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Let $A$ and $B$ denote the statements

$A: \cos \alpha+\cos \beta+\cos \gamma=0$
$B: \sin \alpha+\sin \beta+\sin \gamma=0$

If $\cos (\boldsymbol{\beta}-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha-\beta)=-\frac{3}{2}$, then

1) $\mathrm{A}$ is true and $\mathrm{B}$ is false
2) $\mathrm{A}$ is false and $\mathrm{B}$ is true
3) both $\mathrm{A}$ and $\mathrm{B}$ are true
4) both $\mathrm{A}$ and $\mathrm{B}$ are false
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(3)

Explanation

$\cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha-\beta)=-\frac{3}{2}$

$\Rightarrow 2[\cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha-\beta)]+3=0$

$\Rightarrow 2[\cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha-\beta)]+$

$\sin ^{2} \alpha+\cos ^{2} \alpha+\sin ^{2} \beta+\cos ^{2} \beta+\sin ^{2} \gamma+\cos ^{2} \gamma=0$

$\Rightarrow(\sin \alpha+\sin \beta+\sin \gamma)^{2}+(\cos \alpha+\cos \beta+\cos \gamma)^{2}=0$

by Diamond (74,866 points)

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