Question

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

Then, find

$\sum\rm\cos (\sf x)$

||QUALITY ANSWER REQUIRED||​

Answer 1

ANSWER:

\sum \rm cos(x) = cos(x) + cos(y) + cos(z)∑cos(x)=cos(x)+cos(y)+cos(z) = 0

GIVEN:

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

2

3

TO FIND:

\rm\sum cos(x)∑cos(x) = ??

FORMULAE:

cos(x - y) = cos x cos y + sin x sin y

A² + B² + C² + 2AB + 2BC +2CA = (A + B + C)²

sin² A + cos ² B = 1

EXPLANATION:

EXPAND \sf cos(x-y)+cos(y-z) +cos(x-z)cos(x−y)+cos(y−z)+cos(x−z)

2 (cos x cos y + sin x sin y + cos y cos z + sin y sin z + cos x cos z + sin x sin z) = -3

3 + 2 (cos x cos y + cos y cos z + cos x cos z) + 2(sin x sin y + sin y sin z + sin x sin z) = 0

(sin² x + cos² y) + (sin² y + cos² z) + (sin² x + cos² z) + 2 (cos x cos y + cos y cos z + cos x cos z) + 2(sin x sin y + sin y sin z + sin x sin z) = 0

(sin² x + sin² y + sin² z) + 2(sin x sin y + sin y sin z + sin x sin z) + (cos² x + cos² y + cos² z) + 2 (cos x cos y + cos y cos z + cos x cos z) = 0

(sin x + sin y + sin z)² + (cos x + cos y + cos z)² = 0

(sin x + sin y + sin z)² = (cos x + cos y + cos z)² = 0

sin x + sin y + sin z = cos x + cos y + cos z = 0

HENCE \sum \rm cos(x) = cos(x) + cos(y) + cos(z)∑cos(x)=cos(x)+cos(y)+cos(z) = 0

ADDITIONAL INFORMATION:

(sin x + sin y + sin z)² = (cos x + cos y + cos z)² = 0.

THIS IS BECAUSE WHEN TWO SQUARE TERMS ARE ADDED THEIR SUM ALWAYS GIVES A POSITIVE NUMBER. WE NEEDED ZERO AND THE ONLY POSSIBILITY IS THE BOTH TERMS SHOULD BE ZERO.