Question

In $\cos x+\cos y=a$ , $\cos 2x+\cos 2y=b$ , $\cos 3x+\cos 3y=c$ , then find the value of $a\left( 2{{a}^{2}}-3 \right)$
$\begin{aligned} & \text{a) ab - c} \\\ & \text{b) 2ab - c} \\\ & \text{c) 3ab - c} \\\ & \text{d) 3ac - b} \\\ \end{aligned}$

Answer 1

Now we are given with $\cos 2x+\cos 2y=b$ and we know $\cos 2\theta =2{{\cos }^{2}}\theta -1$ hence using this we will find the value of ${{\cos }^{2}}x+{{\cos }^{2}}y$ . Now we know the value of ${{\cos }^{2}}x+{{\cos }^{2}}y$ and also $\cos x+\cos y=a$ . Hence we will use formula ${{\left( a+b \right)}^{2}}-2ab={{a}^{2}}+{{b}^{2}}$ to find the value of $\cos x\cos y$ . Now consider the equation $\cos 3x+\cos 3y=c$ and we know that $\cos 3\theta =4{{\cos }^{3}}\theta -3\cos \theta$ to this equation we will use the formula ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$ and then substitute the values of $\cos x+\cos y$ , ${{\cos }^{2}}x+{{\cos }^{2}}y$ and $\cos x\cos y$ . Hence we will find a relation in a, b and c. we will rearrange and simplify the equation to arrive at final equation

Complete step-by-step answer:
Now we are given that $\cos x+\cos y=a$
Let $\cos x+\cos y=a..................(1)$
Now consider the equation.
$\cos 2x+\cos 2y=b$
Now we know that $\cos 2\theta =2{{\cos }^{2}}\theta -1$
Hence we have.
$\begin{aligned} & 2{{\cos }^{2}}x-1+2{{\cos }^{2}}y-1=b \\\ & \Rightarrow 2{{\cos }^{2}}x-2+2{{\cos }^{2}}y=b \\\ \end{aligned}$
Now taking – 2 to RHS we get
$2{{\cos }^{2}}x+2{{\cos }^{2}}y=b+2$
Now dividing the whole equation by 2 we get
${{\cos }^{2}}x+{{\cos }^{2}}y=\left( \dfrac{b+2}{2} \right)....................(2)$
Now we know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
Taking 2ab to the RHS we get the formula as ${{\left( a+b \right)}^{2}}-2ab={{a}^{2}}+{{b}^{2}}$ .
Now using this formula to equation (2) we get
${{\left( \cos x+\cos y \right)}^{2}}-2\cos x\cos y=\left( \dfrac{b+2}{2} \right)$
Now we know from equation (1) that $\cos x+\cos y=a$ hence using this substitution we get.
${{a}^{2}}-2\cos x\cos y=\left( \dfrac{b+2}{2} \right)$
Now taking $2\cos x\cos y$ to RHS we get
${{a}^{2}}-\left( \dfrac{b+2}{2} \right)=2\cos x\cos y$
Dividing the whole equation by 2 we get
$\cos x\cos y=\dfrac{{{a}^{2}}-\left( \dfrac{b+2}{2} \right)}{2}$
$\Rightarrow \cos x\cos y=\dfrac{{{a}^{2}}}{2}-\dfrac{b+2}{4}................(3)$
Now consider the equation $\cos 3x+\cos 3y=c$
We know that $\cos 3\theta =4{{\cos }^{3}}\theta -3\cos \theta$
Using this formula in the above equation we get.
$4{{\cos }^{3}}x-3\cos x+4{{\cos }^{3}}y-3\cos y=c$
Now rearranging the terms we get
$4{{\cos }^{3}}x+4{{\cos }^{3}}y-3\cos x-3\cos y=c$
Now let us take 4 and – 3 common from the equation.
$4\left( {{\cos }^{3}}x+{{\cos }^{3}}y \right)-3\left( \cos x+\cos y \right)=c$
Now we know that ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$ hence we will use this formula in the above equation to get
$4\left[ \left( \cos x+\cos y \right)\left( {{\cos }^{2}}x-\cos x\cos y+{{\cos }^{2}}y \right) \right]-3\left( \cos x+\cos y \right)=c$
Rearranging the terms in the above equation we get
$4\left[ \left( \cos x+\cos y \right)\left( \left( {{\cos }^{2}}x+{{\cos }^{2}}y \right)-\left( \cos x\cos y \right) \right) \right]-3\left( \cos x+\cos y \right)=c$
Now let us substitute the values from equation (1), equation (2) and equation (3).
Hence we get,
$4\left[ \left( a \right)\left( \left( \dfrac{b+2}{2} \right)-\left( \dfrac{{{a}^{2}}}{2}-\dfrac{b+2}{4} \right) \right) \right]-3\left( a \right)=c$
Now we have an equation in a, b and c. to arrive at final answer we will have to just simplify the equation.

& \Rightarrow 4\left[ \left( a \right)\left( \dfrac{b+2}{2}-\dfrac{{{a}^{2}}}{2}+\dfrac{b+2}{4} \right) \right]-3\left( a \right)=c \\\ & \Rightarrow 4\left[ \left( a \right)\left( \dfrac{2b+4}{4}-\dfrac{{{a}^{2}}}{2}+\dfrac{b+2}{4} \right) \right]-3\left( a \right)=c \\\ & \Rightarrow 4\left[ \left( a \right)\left( \dfrac{2b+4+b+2}{4}-\dfrac{{{a}^{2}}}{2} \right) \right]-3\left( a \right)=c \\\ & \Rightarrow 4\left[ \left( a \right)\left( \dfrac{3b+6}{4}-\dfrac{{{a}^{2}}}{2} \right) \right]-3\left( a \right)=c \\\ & \Rightarrow 4\left[ \left( a \right)\left( \dfrac{3\left( b+2 \right)}{4}-\dfrac{{{a}^{2}}}{2} \right) \right]-3\left( a \right)=c \\\ \end{aligned}$$ Now taking 4 inside the bracket we get $$\left[ \left( a \right)\left( 3\left( b+2 \right)-2{{a}^{2}} \right) \right]-3\left( a \right)=c$$ $$\begin{aligned} & \Rightarrow \left[ \left( a \right)\left( 3\left( b+2 \right)-2{{a}^{2}} \right) \right]-3\left( a \right)=c \\\ & \Rightarrow 3ab+6a-2{{a}^{3}}-3a=c \\\ & \Rightarrow 3ab-2{{a}^{3}}+3a=c \\\ & \Rightarrow 3ab-a\left( 2{{a}^{2}}-1 \right)=c \\\ \end{aligned}$$ Rearranging terms by shifting $a\left( 2{{a}^{2}}-1 \right)$ to RHS and c to LHS we get $$3ab-c=a\left( {{a}^{2}}-1 \right)$$ Hence we finally get $$a\left( {{a}^{2}}-1 \right)=3ab-c$$ **Note:** Note that the formula for ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$ and not $\left( a+b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$ . Also once we get the equation $$4\left[ \left( a \right)\left( \left( \dfrac{b+2}{2} \right)-\left( \dfrac{{{a}^{2}}}{2}-\dfrac{b+2}{4} \right) \right) \right]-3\left( a \right)=c$$ while simplifying the equation note that we need to find the value of $$a\left( {{a}^{2}}-1 \right)$$ and hence evaluate accordingly.