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 September 26th, 2014, 05:40 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,659 Thanks: 2635 Math Focus: Mainly analysis and algebra Perhaps you were supposed to use $$1 + \cot^2 x = \csc^2 x$$
September 26th, 2014, 09:27 AM   #3
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 Originally Posted by v8archie Perhaps you were supposed to use $$1 + \cot^2 x = \csc^2 x$$
Nope, we have not taken this. As I have told you, my answer was correct (I think?) But it made me very confused (read the questions I listed).

Last edited by skipjack; September 27th, 2014 at 08:02 AM.

 September 27th, 2014, 09:07 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,641 Thanks: 2083 Suppose the opposite and adjacent sides are of length 1 and √3 respectively, so that the hypotenuse has length 2. If θ is the angle between the hypotenuse and the side of length √3, csc(θ) = 2. That implies that θ = arccsc(2) = 30°. In the above, I first wrote an equation that involved csc, then I used arccsc to solve that equation. In a sense, I "used" both functions. As I happen to know that csc(30°) = 2, I could have deduced directly that θ = 30°. However, such knowledge is equivalent to knowing that arccsc(2) = 30°, so I would still implicitly have been using arccsc. If you know the formula θ = arccsc(hyp/opp), you can proceed similarly, but without having to involve the csc function. One can sometimes find an angle without use of any inverse trigonometric function (even implicit use). For example, each angle of an equilateral triangle must be 60° because the three angles are the same and must total 180°. In general, though, you would need to use an inverse trigonometric function. I think the wording of the question was a bit unusual. It was presumably because a particular method was wanted. If any method of solution had been allowed, you could have used the arctan function, and this would have been simpler than calculating the hypotenuse and then using arccsc. In general, you can't use just the six trigonometric functions to find an angle. However, once you know two angles of a triangle, you can, of course, calculate the third angle without further use of trigonometry. Last edited by skipjack; October 10th, 2014 at 05:55 AM.
 October 1st, 2014, 07:36 AM #5 Newbie   Joined: Jan 2014 Posts: 14 Thanks: 0 Another question of a kind: Define Sin without right triangle
October 1st, 2014, 08:34 AM   #6
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Quote:
 Originally Posted by mac_alleb Another question of a kind: Define Sin without right triangle
the measurement of one of the non-right angles (q) the length of the side adjacent to that angle. the length of the triangle's hypotenuse. Futhermore, Definition I gives exact equations that describe each of these relations: sin(q) = opposite / hypotenuse.

 October 1st, 2014, 09:26 AM #7 Newbie   Joined: Jan 2014 Posts: 14 Thanks: 0 Arctan as well as all trig functions possibly has defs throw rows and polynomial. But the very question is How to connect it with triangle def. More future Sin := f(x,y)???
 October 1st, 2014, 04:40 PM #8 Global Moderator   Joined: Dec 2006 Posts: 20,641 Thanks: 2083 One could define sin(A) as a/d, where d is the diameter of the circumcircle of the triangle ABC with side of length a opposite angle A.
 October 9th, 2014, 06:11 AM #9 Newbie   Joined: Sep 2014 From: Somewhere Posts: 3 Thanks: 0 Thanks for the explanation Your efforts are highly appreciated.
 October 13th, 2014, 02:45 PM #10 Newbie   Joined: Oct 2014 From: Ohio Posts: 9 Thanks: 4 You have to use the inverse trig functions to find angles. You have not properly understood what your teacher was doing in class. Last edited by skipjack; October 13th, 2014 at 08:25 PM.

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