- Thread starter hndalama
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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+x^{2}) & cos(y) = 1/√(1+x^{2})

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?

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What didI don't understand how you're getting fromx = tan(y)tosin(y) = x/√(1+x^{2}) & cos(y) = 1/√(1+x^{2})Can you please explain?

Please be complete. Thank you!

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The second one works inI 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?

What did

Please be complete. Thank you!

In part a, it was simply a matter of applying soh cah toa to find siny and cosy in terms of x but this doesn't apply to arccotx. the question is also asking us to express arccotx

I tried using the theorem tany = 1/coty but using this I could only derive y=arccot(1/x). This is where I am stuck, How can I use the triangle to find 'arccotx in terms of y?'

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So I understand why it’s so difficult. You are right. Using the CAST diagram with the triangle and Pythagoras’ might not work this time.

I found out that by definition, arccotx is (pi/2) minus arctanx. So you must show that somehow arctanx is y. So then, if x=tany, then multiplying both sides by tan^-1 you get tan^-1(x)=y. So tan inverse equals y. Here we see that tan^-1(x) is the same as arctanx. Since arccotx=(pi/2) minus arctanx, arccotx=(pi/2) minus y.

I know it’s 2 years late but I am studying the subject and was just struggling with this question as well. Hopefully you or someone else finds this table helpful;

I found it at https://brownmath.com/twt/inverse.htm

nverse trig functions:

function | derived from | domain | range |
---|---|---|---|

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] |