How do you find the exact value of #Sin^-1(cos(pi/4))#?

Question

Carter Anderson · Accepted Answer

#text{Arc}text{sin}(cos (pi/4) ) = pi/4 #

#arcsin(cos (pi/4)) = pi/4 + pi k, quad# integer #k# Oh my, #pi/4# again. What a surprise. Question writers, questions like this are your big chance to get away from the two cliche triangles of trig. I prefer the notation #text{Arc}text{sin}(x)# for the principal value and #arcsin(x)# for all values. Let's run the problem both ways. Of course, #sin(pi/4) = cos (pi/4) = 1/sqrt{2} # #text{Arc}text{sin}(cos (pi/4) ) = text{Arc}text{sin}( 1/sqrt{2} ) = pi/4 # #x = arcsin(cos (pi/4)) # is equivalent to # sin(x) = cos(pi/4)# I always remember #cos x =cos a# has solutions #x=pm a + 2pi k quad# integer #k#. # cos(pi/2 - x) = cos(pi/4)# #pi/2 - x = pm pi/4 + 2pi k# # x = pi/2 pm pi/4 + 2pi k # # x = { pi/4, {3pi}/4 } + 2pi k# We can rewrite that as #x = pi/4 + pi k # #arcsin(cos (pi/4)) = pi/4 + pi k, quad# integer #k#

Carter Anderson · Answer

undefined To find the exact value of sin^-1(cos(pi/4)), you can use the identity sin^-1(cos(x)) = x for 0 ≤ x ≤ π.

Given cos(pi/4) = √2 / 2, sin^-1(√2 / 2) = π/4. Therefore, the exact value of sin^-1(cos(pi/4)) is π/4.