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

2 Answers
Jan 20, 2018

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#

Jan 20, 2018

#=> 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...