Question

If x sin³Ф+y cos³Ф=sinФcosФ and x sinФ=y cosФ,prove that x²+y²=1 [Don't spam]

Answer 1

Given :-

$\bullet\sf\ x\ sin^3\theta+ y\ cos^3\theta= sin\theta\ cos\theta\\ \\ \bullet\sf\ x\ sin\theta= y\ cos\theta$

To Prove:-

$\boxed{\sf x^2+y^2=1}$

Proof :-

$\dashrightarrow\sf x\ sin^3\theta+ y\ cos^3\theta= sin\theta cos\theta\\ \\ \\ \dashrightarrow\sf (x\ sin\theta)sin^2\theta+(y\ cos\theta)cos^2\theta= sin\theta\ cos\theta\\ \\ \\ \dashrightarrow\sf (y\ cos\theta) sin^2\theta+(y\ cos\theta)cos^2\theta=sin\theta\ cos\theta \ \ \ \ \ \big\lgroup \therefore x\ sin\theta= y\ cos\theta\big\rgroup\\ \\ \\ \dashrightarrow\sf (y\ cos\theta)(sin^2\theta+cos^2\theta)=sin\theta\ cos\theta \\ \\ \\ \dashrightarrow\sf (y\ cos\theta)= sin\theta cos\theta \ \ \ \ \ \big\lgroup sin^2\theta+cos^2\theta=1\big\rgroup\\ \\ \\\dashrightarrow\sf y\cancel{cos\theta}=sin\theta \cancel{cos\theta}\\ \\ \\\dashrightarrow {\boxed{\sf y= sin\theta}} \\ \\ \\\dashrightarrow\sf x \cancel{sin\theta}=\cancel{sin\theta}\ cos\theta\ \ \ \ \ \big\lgroup x\ sin\theta= y\ cos\theta \big\rgroup\\ \\ \\\dashrightarrow {\boxed{\sf x= cos\theta}}$

Now , $\it x^2+y^2= 1$

$\dashrightarrow\sf (cos\theta)^2+(sin\theta)^2=1\\ \\ \\ \dashrightarrow\sf cos^2\theta+sin^2\theta=1 \ \ \ \ \ \Big\lgroup sin^2\theta+cos^2\theta=1\Big\rgroup \\ \\ \\\dashrightarrow\sf 1=1 \ \ \ \ \ (Hence \ Proved !)$

Answer 2

Given :-

$$\begin{lgathered}\bullet\sf\ x\ sin^3\theta+ y\ cos^3\theta= sin\theta\ cos\theta\\ \\ \bullet\sf\ x\ sin\theta= y\ cos\theta\end{lgathered}$$

To Prove:-

$$\boxed{\sf x^2+y^2=1}$$

Proof :-

$$\begin{lgathered}\dashrightarrow\sf x\ sin^3\theta+ y\ cos^3\theta= sin\theta cos\theta\\ \\ \\ \dashrightarrow\sf (x\ sin\theta)sin^2\theta+(y\ cos\theta)cos^2\theta= sin\theta\ cos\theta\\ \\ \\ \dashrightarrow\sf (y\ cos\theta) sin^2\theta+(y\ cos\theta)cos^2\theta=sin\theta\ cos\theta \ \ \ \ \ \big\lgroup \therefore x\ sin\theta= y\ cos\theta\big\rgroup\\ \\ \\ \dashrightarrow\sf (y\ cos\theta)(sin^2\theta+cos^2\theta)=sin\theta\ cos\theta \\ \\ \\ \dashrightarrow\sf (y\ cos\theta)= sin\theta cos\theta \ \ \ \ \ \big\lgroup sin^2\theta+cos^2\theta=1\big\rgroup\\ \\ \\\dashrightarrow\sf y\cancel{cos\theta}=sin\theta \cancel{cos\theta}\\ \\ \\\dashrightarrow {\boxed{\sf y= sin\theta}} \\ \\ \\\dashrightarrow\sf x \cancel{sin\theta}=\cancel{sin\theta}\ cos\theta\ \ \ \ \ \big\lgroup x\ sin\theta= y\ cos\theta \big\rgroup\\ \\ \\\dashrightarrow {\boxed{\sf x= cos\theta}}\end{lgathered}$$

Now , $$\it x^2+y^2= 1$$

$$\begin{lgathered}\dashrightarrow\sf (cos\theta)^2+(sin\theta)^2=1\\ \\ \\ \dashrightarrow\sf cos^2\theta+sin^2\theta=1 \ \ \ \ \ \Big\lgroup sin^2\theta+cos^2\theta=1\Big\rgroup \\ \\ \\\dashrightarrow\sf 1=1 \ \ \ \ \ (Hence \ Proved !)\end{lgathered}$$