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Question: If \(\sin \theta \) and \(\cos \theta \) are the roots of the equation \(m{{x}^{2}}+nx+1=0\), then f...

If sin⁡θ\sin \theta and cos⁡θ\cos \theta are the roots of the equation mx2+nx+1=0m{{x}^{2}}+nx+1=0, then find the relation between m and n.

Explanation

Solution

We have the roots of the equation and we know that the sum of the roots of an equation = (-coefficient of x)/ (coefficient of x2) and the product of the roots =(constant)/ (coefficient of x2) if the equation is of the form ax2+bx+c=0a{{x}^{2}}+bx+c=0 . We will find the sum of roots and product of roots and use them to find the relation between m and n.

Complete step by step answer:
We have equation mx2+nx+1=0m{{x}^{2}}+nx+1=0 whose roots are sin⁡θ\sin \theta and cos⁡θ\cos \theta
We know that the sum of roots = −ba\dfrac{-b}{a} ,we have,
⇒sin⁡θ+cos⁡θ=−nm.........(i)\Rightarrow \sin \theta +\cos \theta =\dfrac{-n}{m}.........(i)
We know that the product of roots = ca\dfrac{c}{a}
⇒cos⁡θsin⁡θ=1m.........(ii)\Rightarrow \cos \theta \sin \theta =\dfrac{1}{m}.........(ii)
We will now square both the sides of equation(i), we will get,
⇒(sin⁡θ+cos⁡θ)2=(−nm)2 ⇒sin⁡2θ+cos⁡2θ+2sin⁡θcos⁡θ=n2m2  \begin{aligned} & \Rightarrow {{\left( \sin \theta +\cos \theta \right)}^{2}}={{\left( \dfrac{-n}{m} \right)}^{2}} \\\ & \Rightarrow {{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \cos \theta =\dfrac{{{n}^{2}}}{{{m}^{2}}} \\\ & \\\ \end{aligned}
We know that sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1, we will get,
⇒1+2sin⁡θcos⁡θ=n2m2\Rightarrow 1+2\sin \theta \cos \theta =\dfrac{{{n}^{2}}}{{{m}^{2}}}
We will put the value of sin⁡θcos⁡θ\sin \theta \cos \theta from equation(ii), we will get,
⇒1+2m=n2m2\Rightarrow 1+\dfrac{2}{m}=\dfrac{{{n}^{2}}}{{{m}^{2}}}
We will solve the above equation
⇒1+2m=n2m2 ⇒m+2m=n2m2 ⇒m+2=n2m ⇒m2+2m=n2 ⇒m2−n2+2m=0 \begin{aligned} & \Rightarrow 1+\dfrac{2}{m}=\dfrac{{{n}^{2}}}{{{m}^{2}}} \\\ & \Rightarrow \dfrac{m+2}{m}=\dfrac{{{n}^{2}}}{{{m}^{2}}} \\\ & \Rightarrow m+2=\dfrac{{{n}^{2}}}{m} \\\ & \Rightarrow {{m}^{2}}+2m={{n}^{2}} \\\ & \Rightarrow {{m}^{2}}-{{n}^{2}}+2m=0 \\\ \end{aligned}
We will get the above relation between m and n.

Note:
Alternate Method
We have equation mx2+nx+1=0m{{x}^{2}}+nx+1=0 whose roots are sin⁡θ\sin \theta and cos⁡θ\cos \theta
We know that the roots satisfy the equation
So will put value of x= sin⁡θ\sin \theta , we will get,
⇒msin⁡2θ+nsin⁡θ+1=0..........(i)\Rightarrow m{{\sin }^{2}}\theta +n\sin \theta +1=0..........(i)
We will put value of x= cos⁡θ\cos \theta , we will get,
⇒mcos⁡2θ+ncos⁡2θ+1=0........(ii)\Rightarrow m{{\cos }^{2}}\theta +n{{\cos }^{2}}\theta +1=0........(ii)
We will add equation(i) and equation(ii), we will get
⇒m(sin⁡2θ+cos⁡2θ)+n(sin⁡θ+cos⁡θ)+2=0\Rightarrow m({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )+n(\sin \theta +\cos \theta )+2=0
We know that sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1, we will get,
⇒m+n(sin⁡θ+cos⁡θ)+2=0.........(iii)\Rightarrow m+n(\sin \theta +\cos \theta )+2=0.........(iii)
We know that the sum of roots = −ba\dfrac{-b}{a} ,we have,
⇒sin⁡θ+cos⁡θ=−nm\Rightarrow \sin \theta +\cos \theta =\dfrac{-n}{m}
Putting the value of sin⁡θ+cos⁡θ\sin \theta +\cos \theta in equation(iii), we wiil get,
⇒m+n(−nm)+2=0\Rightarrow m+n\left( \dfrac{-n}{m} \right)+2=0
Solving the above equation, we will get,
⇒m2−n2+2m=0\Rightarrow {{m}^{2}}-{{n}^{2}}+2m=0