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(arcsin x + arccos x)sin(arcsinx+arccosx)
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 xarcsinx and arccos xarccosx are complementary angles because xx is the sine of one and the cosine of the other. Complementary angles of course add to 90^circ90∘ so this expression equals sin90^circ = 1.sin90∘=1.
It's clear in the first quadrant, positive xx, give acute complementary angles which add nicely to 90^circ.90∘. We have to think a little harder about the principal values when xx is negative. Then the arcsinarcsin is in the fourth quadrant with a negative number (between 00 and -90^circ−90∘) and the arccosarccos is in the second quadrant, positive angle between 90^circ90∘ and 180^circ.180∘. They'll add to 90^circ90∘ 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 xarccosx denotes all the angles, countably infinitely many, whose cosine is xx.
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 arccosx as arccos(x/1)arccos(x1), i.e. as a right triangle with adjacent xx and hypotenuse 11. The opposite side is sqrt{1-x^2}√1−x2 which will be the sine given hypotenuse 1.1. 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}sinarccosx=±√1−x2
In the multivalued interpretation, the square roots always come with pm±.
Similar cos arcsin x = pm sqrt{1-x^2}.cosarcsinx=±√1−x2. The pm± signs aren't linked.
Of course cos arccos x = xcosarccosx=x and sin arcsin x = x.sinarcsinx=x. Now we can do this one by the sum angle formula.
sin(arcsin x + arccos x) = sin arcsin x cos arccos x + cos arcsin x sin arccos x sin(arcsinx+arccosx)=sinarcsinxcosarccosx+cosarcsinxsinarccosx
= x^2 + (pm sqrt{1-x^2})(pm sqrt{1-x^2}) =x2+(±√1−x2)(±√1−x2)
= x^2 pm (1-x^2) =x2±(1−x2)
= 1 or 2x^2-1=1or2x2−1
We have this possibility of 2x^2-12x2−1 which is the cosine double angle formula, equal to cos 2 arccos x.cos2arccosx.
Let's try one of the usual cliches as an example. x=-1/sqrt{2}x=−1√2.
The principal values have arccos x=135^circarccosx=135∘ and arcsin x=-45^circarcsinx=−45∘ which add to 90^circ90∘ which has a sine of 1.1.
If we take arccos x = -135^circarccosx=−135∘ then arccos x + arcsin x = -180^circarccosx+arcsinx=−180∘ so a sine of 00 which is indeed the cosine of the double angle of -135^circ.−135∘.
