How would the trigonometric expression sin[(arccos (0)-arcsin(x)]be written as an algebraic expression?

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Answer 1 · 2018-06-03T21:50:59

# sin[arccos (0)-arcsin(x)] = sqrt(1 - x^2) #

We seek:

# S = sin[arccos (0)-arcsin(x)] #

Using #arccos0=pi/2#, and the sine difference formula:

# sin(A-B) = sinAcosB - cosAsinB #

We can write:

# S = sin(pi/2) \ cos(arcsinx) - cos(pi/2) \ sin(arcsinx) #

# \ \ = 1 * cos(arcsinx) - 0 #

# \ \ = cos(arcsinx) #

And utilising the pythagorean identity:

# sin^2A + cos^2A -= 1=> cosA = sqrt(1-sin^2A)#

we have:

# S = sqrt(1 - sin^2(arcsinx)) #

# \ \ = sqrt(1 - x^2) #