Question

If \[x = a\sin \theta + b\cos \theta \] and \[y = a\cos \theta - b\sin \theta \] then find the value of \[{x^2} + {y^2}\].

Answer 1

Hint: Here we will find the value of \[{x^2}\] by squaring the equation of \[x\] on both sides. Then we will find the value of \[{y^2}\] by squaring the equation of \[y\] on both sides. Then we will add the value of \[{x^2}\] and \[{y^2}\] to get the value of \[{x^2} + {y^2}\].

Complete step-by-step answer:
The equation of \[x\] is \[x = a\sin \theta + b\cos \theta \].
Now we will find the value of \[{x^2}\] by squaring the equation of \[x\] both sides. Therefore, we get
\[{x^2} = {\left( {a\sin \theta + b\cos \theta } \right)^2}\]
We know the algebraic identity i.e. \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\]. So by using this identity, we get
\[ \Rightarrow {x^2} = {\left( {a\sin \theta } \right)^2} + \left( {2 \times a\sin \theta \times b\cos \theta } \right) + {\left( {b\cos \theta } \right)^2}\]
\[ \Rightarrow {x^2} = {a^2}{\sin ^2}\theta + {b^2}{\cos ^2}\theta + 2ab\sin \theta \cos \theta \]…………………\[\left( 1 \right)\]
Now, the given equation of \[y\] is \[y = a\cos \theta - b\sin \theta \].
Now we will find the value of \[{y^2}\] by squaring the equation of \[y\] both sides. Therefore, we get
\[{y^2} = {\left( {a\cos \theta - b\sin \theta } \right)^2}\]
We know the algebraic identity i.e. \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\]. So by using this identity, we get
\[ \Rightarrow {y^2} = {\left( {a\cos \theta } \right)^2} - \left( {2 \times a\cos \theta \times b\sin \theta } \right) + {\left( {b\sin \theta } \right)^2}\]
\[ \Rightarrow {y^2} = {a^2}{\cos ^2}\theta + {b^2}{\sin ^2}\theta - 2ab\sin \theta \cos \theta \]…………………\[\left( 2 \right)\]
Now we will find the value of \[{x^2} + {y^2}\] by adding the \[{x^2}\] from equation \[\left( 1 \right)\] and \[{y^2}\] from the equation \[\left( 2 \right)\], we get
\[ \Rightarrow {x^2} + {y^2} = \left( {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta + 2ab\sin \theta \cos \theta } \right) + \left( {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta - 2ab\sin \theta \cos \theta } \right)\]
Now we will simplify and solve this equation, we get
\[ \Rightarrow {x^2} + {y^2} = {a^2}\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right) + {b^2}\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right) + 2ab\sin \theta \cos \theta - 2ab\sin \theta \cos \theta \]
We will cancel out the terms and simplify the equation, we get
\[ \Rightarrow {x^2} + {y^2} = {a^2}\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right) + {b^2}\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)\]
We know that \[{\sin ^2}\theta + {\cos ^2}\theta = 1\] and solve the equation. Therefore, we get
\[ \Rightarrow {x^2} + {y^2} = {a^2}\left( 1 \right) + {b^2}\left( 1 \right)\]
\[ \Rightarrow {x^2} + {y^2} = {a^2} + {b^2}\]
Hence, the value of \[{x^2} + {y^2}\] is equal to \[{a^2} + {b^2}\].

Note: To solve this question we need to know some basic trigonometric identities. We use the trigonometric identity \[{\sin ^2}\theta + {\cos ^2}\theta = 1\] which is also known as Pythagorean Identity. All trigonometric identity is derived from basic trigonometric ratios. Also, the trigonometric identities hold true for right-angled triangles.
Here, we have also used algebraic identities to simplify the equation. Algebraic identities are equations where the value of the left-hand side of the equation is identically equal to the value of the right-hand side of the equation.