How do I write sin(arcsinx + arccosx) as an algebraic expression?
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
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°
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)
confirmation by Wolfram:
http://www.wolframalpha.com/input/?i=simplify+sin%28arcsinx+%2B+arccosx%29
But doesn't sinØ also equal cosØ at 225 degrees?
yeah - so?
the expression still evaluates to 1.
To write sin(arcsin(x) + arccos(x)) as an algebraic expression, we can use the trigonometric identities. Let's break down the process step by step:
1. Use the property of the inverse trigonometric functions: arcsin(sin(x)) = x and arccos(cos(x)) = x, for the given domain.
Therefore, we have:
arcsin(x) + arccos(x) = x + x = 2x
2. Apply the sum-to-product identity for sine: sin(A + B) = sin(A)cos(B) + cos(A)sin(B)
In this case, A = B = x, so we can rewrite the expression as:
sin(arcsin(x) + arccos(x)) = sin(2x)
So, the final algebraic expression is sin(2x).