Question

(i) What is the value of \[{\sin ^2}{29^ \circ } + {\sin ^2}{61^ \circ }\] ?
(ii) If ${\text{x = a}}\sin \theta + {\text{b}}\cos \theta $ and ${\text{y = a}}\cos \theta - {\text{b}}\sin \theta $ then find the value of ${{\text{x}}^2}{\text{ + }}{{\text{y}}^2}$?
(iii) If ${\text{x = acos}}\theta $ and${\text{y = b}}\sin \theta $ then find the value of ${{\text{b}}^2}{{\text{x}}^2}{\text{ + }}{{\text{a}}^2}{{\text{y}}^2}$.

Answer 1

Hint: (i) We can write $\cos \theta = \sin \left( {{{90}^ \circ } - \theta } \right)$ and then use trigonometric identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$ and you’ll get the answer. (ii) First square the values of x and y then add them. Use trigonometric identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$ and you’ll get the answer.(iii) You can multiply b to value of x and a to value of y. Square and add them. You’ll get the answer.

Complete step-by-step answer:
(i)We have to find the value of \[{\sin ^2}{29^ \circ } + {\sin ^2}{61^ \circ }\].
\[ \Rightarrow {\sin ^2}{29^ \circ } + {\sin ^2}{61^ \circ } = {\sin ^2}{29^ \circ } + {\sin ^2}\left( {{{90}^ \circ } - {{61}^ \circ }} \right)\]
 We know that $\cos \theta = \sin \left( {{{90}^ \circ } - \theta } \right)$. So we can write $\cos {29^ \circ } = \sin \left( {{{90}^ \circ } - {{61}^ \circ }} \right)$
$ \Rightarrow {\sin ^2}{29^ \circ } + {\sin ^2}{61^ \circ } = {\sin ^2}{29^ \circ } + {\cos ^2}{29^ \circ }$
We know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$.Here$\theta = {29^ \circ }$ , so on putting the value we get-
$ \Rightarrow {\sin ^2}{29^ \circ } + {\sin ^2}{61^ \circ } = {\sin ^2}{29^ \circ } + {\cos ^2}{29^ \circ } = 1$
Answer-Hence the answer is \[{\sin ^2}{29^ \circ } + {\sin ^2}{61^ \circ } = 1\]
(ii)Given, ${\text{x = a}}\sin \theta + {\text{b}}\cos \theta $--- (I)
${\text{y = a}}\cos \theta - {\text{b}}\sin \theta $---- (II)
We have to find the value of ${{\text{x}}^2}{\text{ + }}{{\text{y}}^2}$. So first we square eq.(I) and (II) then add them.
On squaring them, we get-
\[
   \Rightarrow {{\text{x}}^2}{\text{ = }}{\left( {{\text{a}}\sin \theta + {\text{b}}\cos \theta } \right)^2} \\
   \Rightarrow {{\text{x}}^2}{\text{ = }}{{\text{a}}^2}{\sin ^2}\theta + {{\text{b}}^2}{\cos ^2}\theta + {\text{2absin}}\theta {\text{cos}}\theta \\
 \]
As ${\left( {{\text{a + b}}} \right)^2}{\text{ = }}{{\text{a}}^2} + {{\text{b}}^2} + {\text{2ab}}$
And \[{{\text{y}}^2}{\text{ = }}{\left( {{\text{a}}\cos \theta + {\text{bsin}}\theta } \right)^2}\]
As ${\left( {{\text{a + b}}} \right)^2}{\text{ = }}{{\text{a}}^2} + {{\text{b}}^2} + {\text{2ab}}$, on using the formula-
$ \Rightarrow {{\text{y}}^2}{\text{ = }}{{\text{a}}^2}{\cos ^2}\theta + {{\text{b}}^2}{\text{si}}{{\text{n}}^2}\theta + {\text{2absin}}\theta {\text{cos}}\theta $
