eelnedross Posted October 13, 2011 Report Share Posted October 13, 2011 Hey IBS, i need help on this problems:Simplify:a)2sin(2A)cos(2A)b)2cos(3a)sin(3a)c)2cos2(4o)-1d)1-2cos2(3b)e)1-2sin2(5a)f)2sin2(3D)-1I've tried to solve them but i can't work them out. Thank you very much! Reply Link to post Share on other sites More sharing options...
dessskris Posted October 13, 2011 Author Report Share Posted October 13, 2011 I've typed a reply to this post twice and they were not sent due to my bad internet connection and my browser didn't show me what i typed when i clicked back but i shall not give up a)2sin(2A)cos(2A) b)2cos(3a)sin(3a) remember for sin: sin cos + cos sin a)sin(4A) b)sin(6a) c)2cos2(4o)-1 e)1-2sin2(5a) remember for cos: cos cos - sin sin for double angle: cos² - sin² = 2cos² - 1 = 1 - 2sin² c)cos(8o) d)cos(10a) d)1-2cos2(3b) = -(2cos²(3b) - 1) f)2sin2(3D)-1 = -(1 - 2sin²(3D)) I hope you know how to work out d and f then. if you need further explanation please ask! basically they're just making use of the double angle formulae. 1 Reply Link to post Share on other sites More sharing options...
eelnedross Posted October 13, 2011 Report Share Posted October 13, 2011 I've typed a reply to this post twice and they were not sent due to my bad internet connection and my browser didn't show me what i typed when i clicked back but i shall not give up a)2sin(2A)cos(2A) b)2cos(3a)sin(3a) remember for sin: sin cos + cos sin a)sin(4A) b)sin(6a) c)2cos2(4o)-1 e)1-2sin2(5a) remember for cos: cos cos - sin sin for double angle: cos² - sin² = 2cos² - 1 = 1 - 2sin² c)cos(8o) d)cos(10a) d)1-2cos2(3b) = -(2cos²(3b) - 1) f)2sin2(3D)-1 = -(1 - 2sin²(3D)) I hope you know how to work out d and f then. if you need further explanation please ask! basically they're just making use of the double angle formulae. Ahhh i got you! So basically 3a and 2a are like a new variable right? Thank you. I got confused. Reply Link to post Share on other sites More sharing options...
eelnedross Posted October 14, 2011 Report Share Posted October 14, 2011 I need help on this exercise: a)cos2x+5sinx=0 Again.. Our teacher barely explains anything to us and the book ain't helping either. Reply Link to post Share on other sites More sharing options...
genepeer Posted October 14, 2011 Report Share Posted October 14, 2011 (edited) cos2x = 1 - 2sin2xcos(2x) + 5sinx=01 - 2sin2x + 5sinx=0let u = sinx2u2 - 5u - 1 = 0FInd the two roots, u, then solve for x in sinx=u Edited October 14, 2011 by genepeer 1 Reply Link to post Share on other sites More sharing options...
eelnedross Posted October 14, 2011 Report Share Posted October 14, 2011 (edited) cos2x = 1 - 2sin2x cos(2x) + 5sinx=0 1 - 2sin2x + 5sinx=0 let u = sinx 2u2 - 5u - 1 = 0 FInd the two roots, u, then solve for x in sinx=u Thank you! I did that, but the answer still is: 3,33; 6,10 (Radians) And i get -0.19. Edit: Nevermind, i added pi to +0.16 and substracted +0.16 from 2pi and got the results, thank you very much! Edited October 14, 2011 by CSteamless Reply Link to post Share on other sites More sharing options...
Lero Posted October 18, 2011 Report Share Posted October 18, 2011 Need some help with c) but i'll write down the complete question:a) Write down the seventh degree Taylor polynomial for sin(x)b) Use this polynomial to estimate sin(π/12)c) Use the Lagrange error term to determine an upper bound to the error in this approximation. Any help is appreciated. (From series and differential equations option). Reply Link to post Share on other sites More sharing options...
Hexa Posted October 19, 2011 Report Share Posted October 19, 2011 I don't even know where to start with this one: Prove: cosƟ-isinƟ=cis(-Ɵ)O.o Reply Link to post Share on other sites More sharing options...
Lero Posted October 19, 2011 Report Share Posted October 19, 2011 I think I might be oversimplifying it but:cis(-Ɵ) = cos(-Ɵ) + isin(-Ɵ), simply be De Moivre's theorem.We know that cos(-Ɵ) = cos(Ɵ), and sin(-Ɵ) = -sin(Ɵ). This can be observed by imagining movements on the unit circle. Then it follows that cis(-Ɵ) = cos(Ɵ) - isin(Ɵ). By substitution of the above identities. Reply Link to post Share on other sites More sharing options...
