How do I simplify sin(arccos(sqrt(2)/2)-arcsin(2x))?

Question

Josiah Anderson · Accepted Answer

I get #sin ( arccos(sqrt{2}/2) - arcsin(2x)) ##= { 2x \pm sqrt{1 - 4x^2}}/{sqrt{2}} #

We have the sine of a difference, so step one will be the difference angle formula,

#sin(a-b) = sin a cos b - cos a sin b#

#sin ( arccos(sqrt{2}/2) - arcsin(2x)) #

#= sin arccos(sqrt{2}/2) cos arcsin(2x) + cos arccos(sqrt{2}/2) sin arcsin(2x)

Well the sine of arcsine and the cosine of arccosine are easy, but what about the others? Well we recognize #arccos(\sqrt{2}/2)# as #\pm 45^circ#, so

#sin arccos(\sqrt{2}/2)= \pm \sqrt{2}/2#

I'll leave the #pm# there; I try to follow the convention that arccos is all inverse cosines, versus Arccos, the principal value.

If we know the sine of an angle is #2x#, that's a side of #2x# and a hypotenuse of #1# so the other side is #\sqrt{1-4x^2}#.

# cos arcsin(2x) = \pm sqrt{1-4x^2}#

Now,

#sin ( arccos(sqrt{2}/2) - arcsin(2x)) #

#=\pm \sqrt{2}/2 sqrt{1-4x^2} + (sqrt{2}/2)(2x)#

#= { 2x \pm sqrt{1 - 4x^2}}/{sqrt{2}} #

Grace Ashcraft · Answer

undefined To simplify sin(arccos(sqrt(2)/2) - arcsin(2x)), we can use trigonometric identities and properties:

1. Start by simplifying the inner expressions:
   arccos(sqrt(2)/2) = π/4
   arcsin(2x) = sin^(-1)(2x)

2. Now, we have:
   sin(π/4 - sin^(-1)(2x))

3. Apply the difference of angles formula for sine:
   sin(a - b) = sin(a)cos(b) - cos(a)sin(b)

4. Substitute the values:
   sin(π/4)cos(sin^(-1)(2x)) - cos(π/4)sin(sin^(-1)(2x))

5. Utilize trigonometric identities to simplify:
   sin(π/4)cos(sin^(-1)(2x)) - cos(π/4)sin(sin^(-1)(2x))
   = (sqrt(2)/2)(√(1 - (2x)^2)) - (sqrt(2)/2)(2x)

6. Simplify further:
   = (sqrt(2)/2)(√(1 - 4x^2) - 2x)

That's the simplified expression for sin(arccos(sqrt(2)/2) - arcsin(2x)).