If $x=a\sin \theta +b\cos \theta ,y=a\cos \theta -b\sin \theta $ then show that ${{\left( ax+by \right)}^{2}}+{{\left( bx-ay \right)}^{2}}={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}$.
Answer
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Hint:Multiply expression of x and y with a and b. Thus, get the expression for ax, ay, bx, by. Then substitute to the values in the LHS of the expression. Apply basic identities and simplify it.
Complete step-by-step answer:
We have been given an expression of x and y. we need to show that ${{\left( ax+by \right)}^{2}}+{{\left( bx-ay \right)}^{2}}={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}$
Given, $x=a\sin \theta +b\cos \theta $ ………….. (i)
$y=a\cos \theta -b\sin \theta $……………(ii)
Now multiply equation (i) with a and b and form (ii) expressions. Similarly multiply equation (ii) with a and b and form (ii) expression
Equation (i) $x=a\sin \theta +b\cos \theta $
Multiply by a,
$ax=a\left( \sin \theta +b\cos \theta \right)$
$ax={{a}^{2}}\sin \theta +ab\cos \theta $
Multiply by b,
$\begin{align}
& bx=b\left( a\sin \theta +b\cos \theta \right) \\
& bx=ab\sin \theta +{{b}^{2}}\cos \theta \\
\end{align}$
Similarly, on equation (ii) $y=a\cos \theta -b\sin \theta $
Multiply by a,
$ay={{a}^{2}}\cos \theta -ab\sin \theta $
Multiply by b,
$by=ab\cos \theta -{{b}^{2}}\sin \theta $
These we have 4 expressions
$\begin{align}
& ax={{a}^{2}}\sin \theta +ab\cos \theta \\
& bx=ab\sin \theta +{{b}^{2}}\cos \theta \\
& ay={{a}^{2}}\cos \theta -ab\sin \theta \\
& by=ab\cos \theta -{{b}^{2}}\sin \theta \\
\end{align}$
Now let us find the value of
$ax+by=\left( {{a}^{2}}\sin \theta +ab\cos \theta \right)+\left( ab\cos \theta -{{b}^{2}}\sin \theta \right)={{a}^{2}}\sin \theta -{{b}^{2}}\sin \theta +2ab\cos \theta $
Similarly,
$\begin{align}
& bx-ay=\left( ab\sin \theta +{{b}^{2}}\cos \theta \right)-\left( {{a}^{2}}\cos \theta -ab\sin \theta \right) \\
& =ab\sin \theta +{{b}^{2}}\cos \theta -{{a}^{2}}\cos \theta +ab\sin \theta \\
& ={{b}^{2}}\cos \theta -{{a}^{2}}\cos \theta +2ab\sin \theta \\
\end{align}$
We have been asked to prove that,
${{\left( ax+by \right)}^{2}}+{{\left( bx-ay \right)}^{2}}={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}$
Consider the LHS of expression,
$LHS={{\left( ax+by \right)}^{2}}+{{\left( bx-ay \right)}^{2}}$
Now let us substitute the values of (ax + by), (by – ay)
$LHS={{\left( ax+by \right)}^{2}}+{{\left( bx-ay \right)}^{2}}$
$LHS={{\left( {{a}^{2}}\sin \theta -{{b}^{2}}\sin \theta +2ab\cos \theta \right)}^{2}}+{{\left( {{b}^{2}}\cos \theta -{{a}^{2}}\cos \theta +2absin\theta \right)}^{2}}$ ………… (iii)
We know the basic identity, ${{\left( x+y+z \right)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2zx$.
