(alpha+beta)-(sin^2alpha+1-cos^2beta))$

$= \frac{{sin}^{2} (α + β) - (1 + {cos}^{2} β - {sin}^{2} α)}{sin (α + β) - (1 - ({cos}^{2} β - {sin}^{2} α))}$ $=(sin^2(alpha+beta)-(1+cos^2beta-sin^2alpha))/(sin(alpha+beta)-(1-(cos^2beta-sin^2alpha)))$

$= \frac{{sin}^{2} (α + β) - (1 + cos (β + α) cos (β - α))}{sin (α + β) - (1 - cos (β + α) cos (β - α))}$ $=(sin^2(alpha+beta)-(1+cos(beta+alpha)cos(beta-alpha)))/(sin(alpha+beta)-(1-cos(beta+alpha)cos(beta-alpha))$

$= \frac{{sin}^{2} (π) - (1 + cos (π) cos (\frac{π}{3}))}{sin (π) - (1 - cos (π) cos (\frac{π}{3}))}$ $=(sin^2(pi)-(1+cos(pi)cos(pi/3)))/(sin(pi)-(1-cos(pi)cos(pi/3))$

$= \frac{0 - (1 - \frac{1}{2})}{0 - (1 + \frac{1}{2})} = \frac{1}{3}$ $=(0-(1-1/2))/(0-(1+1/2))=1/3$

Formula used

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

$= {cos}^{2} β {cos}^{2} α - {sin}^{2} β {sin}^{2} α$ $=cos^2betacos^2alpha-sin^2betasin^2alpha$

$= {cos}^{2} β (1 - {sin}^{2} α) - (1 - {cos}^{2} β) {sin}^{2} α$ $=cos^2beta(1-sin^2alpha)-(1-cos^2beta)sin^2alpha$

$= {cos}^{2} β - {sin}^{2} α$ $=cos^2beta-sin^2alpha$

Alternative

$\frac{{sin}^{2} (α + β) - {cos}^{2} α - {cos}^{2} β}{sin (α + β) - {sin}^{2} α - {sin}^{2} β}$ $(sin^2(alpha+beta)-cos^2alpha-cos^2beta)/(sin(alpha+beta)-sin^2alpha-sin^2beta)$

$= \frac{{sin}^{2} (π) - {cos}^{2} (\frac{π}{3}) - {cos}^{2} (\frac{2 π}{3})}{sin (π) - {sin}^{2} (\frac{π}{3}) - {sin}^{2} (\frac{2 π}{3})}$ $=(sin^2(pi)-cos^2(pi/3)-cos^2((2pi)/3))/(sin(pi)-sin^2(pi/3)-sin^2((2pi)/3))$

$= \frac{0 - \frac{1}{4} - \frac{1}{4}}{0 - \frac{3}{4} - \frac{3}{4}} = \frac{\frac{1}{2}}{\frac{3}{2}} = \frac{1}{3}$ $=(0-1/4-1/4)/(0-3/4-3/4)=(1/2)/(3/2)=1/3$

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How to solve $\frac{{sin}^{2} (α + β) - {cos}^{2} α - {cos}^{2} β}{sin (α + β) - {sin}^{2} α - {sin}^{2} β}$ $(sin^2(alpha+beta)-cos^2alpha-cos^2beta)/(sin(alpha+beta)-sin^2alpha-sin^2beta)$ if $α = \frac{π}{3}$ $alpha=pi/3$ and $β = 2 \frac{π}{3}$ $beta=2pi/3$ ?

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How to solve $\frac{{sin}^{2} (α + β) - {cos}^{2} α - {cos}^{2} β}{sin (α + β) - {sin}^{2} α - {sin}^{2} β}$ $(sin^2(alpha+beta)-cos^2alpha-cos^2beta)/(sin(alpha+beta)-sin^2alpha-sin^2beta)$ if $α = \frac{π}{3}$ $alpha=pi/3$ and $β = 2 \frac{π}{3}$ $beta=2pi/3$ ?