Question

cos(x-y)+cos(y-z)+cos(z-x) = -3/2 then,
$\sum \cos(x)$
​

Answer 1

Given:

• cos(x-y)+cos(y-z)+cos(z-x) = -3/2

To Find:

• Value of ∑cos(x)

Solution:

We know that,

➳ cos(A-B)= cosAcosB+sinAsinB

➳ sin²x+cos²x= 1

➳ (a+b+c)²= a²+b²+c²+2(ab+bc+ca)

➳ ∑cos(x)= cosx+cosy+cosz

Now,

$\sf{cos(x-y)+cos(y-z)+cos(z-x) = \dfrac{-3}{2}}$

$\sf{2\big(cos(x-y)+cos(y-z)+cos(z-x)\big)= -3}$

$\sf{3+2\big(cos(x-y)+cos(y-z)+cos(z-x)\big)=0}$

$\sf{3+2cos(x-y)+2cos(y-z)+2cos(z-x)=0---(1)}$

Now,

3= 1+1+1

3=(sin²x+cos²x)+(sin²y+cos²y)+(sin²z+cos²z)

3=sin²x+sin²y+sin²z+cos²x+cos²y+cos²z

Also,

$\sf{2cos(x-y)+2cos(y-z)+2cos(z-x)}$

$\sf{=2(cosxcosy+sinxsiny)+2(cosycosz+sinysinz)+2(coszcosx+sinzsinx)}$

$\sf{=2cosxcosy+2sinxsiny+2cosycosz+2sinysinz+2coszcosx+2sinzsinx}$

$\sf{=2sinx siny+2sinysinz+2sinzsinx+2cosxcosy+2cosycosz+2coszcosx}$

$\sf{=2(sinx siny+sinysinz+sinzsinx)+2(cosxcosy+cosycosz+coszcosx)}$

On putting value of 3 and

2cos(x-y)+2cos(y-z)+2cos(z-x) in (1), we get

⟼ sin²x+sin²y+sin²z+cos²x+cos²y+cos²z

+2(sinx siny+sinysinz+sinzsinx)

+2(cosxcosy+cosycosz+coszcosx)=0

For simplicity just break above equation into two parts

$\sf{A=sin^{2}x+sin^{2}y+sin^{2}z+2(sinx siny+sinysinz+sinzsinx)}$

$\sf{B=cos^{2}x+cos^{2}y+cos^{2}z+2(cosxcosy+cosycosz+coszcosx)}$

where, A+B=0

Now,

$\sf{A=(\sin x+\sin y+\sin z)^{2}}$

$\sf{B=(\cos x+\cos y+\cos z)^{2}}$

On combining A and B, we get our original equation

i.e. A+B=0

$\sf{(\sin x+\sin y+\sin z)^{2}+(\cos x+\cos y+\cos z)^{2}=0}$

We know that square of any number is always positive

So, Sum of above two terms is always positive or can be equal to zero

Now, in the above condition sum of the squares will give the result zero only when, both the terms are equal to zero.

∴ (sinx+siny+sinz)²= 0

And

(cosx+cosy+cosz)²= 0

⇒ cosx+cosy+cosz= 0

⇒ ∑cos(x)= 0

Hence, the value of ∑cos(x) is 0.

Answer 2

$\Large\boxed{\sf{\quad\sum\cos x=0\quad}}$

Solution:-

Given,

$\displaystyle\longrightarrow\sf{\cos(x-y)+\cos(y-z)+\cos(z-x)=-\dfrac{3}{2}}$

From this we obtain an equation,

$\displaystyle\longrightarrow\sf{3+2[\cos(x-y)+\cos(y-z)+\cos(z-x)]=0\quad\quad\dots(1)}$

But we see that,

• $\sf{\sin^2x+\cos^2x=1}$

• $\sf{\sin^2y+\cos^2y=1}$

• $\sf{\sin^2z+\cos^2z=1}$

So that,

$\displaystyle\longrightarrow\sf{(\sin^2x+\cos^2x)+(\sin^2y+\cos^2y)+(\sin^2z+\cos^2z)=1+1+1}$

$\displaystyle\longrightarrow\sf{\sin^2x+\sin^2y+\sin^2z+\cos^2x+\cos^2y+\cos^2z=3}$

