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
Verified515.1k+ views
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.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success
Master Class 12 Maths: Engaging Questions & Answers for Success
Master Class 12 Biology: Engaging Questions & Answers for Success
Master Class 12 Physics: Engaging Questions & Answers for Success
Master Class 12 Business Studies: Engaging Questions & Answers for Success
Master Class 12 English: Engaging Questions & Answers for Success
Trending doubts
The gas that burns in oxygen with a green flame is class 12 chemistry CBSE
The probability that a leap year will have only 52 class 12 maths CBSE
Describe the poetic devices used in the poem Aunt Jennifers class 12 english CBSE
And such too is the grandeur of the dooms We have imagined class 12 english CBSE
What does the god that failed refer to class 12 english CBSE
Which country did Danny Casey play for class 12 english CBSE