Now putting the values in ${{\text{x}}^2}{\text{ + }}{{\text{y}}^2}$, we get
$ \Rightarrow {{\text{x}}^2}{\text{ + }}{{\text{y}}^2} = {{\text{a}}^2}{\sin ^2}\theta + {{\text{b}}^2}{\cos ^2}\theta + {\text{2absin}}\theta {\text{cos}}\theta + {{\text{b}}^2}{\sin ^2}\theta + {{\text{a}}^2}{\cos ^2}\theta + {\text{2absin}}\theta {\text{cos}}\theta $
On separating common terms and simplifying we get,
\[
   \Rightarrow {{\text{x}}^2}{\text{ + }}{{\text{y}}^2} = {{\text{a}}^2}\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right) + {{\text{b}}^2}\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right) + {\text{2absin}}\theta {\text{cos}}\theta + {\text{2absin}}\theta {\text{cos}}\theta \\
   \Rightarrow {{\text{x}}^2}{\text{ + }}{{\text{y}}^2} = {{\text{a}}^2} + {{\text{b}}^2} + 4{\text{absin}}\theta {\text{cos}}\theta \\
 \]
[As ${\sin ^2}\theta + {\cos ^2}\theta = 1$]
\[ \Rightarrow {{\text{x}}^2}{\text{ + }}{{\text{y}}^2} = {{\text{a}}^2} + {{\text{b}}^2} + 4{\text{absin}}\theta {\text{cos}}\theta \]
We know that${\text{2sin}}\theta {\text{cos}}\theta = \sin 2\theta $ , so we get,
Answer\[ \Rightarrow {{\text{x}}^2}{\text{ + }}{{\text{y}}^2} = {{\text{a}}^2} + {{\text{b}}^2} + 2{\text{absin2}}\theta \]
(iii)Given, ${\text{x = acos}}\theta $ -- (i)
And${\text{y = b}}\sin \theta $--- (ii)
 We have to find the value of ${{\text{b}}^2}{{\text{x}}^2}{\text{ + }}{{\text{a}}^2}{{\text{y}}^2}$.
On multiplying eq. (i) with b and eq. (ii) with a and squaring both eq., we get
$ \Rightarrow {\left( {{\text{bx}}} \right)^2}{\text{ = }}{\left( {{\text{ab}}\cos \theta } \right)^2} \Rightarrow {{\text{b}}^2}{{\text{x}}^2}{\text{ = }}{{\text{a}}^2}{{\text{b}}^2}{\text{ co}}{{\text{s}}^2}\theta $ --- (I)
$ \Rightarrow {\left( {{\text{ay}}} \right)^2} = {\left( {{\text{ba}}\sin \theta } \right)^2} \Rightarrow {{\text{a}}^2}{{\text{y}}^2}{\text{ = }}{{\text{a}}^2}{{\text{b}}^2}{\text{ si}}{{\text{s}}^2}\theta $ -- (II)
On adding eq. (I) and (II), we get-
\[ \Rightarrow {{\text{b}}^2}{{\text{x}}^2}{\text{ + }}{{\text{a}}^2}{{\text{y}}^2} = {{\text{a}}^2}{{\text{b}}^2}\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right){\text{ }}\]
We know that ${\cos ^2}\theta + {\sin^2}\theta = 1$. So we get,
Answer$ \Rightarrow {{\text{b}}^2}{{\text{x}}^2}{\text{ + }}{{\text{a}}^2}{{\text{y}}^2} = {{\text{a}}^2}{{\text{b}}^2}$

Note: On solving question (ii) student may go wrong if they leave the answer as \[ \Rightarrow {{\text{x}}^2}{\text{ + }}{{\text{y}}^2} = {{\text{a}}^2} + {{\text{b}}^2} + 4{\text{absin}}\theta {\text{cos}}\theta \] which is wrong as the equation can be further simplified by using the formula ${\text{2sin}}\theta {\text{cos}}\theta = \sin 2\theta $. We have to write the answers of the questions in a simplified form.