Continue....a.) Given that y=arctan(x), express sin(y) + cos(y) in terms of x
y = tan-1(x) → x = tan(y) → sin(y) = x/√(1+x2) & cos(y) = 1/√(1+x2)
b.) Given that x =tany, find arccotx in terms of y
The book gives the solutions but can someone explain how to solve these questions?
What did you get when you drew the right triangle, labelled the one angle as "y", noted that x = x/1, labelled the two appropriate sides of the triangle, applied the Pythagorean Theorem, etc, etc? What have you done? What has this led? Where are you stuck?I don't understand how you're getting from x = tan(y) to sin(y) = x/√(1+x2) & cos(y) = 1/√(1+x2)
Can you please explain?
The second one works in exactly the same way as the first one. Did you even try to do it? If so:I didn't realise I could draw a right triangle from the information given. Knowing that now, part a was simpler than I thought. But how can part b be solved using the triangle?
|Arcsin||inverse of sine function||[−1, +1]||Q IV, I: [−π/2, +π/2]|
|Arccos||Arccos x = π/2 − Arcsin x||[−1, +1]||Q I, II: [0, π]|
|Arctan||inverse of tangent function||all reals||Q IV, I: (−π/2, +π/2)|
|Arccot||Arccot x = π/2 − Arctan x||all reals||Q I, II: (0, π)|
|Arcsec||Arcsec x = Arccos(1/x)||(−∞, −1] and [1, ∞)||Q I, II: [0, π]|
|Arccsc||Arccsc x = Arcsin(1/x)||(−∞, −1] and [1, ∞)||Q IV, I: [−π/2, +π/2]|