NOW QUICKLY U MAGGOTZ Joking, but plz hurry simplify

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highschoolboy

01.09.2021

$\cos(\arccos5x - \arcsin 5x) = 10x\sqrt{1 - 25x^2}$ ======we can let $u=\arccos (5x)$ . it follows that, taking the cosine of both sides, we cancel out arccos $\begin{aligned} \arccos 5x & = u \\ 5x & = \cos u \end{aligned}$ because they're inverses of each other. let's get something with arcsine in it. $\cos u$ can be turned to sine using this identity: $\cos u = \sin\left( \frac{\pi}{2} - u}\right)$ since $\cos u = 5x$ it follows that $\sin\left( \frac{\pi}{2} - u}\right) = 5x$ then we can take the arcsine of both sides to cancel out the sine. $\begin{aligned} \sin\left( \tfrac{\pi}{2} - u}\right) & = 5x \\ \tfrac{\pi}{2} - u & = \arcsin(5x) \end{aligned}$ since $\arccos (5x) = u$ and $\arcsin(5x) = \tfrac{\pi}{2} - u$ , we can substitute into our original expression to take out all the arc stuff: $\cos\big(\arccos(5x) - \arcsin(5x)\big)$ $\begin{aligned} & = \cos\left(u - \left(\tfrac{\pi}{2} - u\right)\right) \\ & = \cos\left(u - \tfrac{\pi}{2} +u\right) \\ & = \cos\left(2u - \tfrac{\pi}{2}\right) \end{aligned}$ cosine subtraction identity, then sine double angle identity $\begin{aligned} \cos(a-b) & = \cos(a)\cos(b) + \sin(a)\sin(b) \\ \cos\left(2u - \tfrac{\pi}{2}\right) & = \cos(2u)\cos\left(\tfrac{\pi}{2}\right) + \sin(2u)\sin\left(\tfrac{\pi}{2}\right) \\ & = 0 + \sin(2u) \cdot (1) \\ & = \sin(2u) \\ & = 2\sin (u)\cos(u) \\ & = 2\sin (\arccos 5x)\cos(\arccos 5x) \\ & = 2\sin (\arccos 5x) \cdot 5x \\& = 10x\sin (\arccos 5x) \end{aligned}$ we want to deal with $\sin (\arccos 5x)$ . we can use the pythagorean identity, with $\arccos 5x$ as the angle. then we can cancel out cosine and isolate sine. $\begin{aligned} \sin^2 (a) + \cos^2 (a) & = 1 \\ \sin^2(\arccos(5x)) + \cos^2 (\arccos(5x)) & =1 \\ \sin^2(\arccos(5x)) + \left[\cos \big(\arccos(5x)\big)\right]^2 & =1\\ \sin^2(\arccos(5x)) + \left[5x\right]^2 & =1\\ \sin^2(\arccos(5x)) & = 1 - \left(5x\right)^2 \\ \sin(\arccos(5x)) & = \pm \sqrt{1 - 25x^2} \end{aligned}$ how do we deal with this plus or minus sign? the function $y = \arccos(x)$ has a range of $[0,\pi]$ (this is by definition). the function $y = \arccos(5x)$ is a horizontal compression by a factor of 5 of the original; it does not affect the range. since $\arccos(5x) \in [0,\pi]$ , the value of $\arccos(5x)$ is located in quadrants 1 and 2. remember, $\arccos(5x)$ is the angle used in sine. so sine is positive or zero because the angle is in quadrant 1 or 2. take the positive square root. $\sin(\arccos(5x)) = \sqrt{1 - 25x^2}$ therefore, from before $\cos\big(\arccos(5x) - \arcsin(5x)\big)$ $\begin{aligned} & = 10x\sin (\arccos 5x)\\ & = 10x\sqrt{1 - 25x^2} \end{aligned}$

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NOW QUICKLY U MAGGOTZ

Joking, but plz hurry simplify

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