Question

If trigonometric equation is given as $\cos x+\cos y=a,\cos 2x+\cos 2y=b,\cos 3x+\cos 3y=c,$ then:
(A). ${{\cos }^{2}}x+{{\cos }^{2}}y=1+\dfrac{b}{2}$
(B). $\cos x\cos y=\dfrac{{{a}^{2}}}{2}-\left( \dfrac{b+2}{4} \right)$
(C). $2{{a}^{3}}+c=3a\left( 1+b \right)$
(D). $a+b+c=3abc$

Answer 1

Hint: First take 2nd equation. put $\cos 2x=2{{\cos }^{2}}x-1$ and simplify it. From this get the value for $\cos x\cos y$ .Now take the 3rd equation. Put $\cos 3x=4{{\cos }^{3}}x-3\cos x$ and simplify. Apply these basic identities and simplify until the terms are only a, b and c.

Complete step-by-step solution -
We have been given three trigonometric equations which are,
$\cos x+\cos y=a$ …………….(1)
$\cos 2x+\cos 2y=b$ ………..(2)
$\cos 3x+\cos 3y=c$ ………….(3)
Let us first take equation (2)
$\cos 2x+\cos 2y=b$ .
We know that $\cos 2x=2{{\cos }^{2}}x-1$ , which is a basic trigonometric identity. Thus substitute this in equation (2)
$2{{\cos }^{2}}x-1+2{{\cos }^{2}}y-1=b$ , let us simplify this.
$\begin{aligned} & 2\left( {{\cos }^{2}}x+{{\cos }^{2}}y \right)-2=b \\\ & 2\left( {{\cos }^{2}}x+{{\cos }^{2}}y \right)=b+2 \\\ & {{\cos }^{2}}x+{{\cos }^{2}}y=\dfrac{b+2}{2}=1+\dfrac{b}{2} \\\ \end{aligned}$
i.e. ${{\cos }^{2}}x+{{\cos }^{2}}y=1+\dfrac{b}{2}$ …………………… (4)
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ .
Thus we can write ${{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab$ . Simplify for ${{\cos }^{2}}x+{{\cos }^{2}}y$we can write,
Where $a=\cos x$ and $b=\cos y$ ,
$\begin{aligned} & \therefore {{\cos }^{2}}x+{{\cos }^{2}}y=\left( \cos x+\cos y \right)-2\cos x\cos y \\\ & \therefore {{\left( \cos x+\cos y \right)}^{2}}-2\cos x\cos y=1+\dfrac{b}{2} \\\ \end{aligned}$
From (1) we know that $\cos x+\cos y=a$ . thus substitute this in the above expression.
$\therefore {{a}^{2}}-2\cos x\cos y=1+\dfrac{b}{2}$ .
Thus, ${{a}^{2}}-\left( \dfrac{b+2}{2} \right)=2\cos x\cos y$
$\therefore \cos x\cos y=\dfrac{{{a}^{2}}}{2}-\left( \dfrac{b+2}{4} \right)$ ………….(5)
Now, we have also been given $\cos 3x+\cos 3y=c$ .We know the formula. $\cos 3x=4{{\cos }^{3}}x-3\cos$ . substitute this in (2) we get,

& 4{{\cos }^{3}}x-3\cos x+4{{\cos }^{3}}y-3\cos y=c \\\ & 4\left[ {{\cos }^{3}}x+{{\cos }^{3}}y \right]-3\left[ \cos x+\cos y \right]=c \\\ \end{aligned}$$ We know that ${{x}^{3}}+{{y}^{3}}=\left( x+y \right)\left( {{x}^{2}}-xy+{{y}^{2}} \right)$ . Similarly for ${{\cos }^{3}}x+{{\cos }^{3}}y=\left( \cos x+\cos y \right)\left[ {{\cos }^{2}}x-\cos x\cos y+{{\cos }^{2}}y \right]$ Let us substitute this in the place of ${{\cos }^{3}}x+{{\cos }^{3}}y$ in the above expression. Hence we get, $4\left[ \left( \cos x+\cos y \right)\left( {{\cos }^{2}}x-\cos x\cos y+{{\cos }^{2}}y \right) \right]-3\left( \cos x+\cos y \right)=c$ Now put $\cos x+\cos y=a$ . ${{\cos }^{2}}x+{{\cos }^{2}}y=\dfrac{b+a}{2}$ and $$\cos x\cos y=\dfrac{{{a}^{2}}}{\alpha }-\left( \dfrac{b+2}{4} \right)$$ Thus we get, $4\left[ \left( a \right)\left[ \left( \dfrac{b+2}{2} \right)-\dfrac{{{a}^{2}}}{\alpha }+\left( \dfrac{b+2}{4} \right) \right] \right]-3a=c$ Let us simplify the above expression. $\begin{aligned} & 4a\left[ \left( \dfrac{b+2}{2} \right)+\left( \dfrac{b+2}{4} \right)-\dfrac{{{a}^{2}}}{\alpha } \right]-3a=c \\\ & 4a\left[ \dfrac{2b+4+b+2}{4}-\dfrac{{{a}^{2}}}{\alpha } \right]-3a=c \\\ & 4a\left[ \left( \dfrac{3b+6}{4} \right)-\dfrac{{{a}^{2}}}{\alpha } \right]-3a=c \\\ & \Rightarrow 3\times 4a\left( \dfrac{b+2}{4} \right)-4a\times \dfrac{{{a}^{2}}}{\alpha }-3a=c \\\ & \Rightarrow 3ab+6a-2{{a}^{3}}-3a=c \\\ & \Rightarrow 3ab+3a-2{{a}^{3}}=c \\\ & \therefore 3a\left( b+1 \right)=2{{a}^{3}}+c \\\ \end{aligned}$ Thus we got the required simplification, $3a\left( b+1 \right)=2{{a}^{3}}+c$ Thus we got, $\begin{aligned} & {{\cos }^{2}}x+{{\cos }^{2}}y=1+\dfrac{b}{2} \\\ & \cos x\cos y=\dfrac{{{a}^{2}}}{\alpha }-\left( \dfrac{b+2}{4} \right) \\\ \end{aligned}$ And $2{{a}^{3}}+c=3a\left( 1+b \right)$ . $\therefore $ Option (a) , (b) and (C) are the correct answers. Note: We have used a lot of trigonometric identities and basic formulas. Thus, it is important that you learn the formulas, so solving problems like these will be easier for you. Just apply the formula and simplify it, you will get the required answer.