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Problem 79 Easy Difficulty

Proof Let $s(x)$ and $c(x)$ be two functions satisfying $s^{\prime}(x)=c(x)$ and $c^{\prime}(x)=-s(x)$ for all $x .$ If $s(0)=0$ and $c(0)=1,$ prove that $[s(x)]^{2}+[c(x)]^{2}=1$




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Video Transcript

So we want to prove that, um, s of X squared plus C of X squared equals one. And to do this, we want to differentiate both sides. So with respect to X, So what will end up getting is two s of X as prime of X plus to see of X see prime of X, and this is gonna be equal to zero because we differentiate both sides. Then we know that this is going to be the same thing as two s of X cfx plus to see of X times negative SFX given what we were already had on DNA. Now, with this, we can recognize that we have to s of X c f x minus two sfx cfx, which is just gonna be equal to zero. So we know, um, that when the derivative of f prime of X equals zero, then the function is constant at X. So what we really just showed is that right here, this is going to equal some constant, Um, and with the information that we have, we can now say we can now substitute s of zero squared plus see of zero squared equals scene. We know that that's going to give us zero squared plus one squared will be equal to C, so C equals one. So no matter what, we're going to get that this is equal to one, which is what we wanted to prove.