cos(x-y)+cos(y-z)+cos(z-x) = -3/2 then,
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Answers
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,
Now,
3= 1+1+1
3=(sin²x+cos²x)+(sin²y+cos²y)+(sin²z+cos²z)
3=sin²x+sin²y+sin²z+cos²x+cos²y+cos²z
Also,
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
where, A+B=0
Now,
On combining A and B, we get our original equation
i.e. A+B=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.
Solution:-
Given,
From this we obtain an equation,
But we see that,
So that,
Hence (1) becomes,
But,
Then (2) becomes,
But we know that,
Hence we see that,
Therefore (3) becomes,
Well, consider the equation for
For it being satisfied, one among
and
should be negative to the other but it's not possible since the square of a real number is always non - negative.
For some non - zero the equation contradicts our assumption that
Also the equation can contradict
for some non - zero
Hence the equation has unique solution.
Therefore, from (4) we get,
Or,
Also,
Or,