Question

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

Answer 1

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.

Answer 2

No, it's not possible

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`