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

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 $3\sin \alpha+ 4\cos \beta=8$