Question

Solve for t algebraically: inverse cos(t) = inverse sin(t).
Where do I start?

Answer 1

since sin(x) = cos(π/2-x)

arccos(t) + arcsin)t) = π/2

In general, draw the triangle. In this case, take cosine of both sides.

cos(arccos(t)) = t

In your triangle, the adjacent side will be t, and the hypotenuse will be 1. The other leg is thus √(1-t^2). So,

sin(arccos(t)) = √(1-t^2)

Now we have

t = √(1-t^2)
t^2 = 1-t^2
2t^2 = 1
t = 1/√2

So, arccos(1/√2) = arcsin(1/√2) = π/4

So, their sum is π/4 + π/4 = π/2

You can sometimes save yourself a lot of work by remembering a few fundamental relationships.