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If $x=rsinCcosD$, $y=rsinCsinD$ and $z=rcosC$, prove that ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{r}^{2}}$
If $x=rsinCcosD$, $y=rsinCsinD$ and $z=rcosC$, prove that ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{r}^{2}}$
Class 10 | Frank | ICSE | Maths | Trigonometric Identities
If $x=rsinCcosD$, $y=rsinCsinD$ and $z=rcosC$, prove that ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{r}^{2}}$

From the question it is given that,

$x=rsinCcosD$

$y=rsinCsinD$

$z=rcosC$

We have to prove that, ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{r}^{2}}$

First we consider Left Hand Side (LHS),
${{x}^{2}}+{{y}^{2}}+{{z}^{2}}$

$={{\left( r\sin C\cos D \right)}^{2}}+{{\left( r\sin C\sin D \right)}^{2}}+{{\left( r\cos C \right)}^{2}}$

$={{r}^{2}}{{\sin }^{2}}C{{\cos }^{2}}D+{{r}^{2}}{{\sin }^{2}}C{{\sin }^{2}}D+{{r}^{2}}{{\cos }^{2}}C$

Taking common terms outside we get,

$={{r}^{2}}{{\sin }^{2}}C\left( {{\cos }^{2}}D+{{\sin }^{2}}D \right)+{{r}^{2}}{{\cos }^{2}}C$

$={{r}^{2}}\left( {{\sin }^{2}}C+{{\cos }^{2}}C \right)$

We know that, ${{\sin }^{2}}C+{{\cos }^{2}}C=1$

$={{r}^{2}}$

Then, Right Hand Side $={{r}^{2}}$

Therefore, LHS = RHS

Hence it is proved that, ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{r}^{2}}$

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