Date: 08/01/2003 at 09:37:49 From: Umair Subject: Trigonometry I couldn't solve this question and hoped that you could provide me with some help. Thanks. 2tan x + sin^2 x * sec x = 1 + sec x
Date: 08/01/2003 at 17:17:49 From: Doctor Ian Subject: Re: Trigonometry Hi Umair, Usually the first thing to do is to replace all other functions with sines and cosines. For example, tan(x) is the same as sin(x)/cos(x), so I would rewrite the equation as 2 sin(x) ---------- + sin^2(x) sec(x) = 1 + sec(x) cos(x) Try that with sec(x), and let me know if you're still stuck. - Doctor Ian, The Math Forum http://mathforum.org/dr.math/
Date: 08/01/2003 at 18:27:47 From: Umair Subject: Trigonometry Thanks Dr. Ian. I applied your advice to the equation: 2tan(x) + sin(2x)sec(x) = 1 + sec(x) I ended up with: 2sin(x) + sin(2x) = cos(x) + 1 Could you please provide me with some more advice?
Date: 08/01/2003 at 21:50:32 From: Doctor Ian Subject: Re: Trigonometry Hi Umair, The next step is usually to see if you can use identities like the ones in our Trigonometry Formulas FAQ, http://mathforum.org/dr.math/faq/formulas/faq.trig.html In particular, you usually want to reduce everything to sines and cosines of a single variable. In this case, there is an identity that you can use to change the argument from (2x) to (x). It is sin(2x) = 2 sin(x)cos(x), Substituting this gives 2sin(x) + 2sin(x)cos(x) = 1 + cos(x) Can you take it from here? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/
Date: 08/01/2003 at 22:02:52 From: Umair Subject: Trigonometry I solved it and got sin(x) = 1/2, so therefore the solutions are pi/3 and 5*pi/6 (30 and 150 degrees). Which according to the question book are right, but there is one more answer according to the book that I can't get, which is also correct but is not a value of x in sin(x) = 1/2. It is pi (180 degrees), so logically the only way I can get this if cos(x) = -1. Once again, I need your help!
Date: 08/01/2003 at 23:28:59 From: Doctor Ian Subject: Re: Trigonometry Your solution is nicely done. But let's look at the original equation again: 2tan(x) + sin(2x) sec(x) = 1 + sec(x) We converted to sines and cosines to get 2sin(x) sin(2x) cos(x) + 1 ------- + ------- = ---------- cos(x) cos(x) cos(x) and expanded sin(2x) to 2sin(x)cos(x) to get 2sin(x) 2sin(x)cos(x) cos(x) + 1 ------- + ------------- = ---------- cos(x) cos(x) cos(x) and multiplied both side of the equation by cos(x) to get 2sin(x) + 2sin(x)cos(x) = cos(x) + 1 which we factored to get 2sin(x)(1 + cos(x)) = (1 + cos(x)) Now, we _can_ divide both sides by (1 + cos(x)), which gives us a simpler equation. But it also eliminates one of the possible solutions, which occurs when (1 + cos(x)) is equal to zero... which happens when x is equal to pi. Look at an equation like x^2 - 3x = 0 One way to 'solve' this is to divide both sides by x, giving you x - 3 = 0 x = 3 This is sort of like dividing 2sin(x)(1 + cos(x)) = (1 + cos(x)) on both sides by (1 + cos(x)) to get 2sin(x) = 1 sin(x) = 1/2 But in each case, you've eliminated a solution by dividing instead of factoring. Let's look at the first one. If instead of dividing, we do this: x^2 - 3x = 0 x(x - 3) = 0 we can see that there are _two_ solutions: x=3, and x=0. We 'lost' the second solution when we divided both sides by x. In the second case, if we hold off dividing immediately, we can look at 2sin(x)(1 + cos(x)) = (1 + cos(x)) and see that this fits the pattern xy = y which has _two_ solutions: x = y, and y = 0. Dividing by y 'loses' this second solution. Does that make sense? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/
Date: 08/02/2003 at 10:00:26 From: Umair Subject: Thank you (Trigonometry) That makes perfect sense. I do appreciate all this help you have given me. Thanks, doc.
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