Question

If $\alpha \text{ and }\beta $ are the solutions of $a\cos \theta +b\sin \theta =c$, then show that $\cos \alpha +\cos \beta =\dfrac{2ac}{{{a}^{2}}+{{b}^{2}}}\text{ and }\cos \alpha \times \cos \beta =\dfrac{\left( {{c}^{2}}-{{b}^{2}} \right)}{\left( {{a}^{2}}+{{b}^{2}} \right)}$.

Answer 1

Hint: In this question, we are given an equation $a\cos \theta +b\sin \theta =c$ and $\alpha \text{ and }\beta $ are its roots. We have to prove that $\cos \alpha +\cos \beta =\dfrac{2ac}{{{a}^{2}}+{{b}^{2}}}\text{ and }\cos \alpha \times \cos \beta =\dfrac{\left( {{c}^{2}}-{{b}^{2}} \right)}{\left( {{a}^{2}}+{{b}^{2}} \right)}$. For this, we will convert the given equation into the form of a quadratic equation in terms of $\cos \theta $. Since values of $\theta $ are $\alpha \text{ and }\beta $, so the roots of the formed equation will be $\cos \alpha \text{ and cos}\beta $. After that, we will apply the sum of roots and product of roots formula to get our required answer. For an equation $a{{x}^{2}}+bx+c$ if $\alpha \text{ and }\beta $ are roots then sum of roots is given by $\dfrac{-b}{a}$ i.e. $\alpha +\beta =\dfrac{-b}{a}$ and product of roots is $\dfrac{c}{a}$ i.e. $\alpha \beta =\dfrac{c}{a}$.

Complete step by step answer:
Here, we are given the equation as $a\cos \theta +b\sin \theta =c$.
Let us convert this equation into a quadratic equation in the form of $\cos \theta $.
\[\Rightarrow a\cos \theta +b\sin \theta =c\cdots \cdots \cdots \cdots \left( 1 \right)\]
Rearranging the terms, we get:
\[\Rightarrow c-a\cos \theta =b\sin \theta \]
Squaring both sides we get:
\[\Rightarrow {{\left( c-a\cos \theta \right)}^{2}}={{\left( b\sin \theta \right)}^{2}}\]
We know, ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ then applying this, we get:
\[\Rightarrow {{c}^{2}}+{{a}^{2}}{{\cos }^{2}}\theta -2ac\cos \theta ={{b}^{2}}{{\sin }^{2}}\theta \]
We know, ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, therefore we can write ${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $. Hence we get:
\[\begin{align}
  & \Rightarrow {{c}^{2}}+{{a}^{2}}{{\cos }^{2}}\theta -2ac\cos \theta ={{b}^{2}}\left( 1-{{\cos }^{2}}\theta \right) \\
 & \Rightarrow {{c}^{2}}+{{a}^{2}}{{\cos }^{2}}\theta -2ac\cos \theta ={{b}^{2}}-{{b}^{2}}{{\cos }^{2}}\theta \\
\end{align}\]
Rearranging the terms, we get:
\[\Rightarrow \left( {{a}^{2}}+{{b}^{2}} \right){{\cos }^{2}}\theta -2ac\cos \theta +\left( {{c}^{2}}-{{b}^{2}} \right)=0\cdots \cdots \cdots \left( 2 \right)\]
Now, we have formed a quadratic equation in terms of $\cos \theta $. Since values of $\theta $ for equation (1) were $\alpha \text{ and }\beta $ so roots of (2) will be in form of $\cos \theta $ and hence $\cos \alpha \text{ and cos}\beta $.
Therefore, $\cos \alpha \text{ and cos}\beta $ are roots of equation (2).
We know, if ${{x}_{1}},{{x}_{2}}$ are roots of equation $a{{x}^{2}}+bx+c=0$ then sum of roots ${{x}_{1}}+{{x}_{2}}$ is given by $\dfrac{-b}{a}$ and product of roots ${{x}_{1}}+{{x}_{2}}$ is given by $\dfrac{c}{a}$.
Comparing equation (2) with $a{{x}^{2}}+bx+c=0$ we get:
$x=\cos \theta ,a={{a}^{2}}+{{b}^{2}},b=-2ac\text{ and }c={{c}^{2}}-{{b}^{2}}$.
Roots of equation (2) are $\cos \alpha \text{ and cos}\beta $ therefore, sum of roots will be $\dfrac{-b}{a}=\dfrac{-\left( -2ac \right)}{{{a}^{2}}+{{b}^{2}}}$.
Hence, \[\Rightarrow \cos \alpha +\text{cos}\beta =\dfrac{2ac}{{{a}^{2}}+{{b}^{2}}}\]
Also, product of roots will be $\dfrac{c}{a}=\dfrac{{{c}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}$.
Hence, \[\Rightarrow \cos \alpha \text{cos}\beta =\dfrac{2ac}{{{a}^{2}}+{{b}^{2}}}\]
Hence proved.

Note: Students should know the formulas of finding sum and product of roots. Students can make mistakes in positive and negative signs while arranging the equation. Students should keep in mind the basic trigonometric identities for solving questions such as ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.