# How do you simplify sin(sin^(-1)x + cos^(-1)x)?

Jun 4, 2018

If we interpret the inverse trig functions as principal values the two angles are complementary, so

$\sin \left(\arcsin x + \arccos x\right) = 1$

Under the multivalued interpretation,

$\sin \left(\arcsin x + \arccos x\right) = 1$ or

$\sin \left(\arcsin x + \arccos x\right) = 2 {x}^{2} - 1 = \cos 2 \arccos x$

#### Explanation:

Two minutes ago or two years ago? Why must Socratic lie to us? And why doesn't Socratic understand about states? It's Pittsburgh, Pennsylvania, USA; most of the other Pittsburgs in the country are spelled without the "h" but you can't count on that in general.

While I'm griping, I'll gripe about the inverse notation. The negative one exponent has been deprecated; the preferred form for this problem is

Simplify $\sin \left(\arcsin x + \arccos x\right)$

There are a few different ways to think about this one. If we are referring to the principal values of the inverse functions, then $\arcsin x$ and $\arccos x$ are complementary angles because $x$ is the sine of one and the cosine of the other. Complementary angles of course add to ${90}^{\circ}$ so this expression equals $\sin {90}^{\circ} = 1.$

It's clear in the first quadrant, positive $x$, give acute complementary angles which add nicely to ${90}^{\circ} .$ We have to think a little harder about the principal values when $x$ is negative. Then the $\arcsin$ is in the fourth quadrant with a negative number (between $0$ and $- {90}^{\circ}$) and the $\arccos$ is in the second quadrant, positive angle between ${90}^{\circ}$ and ${180}^{\circ} .$ They'll add to ${90}^{\circ}$ too so the sine will be one.

We can ask what happens if we intepret the inverse trig functions to be multivalued? That means $\arccos x$ denotes all the angles, countably infinitely many, whose cosine is $x$.

That sounds complicated but in practice it just puts an ambiguity on the sign of some of the other trig functions. We can think of $\arccos x$ as $\arccos \left(\frac{x}{1}\right)$, i.e. as a right triangle with adjacent $x$ and hypotenuse $1$. The opposite side is $\sqrt{1 - {x}^{2}}$ which will be the sine given hypotenuse $1.$ The sign of the sine is unknown because of the multivalued inverse cosine, so we can write

$\sin \arccos x = \pm \sqrt{1 - {x}^{2}}$

In the multivalued interpretation, the square roots always come with $\pm$.

Similar $\cos \arcsin x = \pm \sqrt{1 - {x}^{2}} .$ The $\pm$ signs aren't linked.

Of course $\cos \arccos x = x$ and $\sin \arcsin x = x .$ Now we can do this one by the sum angle formula.

$\sin \left(\arcsin x + \arccos x\right) = \sin \arcsin x \cos \arccos x + \cos \arcsin x \sin \arccos x$

$= {x}^{2} + \left(\pm \sqrt{1 - {x}^{2}}\right) \left(\pm \sqrt{1 - {x}^{2}}\right)$

$= {x}^{2} \pm \left(1 - {x}^{2}\right)$

$= 1 \mathmr{and} 2 {x}^{2} - 1$

We have this possibility of $2 {x}^{2} - 1$ which is the cosine double angle formula, equal to $\cos 2 \arccos x .$

Let's try one of the usual cliches as an example. $x = - \frac{1}{\sqrt{2}}$.

The principal values have $\arccos x = {135}^{\circ}$ and $\arcsin x = - {45}^{\circ}$ which add to ${90}^{\circ}$ which has a sine of $1.$

If we take $\arccos x = - {135}^{\circ}$ then $\arccos x + \arcsin x = - {180}^{\circ}$ so a sine of $0$ which is indeed the cosine of the double angle of $- {135}^{\circ} .$

Jun 9, 2018

$\sin \left(\arcsin x + \arccos x\right) = 1$

#### Explanation:

$\sin \left(\arcsin x + \arccos x\right)$

Set arcsinx=a and arccosx=b. Consequently $x = \sin a = \cos b$ and $\cos a = \sin b = \sqrt{1 - {x}^{2}}$. Hence,

$\sin \left(a + b\right) = \sin a \cdot \cos b + \cos a \cdot \sin b$

=$x \cdot x + \sqrt{1 - {x}^{2}} \cdot \sqrt{1 - {x}^{2}}$

=${x}^{2} + 1 - {x}^{2}$

=$1$