# Pre-Calc

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

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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)

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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

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3. confirmation by Wolfram:

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

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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°

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

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6. yeah - so?

the expression still evaluates to 1.

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