The Attempt at a Solution
No, cos(90°-70°) = cos(20°) = sin(70°).Ah a trig identity I see.
Because f-1(f(x)) = x, as long as x is in the domain of f. That's a basic property of a function and its inverse. It's also true that f(f-1(y)) = y, when y is in the domain of f-1. In short, the composition of a function and its inverse "cancels."Alright. So if cos(90°-x)=sinx, which in my case is 70°, I still do not understand why Tanya then encourages me to use sin^-1(sinθ)=θ.
There's really no need for you to use θ, here, and I think that it is confusing you.How is cos70° to be replaced by sinθ?
Not equivalent to - equal to.By the trig identities, sin(90°-x)=cosx, it is sin20° which is equivalent to cos70°.
Make it simpler by getting rid of θ.Would my equation then be accurately represented by sin^-1(sin20°)=θ and then simply by using cancellation identities reduced to 20°=θ?