Guest hellokitty818 Posted October 19, 2011 Report Share Posted October 19, 2011 How do you show that dT/dt=k(T-5)dT/dt=k(T-5)??? I really do not understand... Reply Link to post Share on other sites More sharing options...
jonathan810 Posted October 19, 2011 Report Share Posted October 19, 2011 For SL, do we have to know about cot, sec etc for trig derivatives? (i.e for the derivative of tan x, do we only use 1/cos2x ? Reply Link to post Share on other sites More sharing options...
Lero Posted October 19, 2011 Report Share Posted October 19, 2011 For SL, do we have to know about cot, sec etc for trig derivatives? (i.e for the derivative of tan x, do we only use 1/cos2x ?SL are not required to know the reciprocal or inverse trig functions. Yes for the derivative of tan(x), 1/(cos(x)^2) should be sufficient. Reply Link to post Share on other sites More sharing options...
Guest hellokitty818 Posted October 19, 2011 Report Share Posted October 19, 2011 For SL, do we have to know about cot, sec etc for trig derivatives? (i.e for the derivative of tan x, do we only use 1/cos2x ?SL are not required to know the reciprocal or inverse trig functions. Yes for the derivative of tan(x), 1/(cos(x)^2) should be sufficient.really?? my friends are learning it....oh wait that was for math II SAT...sorry! Reply Link to post Share on other sites More sharing options...
Lero Posted October 19, 2011 Report Share Posted October 19, 2011 How do you show that dT/dt=k(T-5)dT/dt=k(T-5)??? I really do not understand...I'm not sure what you are asking for precisely? Solve the differential equation?dT/dt = k(T - 5)dT/(T-5) = k dtIntegrate both sides: ln(T - 5) = kt + c.Apply the exponential function to both sides: T - 5 = e^(kt + c).T - 5 = e^(kt)* e^©. Usually textbooks denote the e^© term, which is the constant, by A.So, T = Ae^(kt) + 5.If this was what you were asking for? Reply Link to post Share on other sites More sharing options...
Guest hellokitty818 Posted October 19, 2011 Report Share Posted October 19, 2011 How do you show that dT/dt=k(T-5)dT/dt=k(T-5)??? I really do not understand...I'm not sure what you are asking for precisely? Solve the differential equation?dT/dt = k(T - 5)dT/(T-5) = k dtIntegrate both sides: ln(T - 5) = kt + c.Apply the exponential function to both sides: T - 5 = e^(kt + c).T - 5 = e^(kt)* e^©. Usually textbooks denote the e^© term, which is the constant, by A.So, T = Ae^(kt) + 5.If this was what you were asking for?I am not sure the question says show that dT/dt = k(T - 5) for some constant k Reply Link to post Share on other sites More sharing options...
Lero Posted October 19, 2011 Report Share Posted October 19, 2011 There must be some additional information with that question, you sure that's the whole question? Reply Link to post Share on other sites More sharing options...
jonathan810 Posted October 19, 2011 Report Share Posted October 19, 2011 um... for SL, do we still need to know about the trapezium rule and Newton-Raphson's rules for calculus??? Reply Link to post Share on other sites More sharing options...
iHubble Posted October 20, 2011 Report Share Posted October 20, 2011 (edited) The teacher at the Math Club (and also my EE supervisor) gave us a problem. It goes like this : If three segment of 3, 4 and 5 are respectively starting from each corner of an equilateral triangle and intercepting in a single point somewhere inside the shape, what is the area of the triangle ? Assuming that each side of the triangle is x and the area of an equilateral triangle is ((√3)x2)/4, you can apply the Heron's formula repetitively and then resolve the equation system by equalling the sum of the areas of the three small triangles to the area of an equilateral triangle formula. HOWEVER, this method is extremely long and requires a whole lot of algebra. Do you have any ideas about resolving this problem without a calculator ? I computed the equation and found the area, which is about 19 something (it can also be expressed with square roots and integers). I have drawn the triangle in GeoGebra (a geometry software) to see if my answer was good, and it was, but still. Edited October 20, 2011 by iHubble Reply Link to post Share on other sites More sharing options...
Lero Posted October 20, 2011 Report Share Posted October 20, 2011 um... for SL, do we still need to know about the trapezium rule and Newton-Raphson's rules for calculus??? Never heard of either? Definitely do NOT need to know them. Basically for SL your calculus is composed like so: Differentiation:- power rule, product rule, chain rule, quotient rule, basic trigonometric functions, and basic logarithmic. Integration:- power rule, u-substitution, basic logarithmic and trigonometric. Make sure you know your inflexions, minimum/maximums etc. But certainly no Newton-Raphson's or trapezium method which rely on approximations. Reply Link to post Share on other sites More sharing options...
Guest hellokitty818 Posted October 20, 2011 Report Share Posted October 20, 2011 How do you find the derivative of ln(2y+1)=xe^y?? Reply Link to post Share on other sites More sharing options...
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