How to draw graph of $\arcsin(\cos(x))$ or even $\arcsin(\sin(x))$ without use of graphing calculator , its sort of confusing me from long time. It gives pointed curves when drawing from calculator , why does it looks like that? Why it isn't just linear graph? Can someone shed some light here?
[Math] Graph of $\arcsin(\cos(x))$
graphing-functions
Related Solutions
For fix $k$ you need to find all pairs $x,y$ so that $\ln(x)+\ln(y)=k$. Applying the exponential to both sides we get $e^{\ln(x)+\ln(y)}=e^k$. The left-hand side is equal to $e^{\ln(x)}\cdot e^{\ln(y)}=xy$. Therefore the level surface is equal to the hyperbola $e^k=xy$.
The surface of equation $$z=\sin x+\sin y$$ has the shape of an "egg tray". It has maxima $z=2$ on a periodic grid, and this is why the level curves $z=1$ are regularly spaced approximate circles.
Now if we replace $x$ by $x^2$, we deform space horizontally so that $x^2$ increases faster and faster, giving a "compression effect".
By replacing $y$ with $y^2$, we get the effect on both axis.
Of course, the picture is symmetric by reflection, as the square function is even.
Remains to explain the "spikes" against the axis. If $y$ is small,
$$\sin x^2+\sin y^2=1\to y\approx\sqrt{1-\sin x^2}=\sqrt2\left|\cos\frac{x^2}2\right|$$
shows them.
Best Answer
Hint:
For the first function use the identity $$\cos(x)=\sin\left(x+\dfrac\pi2\right),$$ for the second it is useful to note that $\arcsin(\sin(x))=x$, since $\arcsin x$ and $\sin x$ are inverse functions.
Also be careful about the domains and ranges of those functions, and so $\arcsin(\sin(x))=x$ isn't true for all $x$.