Write sin(sin^-1x-cos^-1x as an algrebraic expression in x.

Question

This is for my trig class, thanks for the help! If you could, could you please show all your work? Thanks again!

Answer 1

Hi Serenity!

 

This is a neat problem!  You have:  sin( sin^-1x - cos^-1x ) .  Whenever you have anything like sin( A - B ), the formula for the Sine of the Difference of Two Angles is often useful:

 

sin ( A - B ) = sin(A) cos(B) - sin(B) cos(A)

 

In this case, we have:

 

sin( sin^-1x - cos^-1x ) = sin( sin^-1x ) cos( cos^-1x ) - sin( cos^-1x ) cos( sin^-1x )

 

Looks messy!  First, note that sin( sin^-1x ) and cos( cos^-1x ) both equal x because you're taking inverse functions.  So it's reduced to:

 

x * x - sin( cos^-1x ) cos( sin^-1x )

 

Now, this next part is tricky to explain on this message board, but it turns out that both cos( sin^-1x ) and sin( cos^-1x ) equal √(1-x²).  This is *much more easily seen* with a diagram, but I've included an algebraic proof on the bottom in lieu of a drawing.  So, inserting √(1-x²) into what we've got so far gives:

 

x * x - √(1-x²) * √(1-x²)

 

x² - (1-x²)

 

2x² - 1

 

Voila.  Now, if you wanna see why sin( cos^-1x ) = √(1-x²) the ugly way without a drawing, proceed below.

 

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We're trying to find the value of A:

 

A = sin( cos^-1x ) .  Let θ = cos^-1x.  Then, substituting, we get A = sin(θ).

 

Let's look at  θ = cos^-1x for a bit.  It means x = cos(θ). 

 

But if A = sin(θ) and x = cos(θ), and since sin²(θ) + cos²(θ) = 1, that means A² + x² = 1.  Thus A = √(1-x²)

 

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Answer 2

sin(sin^-1x-cos^-1x)

= sin(sin^-1x)cos(cos^-1x) - cos(sin^-1x)sin(cos^-1x) 

= x^2 - (1-x^2)

= 2x^2 - 1

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Attn:

cos(sin^-1x) = sqrt(1-x^2), since the domain of sin^-1x is in the first and fourth quadrants, where cos x is positive.

sin(cos^-1x) = sqrt(1-x^2), since the domain of cos^-1x is in the first and the second quadrants, where sin x is positive.