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Question: If \(x=a\sin \theta +b\cos \theta ,y=a\cos \theta -b\sin \theta \) then show that \({{\left( ax+by \...

If x=asin⁡θ+bcos⁡θ,y=acos⁡θ−bsin⁡θx=a\sin \theta +b\cos \theta ,y=a\cos \theta -b\sin \theta then show that (ax+by)2+(bx−ay)2=(a2+b2)2{{\left( ax+by \right)}^{2}}+{{\left( bx-ay \right)}^{2}}={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}.

Explanation

Solution

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 (ax+by)2+(bx−ay)2=(a2+b2)2{{\left( ax+by \right)}^{2}}+{{\left( bx-ay \right)}^{2}}={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}
Given, x=asin⁡θ+bcos⁡θx=a\sin \theta +b\cos \theta ………….. (i)
y=acos⁡θ−bsin⁡θy=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=asin⁡θ+bcos⁡θx=a\sin \theta +b\cos \theta
Multiply by a,
ax=a(sin⁡θ+bcos⁡θ)ax=a\left( \sin \theta +b\cos \theta \right)
ax=a2sin⁡θ+abcos⁡θax={{a}^{2}}\sin \theta +ab\cos \theta
Multiply by b,
bx=b(asin⁡θ+bcos⁡θ) bx=absin⁡θ+b2cos⁡θ \begin{aligned} & bx=b\left( a\sin \theta +b\cos \theta \right) \\\ & bx=ab\sin \theta +{{b}^{2}}\cos \theta \\\ \end{aligned}
Similarly, on equation (ii) y=acos⁡θ−bsin⁡θy=a\cos \theta -b\sin \theta
Multiply by a,
ay=a2cos⁡θ−absin⁡θay={{a}^{2}}\cos \theta -ab\sin \theta
Multiply by b,
by=abcos⁡θ−b2sin⁡θby=ab\cos \theta -{{b}^{2}}\sin \theta
These we have 4 expressions
ax=a2sin⁡θ+abcos⁡θ bx=absin⁡θ+b2cos⁡θ ay=a2cos⁡θ−absin⁡θ by=abcos⁡θ−b2sin⁡θ \begin{aligned} & 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{aligned}
Now let us find the value of
ax+by=(a2sin⁡θ+abcos⁡θ)+(abcos⁡θ−b2sin⁡θ)=a2sin⁡θ−b2sin⁡θ+2abcos⁡θax+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,
bx−ay=(absin⁡θ+b2cos⁡θ)−(a2cos⁡θ−absin⁡θ) =absin⁡θ+b2cos⁡θ−a2cos⁡θ+absin⁡θ =b2cos⁡θ−a2cos⁡θ+2absin⁡θ \begin{aligned} & 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{aligned}
We have been asked to prove that,
(ax+by)2+(bx−ay)2=(a2+b2)2{{\left( ax+by \right)}^{2}}+{{\left( bx-ay \right)}^{2}}={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}
Consider the LHS of expression,
LHS=(ax+by)2+(bx−ay)2LHS={{\left( ax+by \right)}^{2}}+{{\left( bx-ay \right)}^{2}}
Now let us substitute the values of (ax + by), (by – ay)
LHS=(ax+by)2+(bx−ay)2LHS={{\left( ax+by \right)}^{2}}+{{\left( bx-ay \right)}^{2}}
LHS=(a2sin⁡θ−b2sin⁡θ+2abcos⁡θ)2+(b2cos⁡θ−a2cos⁡θ+2absinθ)2LHS={{\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, (x+y+z)2=x2+y2+z2+2xy+2yz+2zx{{\left( x+y+z \right)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2zx.
Similarly,
(a2sin⁡θ+(−b2sin⁡θ)+2abcos⁡θ)2 =(a2sin⁡θ)2+(−b2sin⁡θ)2+(2abcos⁡θ)2+2[(a2sin⁡θ)(−b2sin⁡θ)] +2[(−b2sin⁡θ)(2abcos⁡θ)]+2[(2abcos⁡θ)(a2sin⁡θ)] \begin{aligned} & {{\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{aligned}
a4sin⁡2θ+b4sin⁡2θ+4a2b2cos⁡2θ−2a2b2sin⁡2θ−4ab3sin⁡θcos⁡θ+4a3bsin⁡θcos⁡θ{{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,
(b2cosθ+(−a2cosθ)+2absinθ)2 =(b2cos⁡θ)2+(−a2cos⁡θ)2+(2absinθ)2+[2(b2cos⁡θ)(−a2cos⁡θ)] +[2(−a2cos⁡θ)(2absin⁡θ)]+[2(+b2cos⁡θ)(2absin⁡θ)] \begin{aligned} & {{\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{aligned}
=b4cos⁡2θ+a4cos⁡2θ+4a2b2sin⁡2θ−2a2b2cos⁡2θ−4a3bsin⁡θcos⁡θ+4ab3sin⁡θcos⁡θ={{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)
∴LHS=(a2sin⁡θ−b2sin⁡θ+2abcos⁡θ)2+(b2cosθ−a2cosθ+2absinθ)2\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)
∴LHS=a4sin⁡2θ+b4sin⁡2θ+4a2b2cos⁡2θ−2a2b2sin⁡2θ−4ab3sin⁡θcos⁡θ+4a3bsin⁡θcos⁡θ +b4cos⁡2θ+a4cos⁡2θ+4a2b2sin⁡2θ−2a2b2cos⁡2θ−4a3bsin⁡θcos⁡θ+4ab3sin⁡θcos⁡θ \begin{aligned} & \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{aligned}
Cancel (4ab3sin⁡θcos⁡θ),(4a3bsin⁡θcos⁡θ)\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,
∴LHS=a4sin⁡2θ+b4sin⁡2θ+4a2b2cos⁡2θ−2a2b2sin⁡2θ+b4cos⁡2θ+a4cos⁡2θ+4a2b2sin⁡2θ−2a2b2cos⁡2θ\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 (sin⁡2θ+cos⁡2θ)\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)
We know that sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1
∴LHS=a4(sin⁡2θ+cos⁡2θ)+b4(sin⁡2θ+cos⁡2θ)+4a2b2(sin2θ+cos⁡2θ)−2a2b2(sin⁡2θ+cos⁡2θ)\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)
=(a4×1)+(b4×1)+(4a2b2×1)−(2a2b2×1) =a4+b4+4a2b2−2a2b2 =a4+b4+2a2b2 =(a2)2+(b2)2+2(a)2(b)2 \begin{aligned} & =\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{aligned}
[∵(a+b)2=a2+b2+2ab]\left[ \because {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \right]
LHS=(a2+b2)2LHS={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}
Thus, we got LHS = RHS = (a2+b2)2{{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}
Hence, we proved that,
(ax+by)2+(bx−ay)2=(a2+b2)2{{\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 (x+y+z)2=x2+y2+z2+2xy+2yz+2zx{{\left( x+y+z \right)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2zx is used for simplification.