Question

How do you write the trigonometric expression `cos(arccosx+arcsinx)` as an algebraic expression?

Answer 1

The answer is `=0`

Explanation:

Let `y=arccosx`, `=>`, `cosy=x`

Let `z=arcsinx`, `=>`, `sinz=x`

Therefore,

`cos(arccosx+arcsinx)=cos(y+z)`

`=cosycosz-sinysinz`

`=x*sqrt(1-x^2)-xsqrt(1-x^2)`

`=0`

Answer 2

`=> cos( color(red)(arccosx) + color(blue)(arcsinx )) = 0`

`AA |x| <=1`

• `" for all " x " that is in the interval " -1<= x <= 1`

Explanation:

The first thing we need to consider is our additional formulae for trig:

`cos(A + B ) = cosAcosB - sinAsinB`

So for this problem we see that `color(red)(A = arccosx` and `color(blue)(B = arcsinx`

`=> cos( color(red)(arccosx) + color(blue)(arcsinx )) =`

`cos( color(red)(arccosx) ) * cos color(blue)((arcsinx )) - sin color(red)( (arccosx) ) * sin color(blue)( (arcsinx) )`

We know:

`cos(arccosx) = x " and " sin(arcsinx) = x`

Now to find `cos(arcsinx) " and " sin(arccosx ) :`

`"let " color(orange)(theta_1 = arcsinx`

`=> sin theta_1 = x`

Use `color(green)( cos^2 x + sin^2 -= 1`

`costheta_1 = sqrt(1-sin^2 theta_1 ) = sqrt(1-x^2 )`

`=> cos(theta_1) = cos(arcsinx ) = sqrt(1-x^2 )`

`"let " color(orange)(theta_2 = arccosx`

`=> costheta_2 = x`

`=> sintheta_2 = sqrt(1-x^2 )`

`=> sin(arccosx) = sqrt(1-x^2 )`

Hence this:

`cos( color(red)(arccosx) ) * cos color(blue)((arcsinx )) - sin color(red)( (arccosx) ) * sin color(blue)( (arcsinx) )`

Becomes:

`xsqrt(1-x^2) - xsqrt(1-x^2) = color(red)(0`

Note: This only holds for `|x| <= 1`

As that is values of `x` that `arcsinx` and `arccosx` are valid for...