Question

If $x{\sin }^3\left(\theta \right)+y{\cos }^3\left(\theta \right)=\sin \left(\theta \right)\cos \left(\theta \right)$ and$x\sin \left(\theta \right)=y\cos \left(\theta \right)$, prove that $x^2+y^2=1$.

Answer 1

Prove the result $x^2+y^2=1$

Given: $x{\sin }^3\left(\theta \right)+y{\cos }^3\left(\theta \right)=\sin \left(\theta \right)\cos \left(\theta \right).........\left(i\right)$

and $x\sin \left(\theta \right)=y\cos \left(\theta \right).......\left(ii\right)$

Now, from $\left(i\right)$ we get

$$\begin{gathered} x{\sin }^3\left(\theta \right)+y{\cos }^3\left(\theta \right)=\sin \left(\theta \right)\cos \left(\theta \right) \\ \Rightarrow xsin\left(\theta \right)si{n}^2\left(\theta \right)+yco{s}^3\theta =sin\theta cos\theta \\ \Rightarrow ycos\left(\theta \right)si{n}^2\left(\theta \right)+yco{s}^3\theta =sin\theta cos\theta \mathbf{}\mathbf{\left[}\mathit{F}\mathit{r}\mathit{o}\mathit{m}\mathbf{\right(}\mathit{i}\mathit{i}\mathbf{\left)}\mathbf{\right]} \\ \Rightarrow ycos\left(\theta \right)\left[si{n}^2\right(\theta )+co{s}^2\theta ]=sin\theta cos\theta \\ \Rightarrow ycos\left(\theta \right)=sin\theta cos\theta \mathbf{}\mathbf{\left[}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathbf{\right(}\mathit{\theta }\mathbf{)}\mathbf{+}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\theta }\mathbf{=}\mathbf{1}\mathbf{]} \\ \Rightarrow ycos\left(\theta \right)=sin\theta \\ \Rightarrow y=sin\theta ........\left(iii\right) \end{gathered}$$

From $\left(ii\right)$ we get

$$\begin{gathered} \Rightarrow x\sin \theta =y\cos \theta \\ \Rightarrow x\sin \theta =\sin \theta \cos \theta \\ \Rightarrow x=\cos \theta ..........\left(iv\right) \end{gathered}$$

Now using $\left(iii\right)\&\left(iv\right)$ , we get;

$$\begin{gathered} x^2+y^2={\sin }^2\left(\theta \right)+{\cos }^2\left(\theta \right) \\ =1 \end{gathered}$$

Therefore $x^2+y^2=1$

Hence proved.