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 June 19th, 2010, 04:42 PM #1 Member   Joined: Jun 2010 From: USA, Texas Posts: 85 Thanks: 0 Some Questions Already well I came across some questions that I wasn't really sure how to do. So bare with me because there's quite a bit, but I could really use some help understanding them and how to do them. ----------- Simplify expressions: 1.) (sin²t + cot²t-1) / cos²t Okay I know I'm suppose to separate it into two different fractions: (sin²t / cos²t) + (cot²t-1 / cos²t) But what next? The -1 throws me off too. ----------- Solve Equation: 2.) 2sinxcotx - cotx = 1-2sinx ------------ 3.) sin3x - sinx = 0 ------------ 4.) (sin14x) / (sin13x + sinx) Okay so I should separate this into two fractions too right? (sin14x / sin13x) + (sin14x / sinx) ------------- 5.) tanx - 3cotx = 0 ------------ 6.) (1 - cosx / sinx) + (sinx / 1-cosx) ------------ 7.) Graphing rectangular (Hint: Convert to polar coordinates) (x²+y²) = 36x²y² ------------ 8.) Find the compnent of u along v u = <9, -7> , v = <-3,8> On this, I don't really understand what the question is asking, am I suppose to multiply u and v together? ----------- 9.) A pilot is taking his jet going East at 410 mi/h with the wind blowing North at 40 mi/h. Find the velocity of the jet as a vector. ---------- 10.) (sinx + tanx)(cosx - cotx) ----------- 11.) 4cos²x - 4cosx+1 = 0 --------- 12.) [sin(p+y) - sin(p-y)] / [cos(P+y) + cos(p-y)] ---------- 13.) (sin14x / sin13x) + (sin14x / sinx) ---------- That's about it, sorry if this seems like a lot, as you can see, I have a problem with Trigonometry and simplifying expressions and solving them with k equaling any number :/....
 June 19th, 2010, 05:49 PM #2 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Some Questions Any chance you could number those questions? That would make it easier to offer assistance from my phone! For starters, there was an expression involving sines (I think) where you essentially wrote a/(b+c) = a/b + a/c. This is not correct. Tan -3cot = 0. Tan = 3cot tantan = 3 cottan tan^2 = 3 tan = ...
 June 19th, 2010, 10:34 PM #3 Member   Joined: Jun 2010 From: USA, Texas Posts: 85 Thanks: 0 Re: Some Questions Okay I numbered them. Question about your answer, how does cotx x tanx cancel out? Is it because of Tangent and Cotangent Identities Rules? And at Tan²x=3, does that break down even further?
 June 20th, 2010, 02:29 AM #4 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Some Questions They ate reciprocals, so when you multiply them, they cancel (or equal one, more precisely). This can also be seen by expressing them both in terms of sine and cosine. 2. 2sinxcotx - cotx = 1-2sinx Factor a cot from the left, and a (-1) from the right. cotx(2sinx - 1) = (-1)(2sinx - 1). You can "cancel" the (2sinx -1), but have to consider that a solution(s) exists when this expression = 0. (another way to see this would be to move everything to one side, then factor by grouping. Either way, you'll get) cotx = -1 or sinx = 1/2 4. Is the one with the incorrect step. 5.Tan^2=3 tan = +/- sqrt(3). From the square root property. This is solvable by using the unit circle. Should be four solutions in a period of 2pi. 6. I assume you mean (1 - cosx) / sinx + sinx / (1-cosx) when you see the expression: one, plus or minus a sine or cosine (NOT sine or cosine SQUARED), you should try using the conjugate of that term, then the fundamental trig identity. So in this case, look at the second term sin/(1-cos). Since the expression is in the denominator, we'll start here. Multiply top and bottom by 1+cos and see what happens. You should end up with 2/sinx, if I haven't totally screwed this up! 11.4cos²x - 4cosx+1 = 0 do you know how to factor 4u^2 -4u +1 ??
 June 20th, 2010, 08:29 AM #5 Member   Joined: Jun 2010 From: USA, Texas Posts: 85 Thanks: 0 Re: Some Questions Thanks for your help, mind showing what to do on 4 too since I didn't do the step right. As for 11, 4u² - 4u + 1 = 0 would factor out to: 4u² = 4u - 1 squareroot(4u²) = squareroot(4u - 1) 4u = squareroot(4u-1) 4u / 4 = squareroot(4u-1) / 4 u = squareroot(4u-1) / 4 Now to do with trig...
June 20th, 2010, 09:33 AM   #6
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Joined: Nov 2009
From: Northwest Arkansas

Posts: 2,766
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Re: Some Questions

Quote:
 Originally Posted by RMG46 Thanks for your help, mind showing what to do on 4 too since I didn't do the step right. As for 11, 4u² - 4u + 1 = 0 would factor out to: 4u² = 4u - 1 squareroot(4u²) = squareroot(4u - 1) 4u = squareroot(4u-1) 4u / 4 = squareroot(4u-1) / 4 u = squareroot(4u-1) / 4 Now to do with trig...
I'm going to say this plainly because you need to recognize a certain problem:
you do not know the difference between solving a LINEAR equation and solving a QUADRATIC equation.
To solve linear (in x), get all the x's on one side and everything else on the other. Then divide by the coefficient of x.

To solve quadratic, put it in standard form, then:
1. Factor, zero product property,
2. Complete the square (don't do this), or

You tried to isolate x^2 like you would for x(^1) in a linear equation. Big no-no!!!
You should get (2u-1)(2u-1) = 0.

For 4.... I thought I knew! I'll get back to you.

 June 20th, 2010, 09:43 AM #7 Member   Joined: Jun 2010 From: USA, Texas Posts: 85 Thanks: 0 Re: Some Questions Oops my bad -.- So: 4cos²x - 4cosx+1 = 0 = (2cosx-1)(2cosx-1)
 June 20th, 2010, 09:44 AM #8 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Some Questions When you "FOIL" that out, the first term isn't right! The 4's should be 2's...
 June 20th, 2010, 09:55 AM #9 Member   Joined: Jun 2010 From: USA, Texas Posts: 85 Thanks: 0 Re: Some Questions Sorry accident typo... As for number 6, I broke it down to: sinx + sinx cosx / sin²x Reducing down to: sin2x / sin²x That right so far? And does the sin2x = sin²x = sinx² = 2sinx? Or is it: sin2x = 2sinx //// sin²x = sinx²
 June 20th, 2010, 12:38 PM #10 Member   Joined: Jun 2010 From: USA, Texas Posts: 85 Thanks: 0 Re: Some Questions Someone mind doing 7, 8, and 9? Those don't use identities so I don't know where to being on them.. :[

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