Similarly,
$\begin{align}
& {{\left( {{a}^{2}}\sin \theta +\left( -{{b}^{2}}\sin \theta \right)+2ab\cos \theta \right)}^{2}} \\
& ={{\left( {{a}^{2}}\sin \theta \right)}^{2}}+{{\left( -{{b}^{2}}\sin \theta \right)}^{2}}+{{\left( 2ab\cos \theta \right)}^{2}}+2\left[ \left( {{a}^{2}}\sin \theta \right)\left( -{{b}^{2}}\sin \theta \right) \right] \\
& +2\left[ \left( -{{b}^{2}}\sin \theta \right)\left( 2ab\cos \theta \right) \right]+2\left[ \left( 2ab\cos \theta \right)\left( {{a}^{2}}\sin \theta \right) \right] \\
\end{align}$
${{a}^{4}}{{\sin }^{2}}\theta +{{b}^{4}}{{\sin }^{2}}\theta +4{{a}^{2}}{{b}^{2}}{{\cos }^{2}}\theta -2{{a}^{2}}{{b}^{2}}{{\sin }^{2}}\theta -4a{{b}^{3}}\sin \theta \cos \theta +4{{a}^{3}}b\sin \theta \cos \theta $……(iv)
Similarly,
$\begin{align}
& {{\left( {{b}^{2}}cos\theta +\left( -{{a}^{2}}cos\theta \right)+2absin\theta \right)}^{2}} \\
& ={{\left( {{b}^{2}}\cos \theta \right)}^{2}}+{{\left( -{{a}^{2}}\cos \theta \right)}^{2}}+{{\left( 2absin\theta \right)}^{2}}+\left[ 2\left( {{b}^{2}}\cos \theta \right)\left( -{{a}^{2}}\cos \theta \right) \right] \\
& +\left[ 2\left( -{{a}^{2}}\cos \theta \right)\left( 2ab\sin \theta \right) \right]+\left[ 2\left( +{{b}^{2}}\cos \theta \right)\left( 2ab\sin \theta \right) \right] \\
\end{align}$
$={{b}^{4}}{{\cos }^{2}}\theta +{{a}^{4}}{{\cos }^{2}}\theta +4{{a}^{2}}{{b}^{2}}{{\sin }^{2}}\theta -2{{a}^{2}}{{b}^{2}}{{\cos }^{2}}\theta -4{{a}^{3}}b\sin \theta \cos \theta +4a{{b}^{3}}\sin \theta \cos \theta $ …..(v)
$\therefore LHS={{\left( {{a}^{2}}\sin \theta -{{b}^{2}}\sin \theta +2ab\cos \theta \right)}^{2}}+{{\left( {{b}^{2}}cos\theta -{{a}^{2}}cos\theta +2absin\theta \right)}^{2}}$
Now substitute the values of (iv), (v) in equation (iii)
$\begin{align}
& \therefore LHS={{a}^{4}}{{\sin }^{2}}\theta +{{b}^{4}}{{\sin }^{2}}\theta +4{{a}^{2}}{{b}^{2}}{{\cos }^{2}}\theta -2{{a}^{2}}{{b}^{2}}{{\sin }^{2}}\theta -4a{{b}^{3}}\sin \theta \cos \theta +4{{a}^{3}}b\sin \theta \cos \theta \\
& +{{b}^{4}}{{\cos }^{2}}\theta +{{a}^{4}}{{\cos }^{2}}\theta +4{{a}^{2}}{{b}^{2}}{{\sin }^{2}}\theta -2{{a}^{2}}{{b}^{2}}{{\cos }^{2}}\theta -4{{a}^{3}}b\sin \theta \cos \theta +4a{{b}^{3}}\sin \theta \cos \theta \\
\end{align}$
Cancel $\left( 4a{{b}^{3}}\sin \theta \cos \theta \right),\left( 4{{a}^{3}}b\sin \theta \cos \theta \right)$ from the above expression. Thus, LHS becomes,
$\therefore LHS={{a}^{4}}{{\sin }^{2}}\theta +{{b}^{4}}{{\sin }^{2}}\theta +4{{a}^{2}}{{b}^{2}}{{\cos }^{2}}\theta -2{{a}^{2}}{{b}^{2}}{{\sin }^{2}}\theta +{{b}^{4}}{{\cos }^{2}}\theta +{{a}^{4}}{{\cos }^{2}}\theta +4{{a}^{2}}{{b}^{2}}{{\sin }^{2}}\theta -2{{a}^{2}}{{b}^{2}}{{\cos }^{2}}\theta $
Now let us pair the terms with respect to $\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)$
We know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
$\therefore LHS={{a}^{4}}\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)+{{b}^{4}}\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)+4{{a}^{2}}{{b}^{2}}\left( si{{n}^{2}}\theta +{{\cos }^{2}}\theta \right)-2{{a}^{2}}{{b}^{2}}\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)$
$\begin{align}
& =\left( {{a}^{4}}\times 1 \right)+\left( {{b}^{4}}\times 1 \right)+\left( 4{{a}^{2}}{{b}^{2}}\times 1 \right)-\left( 2{{a}^{2}}{{b}^{2}}\times 1 \right) \\
& ={{a}^{4}}+{{b}^{4}}+4{{a}^{2}}{{b}^{2}}-2{{a}^{2}}{{b}^{2}} \\
& ={{a}^{4}}+{{b}^{4}}+2{{a}^{2}}{{b}^{2}} \\
& ={{\left( {{a}^{2}} \right)}^{2}}+{{\left( {{b}^{2}} \right)}^{2}}+2{{\left( a \right)}^{2}}{{\left( b \right)}^{2}} \\
\end{align}$
$\left[ \because {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \right]$
$LHS={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}$
Thus, we got LHS = RHS = ${{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}$
Hence, we proved that,
${{\left( ax+by \right)}^{2}}+{{\left( bx-ay \right)}^{2}}={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}$
$\therefore $ LHS = RHS
Note: We need basic trigonometric and algebraic identities, which should be remembered for solving these types of questions.Multiplying the expressions of x and y with a and b we get values of ax,ab,bx and by. Substituting in the L.H.S and simplifying it we get the required answer.The important algebraic formula ${{\left( x+y+z \right)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2zx$ is used for simplification.
