Question

$\sf If \;cos(x-y)+cos(y-z)\\\sf +cos(x-z)=-\dfrac{3}{2}$
Then find,
$\rm\sum cos(x)$​

Hint : $\sum \rm cos(x) = cos(x) + cos(y) + cos(z)$

Answer 1

ANSWER:

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

GIVEN:

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

TO FIND:

$\rm\sum cos(x)$ = ??

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 $\sf cos(x-y)+cos(y-z) +cos(x-z)$

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 $\sum \rm cos(x) = cos(x) + cos(y) + cos(z)$ = 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.

Answer 2

Answer:

2∑cosxcosy+2∑sinxsiny+3=0

2∑cosxcosy+2∑sinxsiny+1+1+1=0

2∑cosxcosy+2∑sinxsiny+sin

2

x+cos

2

x+sin

2

y+cos

2

y+sin

2

z+cos

2

z=0

[cos

2

x+cos

2

y+cos

2

z+2∑cosxcosy]+[sin

2

x+sin

2

y+sin

2

z+2∑sinxsiny]=0

(cosx+cosy+cosz)

2

+(sinx+siny+sinz)

2

=0

⇒cosx+cosy+cosz=0=sinx+siny+sinz

Hence, option 'D' is correct.

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