How do I write sin(arcsinx + arccosx) as an algebraic expression?

Answer 1

sin(arcsin(x)) = x
cos(arcsin(x)) = √(1-x^2)

so, if
A = arcsinx
B = arccosx

ain(A+B) = sinAcosB + cosAsinB
= x√(1-x^2) + √(1-x^2)*x
= 2x√(1-x^2)

Answer 2

arcsinx is an angle Ø so that sinØ = x/1 = x
arccosx is an angle Ø so that cosØ = x/1
thus sinØ = cosØ
and Ø = 45°

sin(arcsinx + arccosx)
= sin(45° + 45°)
= sin 90°
= 1

Answer 3

confirmation by Wolfram:

http://www.wolframalpha.com/input/?i=simplify+sin%28arcsinx+%2B+arccosx%29

Answer 4

sinAcosB + cosAsinB
= x*x + √(1-x^2)√(1-x^2)
= x^2 + 1-x^2
= 1

of course- since if you draw the triangle, you are taking the sine of two angles which add to 90°

Answer 5

But doesn't sinØ also equal cosØ at 225 degrees?

Answer 6

yeah - so?

the expression still evaluates to 1.