Complete step-by-step answer:
We have been given an expression of x and y. we need to show that ${{\left( ax+by \right)}^{2}}+{{\left( bx-ay \right)}^{2}}={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}$
Given, $x=a\sin \theta +b\cos \theta $ ………….. (i)
$y=a\cos \theta -b\sin \theta $……………(ii)
Now multiply equation (i) with a and b and form (ii) expressions. Similarly multiply equation (ii) with a and b and form (ii) expression
Equation (i) $x=a\sin \theta +b\cos \theta $
Multiply by a,
$ax=a\left( \sin \theta +b\cos \theta \right)$
$ax={{a}^{2}}\sin \theta +ab\cos \theta $
Multiply by b,
$\begin{align}
& bx=b\left( a\sin \theta +b\cos \theta \right) \\
& bx=ab\sin \theta +{{b}^{2}}\cos \theta \\
\end{align}$
Similarly, on equation (ii) $y=a\cos \theta -b\sin \theta $
Multiply by a,
$ay={{a}^{2}}\cos \theta -ab\sin \theta $
Multiply by b,
$by=ab\cos \theta -{{b}^{2}}\sin \theta $
These we have 4 expressions
$\begin{align}
& ax={{a}^{2}}\sin \theta +ab\cos \theta \\
& bx=ab\sin \theta +{{b}^{2}}\cos \theta \\
& ay={{a}^{2}}\cos \theta -ab\sin \theta \\
& by=ab\cos \theta -{{b}^{2}}\sin \theta \\
\end{align}$
Now let us find the value of
$ax+by=\left( {{a}^{2}}\sin \theta +ab\cos \theta \right)+\left( ab\cos \theta -{{b}^{2}}\sin \theta \right)={{a}^{2}}\sin \theta -{{b}^{2}}\sin \theta +2ab\cos \theta $
Similarly,
$\begin{align}
& bx-ay=\left( ab\sin \theta +{{b}^{2}}\cos \theta \right)-\left( {{a}^{2}}\cos \theta -ab\sin \theta \right) \\
& =ab\sin \theta +{{b}^{2}}\cos \theta -{{a}^{2}}\cos \theta +ab\sin \theta \\
& ={{b}^{2}}\cos \theta -{{a}^{2}}\cos \theta +2ab\sin \theta \\
\end{align}$
We have been asked to prove that,
${{\left( ax+by \right)}^{2}}+{{\left( bx-ay \right)}^{2}}={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}$
Consider the LHS of expression,
$LHS={{\left( ax+by \right)}^{2}}+{{\left( bx-ay \right)}^{2}}$
Now let us substitute the values of (ax + by), (by – ay)
$LHS={{\left( ax+by \right)}^{2}}+{{\left( bx-ay \right)}^{2}}$
$LHS={{\left( {{a}^{2}}\sin \theta -{{b}^{2}}\sin \theta +2ab\cos \theta \right)}^{2}}+{{\left( {{b}^{2}}\cos \theta -{{a}^{2}}\cos \theta +2absin\theta \right)}^{2}}$ ………… (iii)
We know the basic identity, ${{\left( x+y+z \right)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2zx$.
Similarly,
$\begin{align}
& {{\left( {{a}^{2}}\sin \theta +\left( -{{b}^{2}}\sin \theta \right)+2ab\cos \theta \right)}^{2}} \\
& ={{\left( {{a}^{2}}\sin \theta \right)}^{2}}+{{\left( -{{b}^{2}}\sin \theta \right)}^{2}}+{{\left( 2ab\cos \theta \right)}^{2}}+2\left[ \left( {{a}^{2}}\sin \theta \right)\left( -{{b}^{2}}\sin \theta \right) \right] \\
& +2\left[ \left( -{{b}^{2}}\sin \theta \right)\left( 2ab\cos \theta \right) \right]+2\left[ \left( 2ab\cos \theta \right)\left( {{a}^{2}}\sin \theta \right) \right] \\
\end{align}$
${{a}^{4}}{{\sin }^{2}}\theta +{{b}^{4}}{{\sin }^{2}}\theta +4{{a}^{2}}{{b}^{2}}{{\cos }^{2}}\theta -2{{a}^{2}}{{b}^{2}}{{\sin }^{2}}\theta -4a{{b}^{3}}\sin \theta \cos \theta +4{{a}^{3}}b\sin \theta \cos \theta $……(iv)
Similarly,
$\begin{align}
& {{\left( {{b}^{2}}cos\theta +\left( -{{a}^{2}}cos\theta \right)+2absin\theta \right)}^{2}} \\
& ={{\left( {{b}^{2}}\cos \theta \right)}^{2}}+{{\left( -{{a}^{2}}\cos \theta \right)}^{2}}+{{\left( 2absin\theta \right)}^{2}}+\left[ 2\left( {{b}^{2}}\cos \theta \right)\left( -{{a}^{2}}\cos \theta \right) \right] \\
