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Reika [66]
3 years ago
14

Is 3 sin(alpha)+4 cos(beta)=8 possible

Mathematics
2 answers:
sdas [7]3 years ago
8 0

Answer:

No, it's not possible

Step-by-step explanation:

We know that for various values of x , -1\leq \sin x\leq 1

and -1\leq \cos x \leq 1

For values of \alpha, -1\leq \sin \alpha\leq 1

On multiplying all sides by 3, we get

-3\leq 3\sin \alpha\leq 3

For values of \beta, -1\leq\cos \beta\leq 1

On multiplying all sides  by 4, we get

-4\leq 4\cos \beta\leq 4

On adding all sides of -3\leq 3\sin \alpha\leq 3\,\,,\,\,-4\leq 4\cos \beta\leq 4 , we get

-3-4\leq 3\sin \alpha+ 4\cos \beta\leq 3+4\\-7\leq 3\sin \alpha+ 4\cos \beta\leq 7

Therefore, maximum value of 3\sin \alpha+ 4\cos \beta is 7.

So, it's not possible that 3\sin \alpha+ 4\cos \beta=8

babunello [35]3 years ago
3 0
No because the maximum amount for sin and cos is 1.if alpha is 90 and beta is 0 the answer is 7 so it's not right.
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