Pre-Calculus Questions (Trig kinda stuff) help!?
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The crux of this problem is for you to realize that the given angles are "complementary pairs", which add up to 90 degrees.\n\nYou are expected to know that the "co-function" of angle A gives the same value as the "function" of angle A's complement.\n\nSpecifically:\n\n cos(A) = sin(90-A)\n\nA graphical way of looking at this statement is that "cosine and sine are the same function, but offset by 90 degrees!"  We call that "out of phase by pi/2 or 90 degrees."\n\nThe same works for tangent and its co-function:\n\n tan(A) = cot(90-A)\n\nSo the answer to the second problem is 0.\n\nIf one of the fuctions in the first problem had been sin, the quotient would have been 1.  However, that's not the case, and I leave the upper problem as an exercise for another Answerer.\n\n Hope this helps!
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θ = α + β  where α = tan^-1 (1/3) and β = tan^-1 (1/2)\nie tanα = 1/3 and tanβ = 1/2\n\ntanθ = tan(α + β)\n= (tanα + tanβ)/(1 - tanαtanβ)\n= (1/3 + 1/2)/(1 - 1/3 * 1/2)\n= (5/6)/(5/6)\n= 1\nSo θ = ....., π/4, 5π/4, .....\nSo cosθ = ±1/√2 = ±√2/2
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I'm no calculus genius or anything but...\n∫dx/(sec^2)(5x)\n∫(cos^2)(5x)dx\n∫(1+cos(2*5x))/2 dx\n∫1/2dx + ∫ cos(10x) dx\nx/2 + sin(10x)/20
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∫x³/√(x² + 9)dx Let x = 3 tanθ therefore dx = 3 sec²θ . dθ\n=∫27 tan³θ/√(9 tan²θ + 9) . 3 sec²θdθ\n= 27 tan³θ/√(tan²θ + 1) . sec²θ . dθ (tan²θ + 1 = sec²θ)\n= 27∫tan³θ/secθ . sec²θ . dθ \n= 27∫tan³θsecθ . dθ \n\n= 27∫tanθsecθ (sec²θ + 1) . dθ \n= 27∫(sec²θ. tanθsecθ + tanθsecθ) . dθ (NOTE dsecθ/dθ = tanθsecθ)\n= 27( ⅓sec³θ + secθ) + c\nIf x = 3 tanθ then tanθ = x/3 so cosθ = 3/√(x² + 9)\nhence secθ = √(x² + 9)/3\n\nThus required integral = ⅓(x² + 9)^(3/2) + 9(x² + 9)^½ + c
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Hello...\n\nWell the first part is a bit tricky to explain and I'm not sure what range of x you want your answer in, because with inequalities and trig functions the solution lies in many intervals. So I'll explain the second one first\n\nIt's actually not that bad, you just need to be familiar with some trig identities\n\nIf you recall, I'm sure you learned that 2sinxcosx = sin(2x)\n\ntherefore, you can say that sin(2x)=sqrt(5)/3, and solve for x by taking the inverse sin of both sides.\n\nNow for the first problem, you can work the inequality a bit further and get cosx greater than or equal to -1/2\n\nIf you're looking for a simple interval, then x is between -2pi/3 and 2pi/3. But the actual answer also includes (4pi/3,8pi/3), (10pi/3,14pi/3)....\n\nThe best way to do the inequality is to draw the cosine function on your paper, draw y=-1/2 (horizontal line), and you can find the first intersection by taking the inverse cosine of (-1/2), which is 2pi/3. You'd have to study the symmetry of the cosine function, so since you know that it's symmetric about x=pi, you'll figure that the next point is 4pi/3 and so on and so forth and you'll see a pattern really quickly. \n\nThis may be more complicated than what you're looking for but I hope it helps you out. Good luck on your test
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A'ighty, first write sec(x) = 1/cos(x).  Now take the derivative using the Quotient Rule:\n\nd/dx (1/cos(x)) = (cos(x)*d/dx(1) - 1*d/dx(cos(x))) / cos^2(x)\n\n                        = (0 - (-sin(x))) / cos^2(x)\n\n                        = sin(x) / cos^2(x)\n\n                        = (1/cos(x)) * (sin(x) / cos(x))\n\n                        = sec(x)*tan(x).\n\nAnd there you go.  ^_^
