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Question: If \( x = a\cos \theta - b\sin \theta \) and \( y = a\sin \theta + b\cos \theta \) then prove that \...

If x=acos⁡θ−bsin⁡θx = a\cos \theta - b\sin \theta and y=asin⁡θ+bcos⁡θy = a\sin \theta + b\cos \theta then prove that x2+y2=a2+b2{x^2} + {y^2} = {a^2} + {b^2}

Explanation

Solution

Hint : To prove the given expression we need to find the square of each x and y separately then add them and simplify till RHS will not be achieved. We make use appropriate trigonometric identities.

Complete step-by-step answer :
Given x=acos⁡θ−bsin⁡θx = a\cos \theta - b\sin \theta and y=asin⁡θ+bcos⁡θy = a\sin \theta + b\cos \theta we have to prove x2+y2=a2+b2{x^2} + {y^2} = {a^2} + {b^2}
So in LHS it is given that we need to find x2+y2{x^2} + {y^2}
So taking the value of x and squaring both side we get,
x2=a2cos⁡2θ+b2sin⁡2θ−2absin⁡θcos⁡θ{x^2} = {a^2}{\cos ^2}\theta + {b^2}{\sin ^2}\theta - 2ab\sin \theta \cos \theta
Similarly to find the value of y2{y^2} we will squaring the value of y on both side we get,
Now to get the value of LHS just add the value of x2{x^2} and y2{y^2}
On adding we get,
⇒x2+y2=a2cos⁡2θ+b2sin⁡2θ−2absin⁡θcos⁡θ+a2sin⁡2θ+b2cos⁡2θ+2absin⁡θcos⁡θ\Rightarrow {x^2} + {y^2} = {a^2}{\cos ^2}\theta + {b^2}{\sin ^2}\theta - 2ab\sin \theta \cos \theta + {a^2}{\sin ^2}\theta + {b^2}{\cos ^2}\theta + 2ab\sin \theta \cos \theta
Here the 2ab term will get cancelled out
So finally we get,
⇒x2+y2=a2cos⁡2θ+b2sin⁡2θ+a2sin⁡2θ+b2cos⁡2θ\Rightarrow {x^2} + {y^2} = {a^2}{\cos ^2}\theta + {b^2}{\sin ^2}\theta + {a^2}{\sin ^2}\theta + {b^2}{\cos ^2}\theta
⇒x2+y2=a2(cos⁡2θ+sin⁡2θ)+b2(sin⁡2θ+cos⁡2θ)\Rightarrow {x^2} + {y^2} = {a^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) + {b^2}\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)
Now taking a2{a^2} and b2{b^2} we get
⇒x2+y2=a2(cos⁡2θ+sin⁡2θ)+b2(sin⁡2θ+cos⁡2θ)\Rightarrow {x^2} + {y^2} = {a^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) + {b^2}\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)
We know the value of sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1
Applying this result on the above we get
⇒x2+y2=a2+b2\Rightarrow {x^2} + {y^2} = {a^2} + {b^2}
LHS=RHS
Hence it is proved.
So, the correct answer is “ x2+y2=a2+b2{x^2} + {y^2} = {a^2} + {b^2} ”.

Note : Students also initiate from RHS to LHS means first find the value a2{a^2} and b2{b^2} then put it in RHS and solved such that you will get LHS. Student can go for this way but this is quite difficult and opposite process. So try to go from LHS to RHS.