If x{sin ^3}alpha + y{cos ^3}alpha = sin alph | vedclass

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Question

(c) We have $x\,{\sin ^3}\alpha + y\,{\cos ^3}\alpha = \sin \,\alpha \,\cos \,\alpha $&hellip;..$(i)$

and $x\,\sin \,\alpha - y\,\cos \,\alpha = 0$

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(c) We have $x\,{\sin ^3}\alpha + y\,{\cos ^3}\alpha = \sin \,\alpha \,\cos \,\alpha $&hellip;..$(i)$

and $x\,\sin \,\alpha - y\,\cos \,\alpha = 0$&hellip;..$(ii)$

Now from $(ii)$, $x\,\sin \,\alpha = y\,\cos \,\alpha $

Putting in $(i),$ we get

$ \Rightarrow \,\,y\,\cos \alpha \,{\sin ^2}\alpha + y\,{\cos ^3}\alpha = \sin \,\alpha \,\cos \,\alpha $

$ \Rightarrow \,\,y\,\cos \alpha \,\left\{ {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right\} = \sin \,\alpha \,\cos \,\alpha $

$ \Rightarrow \,\,y\,\cos \,\alpha = \sin \,\alpha \,\cos \,\alpha \, $

$\Rightarrow \,\,y = \sin \,\alpha $ and $x = \cos \,\alpha $

Hence, ${x^2} + {y^2} = {\sin ^2}\alpha + {\cos ^2}\alpha = 1.$

If $x{\sin ^3}\alpha + y{\cos ^3}\alpha = \sin \alpha \cos \alpha $ and $x\sin \alpha - y\cos \alpha = 0,$ then ${x^2} + {y^2} = $

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