# Equivalence of $\sin(\arccos(1-x/5))$

On my calculator $5\sin(\arccos(1-x/5))$ gives 0 when x= 0 and 3 when x=1 and same results gives $\sqrt{x(10-x)}$ ; therefore $5\sin(\arccos(1-x/5))-\sqrt{x(10-x)}$ gives always 0.

• can you please tell me if $5\sin(\arccos(1-x/5))$ can be simplified?
• 1) No: it cannot be simplified. 2) You wrote $cos^1$ instead of $\arccos$. Oct 19, 2015 at 8:54
• $x=0$ and $x=1$ is far from being always.
– A.Γ.
Oct 19, 2015 at 8:54
• Use $\sin\alpha=\pm\sqrt{1-\cos^2\alpha}$ to simplify. Take $\alpha=\arccos(1-x/5)$.
– A.Γ.
Oct 19, 2015 at 9:00

Some two functions $f(x)$ and $g(x)$ are generally not the same just because they give the same result at two distinct points. It just means that their graphs intersect at that point. In the case of $f(x)=5\mathrm{sin}(\mathrm{arccos}(1-x/5))$ and $g(x)=\sqrt{x*(10-x)}$, however, you can prove analytically that they are the same.
Regarding the simplification results, a simple Wikipedia search yields $f(x)=5\sqrt{1-(1-x/5)^2}=\sqrt{10x-x^2}=g(x)$, so in this case you got lucky, but remember in the future that numerical coincidences are not a proof.
For a suitable $x$ range you have $\cos(\arccos(1-x))=1-x$ and therefore with $\sin^2 x +\cos^2 x = 1$ $$\sin(\arccos(1-x))=\sqrt{1-(\cos(\arccos(1-x)))^2}=\sqrt{1-(1-x^2)}=\sqrt{x(2-x)}$$ For your value $x=\frac{1}{5}$ this will be $\sqrt{\frac{1}{5}(2-(\frac{1}{5}))} = \frac{3}{5}$ and for $x=0$ it will be zero.
For any $u ,\sin(\arccos(u))= \sqrt{1-u^2}.$