JEE Novice - (35)
sin x + sin y + sin z = 3 cos x + cos y + cos z = ?
Assume x , y , z are real numbers.
This question is a part of JEE Novices .
The answer is 0.
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2 solutions
Sorry i read ur solution wrong in a hurry
This is not a solution bro
For real x , y , z , there is only one possibility and that is sin x + sin y + sin z = 3 that is when sin x = sin y = sin z = 1 are x = y = z = 2 ( 2 k − 1 ) π , where k is an integer. Also when cos x = cos y = cos z = 0 , ⇒ cos x + cos y + cos z = 0 .
Moderator note:
Why is there only one possible answer?
Even though it's quite obvious, you should mention in your solution that the function f : R ↦ R defined by f ( x ) = sin ( x ) is bounded on [ − 1 , 1 ] with f ( π / 2 ) = 1 for completeness.
In other words, the range of the trigonometric sine function is [ − 1 , 1 ] in real domain.
sinx=siny=sinz=1 to satisfy the eqn. hence cosx+cosy+cosz=0 (cosine function will be zero )