Hence (1) becomes,

$\displaystyle\begin{aligned}\longrightarrow\ \ &\sf{\sin^2x+\sin^2y+\sin^2z+\cos^2x+\cos^2y+\cos^2z}&\\+\, &\sf{2\left[\cos(x-y)+\cos(y-z)+\cos(z-x)\right]}&\sf{=0\quad\quad\dots(2)}\end{aligned}$

But,

• $\sf{\cos(x-y)=\cos x\cos y+\sin x\sin y}$

• $\sf{\cos(y-z)=\cos y\cos z+\sin y\sin z}$

• $\sf{\cos(z-x)=\cos z\cos x+\sin z\sin x}$

Then (2) becomes,

$\displaystyle\begin{aligned}\longrightarrow\ \ &\sf{\sin^2x+\sin^2y+\sin^2z+\cos^2x+\cos^2y+\cos^2z}&\\+\, &\sf{2[\cos x\cos y+\sin x\sin y+\cos y\cos z}&\\+\, &\sf{\sin y\sin z+\cos z\cos x+\sin z\sin x]}&\sf{=0}\end{aligned}$

$\displaystyle\begin{aligned}\longrightarrow\ \ &\sf{\sin^2x+\sin^2y+\sin^2z+\cos^2x+\cos^2y+\cos^2z}&\\+\, &\sf{2[\sin x\sin y+\sin y\sin z+\sin z\sin x]}&\\+\, &\sf{2[\cos x\cos y+\cos y\cos z+\cos z\cos x]}&\sf{=0}\end{aligned}$

$\displaystyle\begin{aligned}\longrightarrow\ \ &\sf{\sin^2x+\sin^2y+\sin^2z+2[\sin x\sin y+\sin y\sin z+\sin z\sin x]}&\\+\, &\sf{\cos^2x+\cos^2y+\cos^2z+2[\cos x\cos y+\cos y\cos z+\cos z\cos x]}&\sf{=0\quad\quad\dots(3)}\end{aligned}$

But we know that,

• $\sf{a^2+b^2+c^2+2(ab+bc+ca)=(a+b+c)^2}$

Hence we see that,

• $\sf{\sin^2x+\sin^2y+\sin^2z+2(\sin x\sin y+\sin y\sin z+\sin z\sin x)=(\sin x+\sin y+\sin z)^2}$

• $\sf{\cos^2x+\cos^2y+\cos^2z+2(\cos x\cos y+\cos y\cos z+\cos z\cos x)=(\cos x+\cos y+\cos z)^2}$

Therefore (3) becomes,

$\displaystyle\longrightarrow\sf{(\sin x+\sin y+\sin z)^2+(\cos x+\cos y+\cos z)^2=0\quad\quad\dots(4)}$

Well, consider the equation $\sf{a^2+b^2=0}$ for $\sf{a,\ b\in\mathbb{R}.}$ For it being satisfied, one among $\sf{a^2}$ and $\sf{b^2}$ should be negative to the other but it's not possible since the square of a real number is always non - negative.

$\displaystyle\longrightarrow\sf{a^2+b^2=0}$

$\displaystyle\longrightarrow\sf{a^2=-b^2}$

$\displaystyle\longrightarrow\sf{a=\pm\sqrt{-b^2}}$

$\displaystyle\longrightarrow\sf{a=\pm ib}$

For some non - zero $\sf{b,}$ the equation contradicts our assumption that $\sf{a\in\mathbb{R}.}$ Also the equation can contradict $\sf{b\in\mathbb{R}}$ for some non - zero $\sf{a.}$ Hence the equation has unique solution.

$\displaystyle\longrightarrow\sf{a=b=0}$

Therefore, from (4) we get,

$\displaystyle\longrightarrow\sf{\underline{\underline{\cos x+\cos y+\cos z=0}}}$

Or,

$\displaystyle\longrightarrow\sf{\underline{\underline{\sum\cos x=0}}}$

Also,

$\displaystyle\longrightarrow\sf{\sin x+\sin y+\sin z=0}$

Or,

$\displaystyle\longrightarrow\sf{\sum\sin x=0}$