Question

if sinx +sinx+sinz=-3 , prove that cosx + cosy+cosz=0

Answer 1

Sin x + Sin y + Sin z = -3

Sin x = Sin y = Sin z = -1

Now,
$\cos(x) = \sqrt{1 - {sin}^{2} x} = \sqrt{1 - ({ - 1})^{2} } = 0 \\ similarly \: \cos \: y \: = \cos \: z = 0 \\ \\ hence \: \: \cos(x ) + \cos(y) + \cos(z) = 0 + 0 + 0 = 0$

Answer 2

Dear here is ur solution

It is given that sinx +siny+sinz=-3

We can write it as

sinx+Siny+sinz=-1-1-1

We can say that sinx=-1 &Siny=-1 &sinz=-1

So we find out value of x, y&z

so value of x, y&z=270°

We put these values in question

Cos270°+cos270°+cos270°=0

I hope it helps you