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

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Is 3 sin(alpha)+4 cos(beta)=8 possible - id3482152436, SimplyGenesis762

silverdays566

01.01.2022

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

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