& +\left[ 2\left( -{{a}^{2}}\cos \theta \right)\left( 2ab\sin \theta \right) \right]+\left[ 2\left( +{{b}^{2}}\cos \theta \right)\left( 2ab\sin \theta \right) \right] \\
\end{align}$
$={{b}^{4}}{{\cos }^{2}}\theta +{{a}^{4}}{{\cos }^{2}}\theta +4{{a}^{2}}{{b}^{2}}{{\sin }^{2}}\theta -2{{a}^{2}}{{b}^{2}}{{\cos }^{2}}\theta -4{{a}^{3}}b\sin \theta \cos \theta +4a{{b}^{3}}\sin \theta \cos \theta $ …..(v)
$\therefore LHS={{\left( {{a}^{2}}\sin \theta -{{b}^{2}}\sin \theta +2ab\cos \theta \right)}^{2}}+{{\left( {{b}^{2}}cos\theta -{{a}^{2}}cos\theta +2absin\theta \right)}^{2}}$
Now substitute the values of (iv), (v) in equation (iii)
$\begin{align}
& \therefore LHS={{a}^{4}}{{\sin }^{2}}\theta +{{b}^{4}}{{\sin }^{2}}\theta +4{{a}^{2}}{{b}^{2}}{{\cos }^{2}}\theta -2{{a}^{2}}{{b}^{2}}{{\sin }^{2}}\theta -4a{{b}^{3}}\sin \theta \cos \theta +4{{a}^{3}}b\sin \theta \cos \theta \\
& +{{b}^{4}}{{\cos }^{2}}\theta +{{a}^{4}}{{\cos }^{2}}\theta +4{{a}^{2}}{{b}^{2}}{{\sin }^{2}}\theta -2{{a}^{2}}{{b}^{2}}{{\cos }^{2}}\theta -4{{a}^{3}}b\sin \theta \cos \theta +4a{{b}^{3}}\sin \theta \cos \theta \\
\end{align}$
Cancel $\left( 4a{{b}^{3}}\sin \theta \cos \theta \right),\left( 4{{a}^{3}}b\sin \theta \cos \theta \right)$ from the above expression. Thus, LHS becomes,
$\therefore LHS={{a}^{4}}{{\sin }^{2}}\theta +{{b}^{4}}{{\sin }^{2}}\theta +4{{a}^{2}}{{b}^{2}}{{\cos }^{2}}\theta -2{{a}^{2}}{{b}^{2}}{{\sin }^{2}}\theta +{{b}^{4}}{{\cos }^{2}}\theta +{{a}^{4}}{{\cos }^{2}}\theta +4{{a}^{2}}{{b}^{2}}{{\sin }^{2}}\theta -2{{a}^{2}}{{b}^{2}}{{\cos }^{2}}\theta $
Now let us pair the terms with respect to $\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)$
We know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
$\therefore LHS={{a}^{4}}\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)+{{b}^{4}}\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)+4{{a}^{2}}{{b}^{2}}\left( si{{n}^{2}}\theta +{{\cos }^{2}}\theta \right)-2{{a}^{2}}{{b}^{2}}\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)$
$\begin{align}
& =\left( {{a}^{4}}\times 1 \right)+\left( {{b}^{4}}\times 1 \right)+\left( 4{{a}^{2}}{{b}^{2}}\times 1 \right)-\left( 2{{a}^{2}}{{b}^{2}}\times 1 \right) \\
& ={{a}^{4}}+{{b}^{4}}+4{{a}^{2}}{{b}^{2}}-2{{a}^{2}}{{b}^{2}} \\
& ={{a}^{4}}+{{b}^{4}}+2{{a}^{2}}{{b}^{2}} \\
& ={{\left( {{a}^{2}} \right)}^{2}}+{{\left( {{b}^{2}} \right)}^{2}}+2{{\left( a \right)}^{2}}{{\left( b \right)}^{2}} \\
\end{align}$
$\left[ \because {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \right]$
$LHS={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}$
Thus, we got LHS = RHS = ${{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}$
Hence, we proved that,
${{\left( ax+by \right)}^{2}}+{{\left( bx-ay \right)}^{2}}={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}$
$\therefore $ LHS = RHS
Note: We need basic trigonometric and algebraic identities, which should be remembered for solving these types of questions.Multiplying the expressions of x and y with a and b we get values of ax,ab,bx and by. Substituting in the L.H.S and simplifying it we get the required answer.The important algebraic formula ${{\left( x+y+z \right)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2zx$ is used for simplification.
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