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First, you need a protractor. Create a right triangle ABC with C as the right triangle. Let angle B be your angle. Make rough measurements of the three sides of the triangle. now:\nsin B = AC/AB\n\nOnce you know its sine, use these formulae for the other trigonometric functions:\ncos B = sqrt(1 - sin² B)\ntan B = sin B/cos B\n\nSpecial cases are:\nsin 0 = 0\nsin 90 = 1\nsin 30 = 1/2\nsin 60 = sqrt3/2\nsin 45 = sqrt2/2\nFrom these, you can compute all angles.\nremember,\nsin B = sin (B - 360n) where n is an integer\nsin B = cos (90 - B)\ncos B = sin (90 - B)\ncos (-B) = cos B\nsin (-B) = -sin B\n\nexample:\ncalculate sin, cos and tan of -150.\nfirst, sin -150 = -sin 150\n-sin 150 = -cos(90 - 150)\n= -cos (-60)\n= - cos 60 = -sin(90 - 60) = - sin 30\n= -1/2\nsin -150 = -1/2\ncos -150 = cos 150 = sin(90 - 150) \n= sin (-60) = -sin 60\n= -sqrt3/2\ntan -150 = sin -150/cos -150\n= (-1/2)/(-sqrt3/2)\n= sqrt3/3\n\nJust use these sign rules:\nif angle is in Q1 = sin cos and tan are +\nif angle is in Q2 = sin is +,cos is -,tan is -\nif angle is in Q3 = sin is -,cos is -,tan is +\nis angle is in Q4 = sin is -,cos is +,tan is -\n\n^_^
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im taking Trig. try finding a homework helper web page. or go to the book that ur learning from web page because the majority of the times they have practice help on there.
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Try this guy:\n\nhttp://dli.taft.cc.ca.us/mathvideo/Calculus/Calculas.htm
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I'm only going to calculate the principal values.\n\nLet x = cos^[-1](sqrt(2) / 2).\nTherefore, cos(x) = sqrt(2) / 2.\nSo, x = pi / 4 radians.\n\nLet y = sin[-1](-1).\nTherefore, sin(y) = -1.\nSo, y = - pi / 2 radians.\n\nYour expression is then : sec(x + y)\n= sec(pi / 4 - pi / 2)\n= sec(- pi / 4)\n= 1 / cos(- pi / 4)\n= 1 / [sqrt(2) / 2]\n= 2 / sqrt(2)\n= sqrt(2)
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I'll give you a hint and let you try to figure out the rest yourself (unless someone else tells you the exact answer). =)\n\nUse your trig identities:\n\ntan A = sinA/cosA\n\ncot A = 1/tanA or cot A = cosA/sinA
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I forgot how to do those, but it's some sort of trig identity isn't it?  Like\n\nsqrt (1-tan [x])\n\nProbably not though.
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Hopefully you'll find this site helpful.\n\nhttp://www.sosmath.com/calculus/calculus.html\n\nThere is a lot of information and example problems on limits and derivatives.  Good luck with your calculus class!
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Problem 1: Find an exponential function of the form f(x)=ba^x+c with y-intercept 2, horizontal asymptote y=-2, that passes through the point P(1,4). \n\nYou find the y-intercept by replacing x with zero, which gives f(x) [which is y on the graph] = 2, so b + c = 2 (since a^0 = 1)\n\nThe horizontal asymptote is y = c, so c = -2.  That means b must be [you figure it out]\nNow it is y = 4a^x - 2.  This must be true when x=1 and y=4.\nSo 4 = [you figure it out; answer deleted due to complaint from below]\n\n\nProb. 3.  What is the amount in the account if $500 is invested for 10 years at the annual rate of 5% compounded continuously? \n  Just plug in the values.\nA = 500 e^(.05 x 10)\n\n[answer deleted]
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I have a bit of calc links from when I took the AP exams.\nhttp://www.jimmyr.com/education.php\n\nunder the section calculus. those sites were really useful.
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I am taking College Algebra with Trigonometry and that is hard, so I know Calculus is real hard. That is a hard question to figure out.
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sec = (1/cos); cot = (1/tan) = (cos/sin)\n\ncos(-pi/3) = 1/2 \n\n1/1/2 = 2 = sec(- pi/3)\n\ncos(-5pi/4) = -sqrt(2)/2 ;      sin(-5pi/4) = sqrt(2)/2\n\n(-sqrt(2)/2)/(sqrt(2)/2) = -1 = cot(- 5pi/4)\n\n2 - (-1) = 3 There's your answer.