A Related Rates dealing with trig functions?
825620
A diagram shows that the UFO is 40008 ground feet from the camera at the time of the 32 degree angle (tan 32 = 25000/g)\n\nThe angle will equal tan^-1 (25000/(40008 - 3000t))\n\nFind the derivative of this and you have your change in camera angle.
276671
(e^(cose^e+sin^e)/2)^pi\nhahaha ..just kidding
862258
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.
509207
Yes. You can use sine or cosine to determine the angle.\n\nSine x = Opposite / Hypotonuse\nSin x=5/13\nSin x=.38\nx=22.6*\n\nCosine x = Adjacent / Hypotonuse\nCos x=12/13\nCos x=.92\nx=22.6*
349408
The answer is B) 1.122\n\nFor the different trigo. functions:\n\nLet:\na = side opposite angle n\nb = side adjacent angle n\nc = hypotenuse of the right triangle\n\nTherefore:\n\ncos n = adj./hyp. = b/c\nsin n = opp./hyp. = a/c\ntan n = opp./adj. = a/b\nsec n = hyp./adj. = c/b\ncsc n = hyp./opp. = c/a\ncot n = adj./opp. = b/a\n\nYou will notice the following:\n\n1) the secant is the reciprocal of cosine.\n2) the cosecant is the reciprocal of sine.\n3) the cotangent is the reciprocal of tangent.\n\nTherefore: \n\ncsc n = 1/sin n = 1/0.891 = 1.122\n\nFor the remaining trigo. functions, solve for n first:\n\nn = arcsin 0.891 = 62.99917663˚ or 62˚59'57.04"\n\nTherefore:\n\n1) cos 62˚59'57.04" = 0.454\n2) tan 62˚59'57.04" = 1.963\n3) sec 62˚59'57.04" = 1/cos 62˚59'57.04" = 2.203\n4) cot 62˚59'57.04" = 1/tan 62˚59'57.04" = 0.510\n\nI hope that the solution was clear enough...
100448
There are 2 x pi radians per revolution, by definition of radian.  So \n\n45 rev./min x 1 min/(60 s) x 2 PI rad/rev =4.71 rad/s
878725
antiderivative of:\n20-12cos2pi(x-0.1)\nis:\n\n20x - 12 sin( 2pi(x-0.1)) / 2pi +C\n=20x - 6 sin( 2pi(x-0.1)) / pi +C\nto double check just take the derivative of the function and see that you get:  20-12cos2pi(x-0.1) '
830426
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!
624769
You can use sin and cosine.
263848
cot = o/a\ncsc = r/o\n\no = opposite side\na = adjacent side\nr = radius\n\ncalculate csc(2x) in terms of o and r.\nSubstitute o/a for cot(x) and substitue the value you determine for csc(2x) \nmultiply these, then convert to a trig function in terms of x.\n\nIf you know trig identities better than I do, you can reduce this expression using identities, without resorting to o, a, and r.  However, either route will work.
1132233
this question will require calcs using logarithm.\nhence will take time
890202
in trigonomertic form\n\nz = r cos t + ir sin t = 1 + i\n\nso r cos t = 1\n r sin t = 1\n\nsquare and add r^2 = 2\nso r = sqrt(2)\n\nnow cos t = 1/sqrt(2) and sint = 1/sqrt(2)\n\nso t = pi/4 (as cos pi/4 = 1/sqrt(2) and sin is also positive so 1st quadrant)\n\nso t = pi/4 and r = sqrt(2)
951198
^ Parasite alert LOL! ^\n\n30 s for C to move from A to B, maybe? I'll assume that. Also, x seems to be a length, not an angle. Maybe a) is asking for the rate of change of z (called z')? We know the angular rate of C about O (y') is pi/30 (answer to b)), and triangle CAO is isosceles, so z'=-0.5*pi/30. \nFor c), circumferential velocity is pi*r/30. Angle CAO is, conveniently, 45 deg, so x' = sqrt(0.5)*circumferential velocity.\nFor d), the area of triangle OAC is .5*AO*altitude, and altitude=r*sin(y). This is a maximum when y (not x) =pi/2.\nSorry, I know you wanted  no answers, but I couldn't see how/where to stop short of most of the answers.
768749
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.  ^_^
104581
It takes practice!  And its like a puzzle.  Just use the main trig identities as reference and a good math tutor.  Unfortunately, this place isn't the best place to ask that kind of question because we are limited by text explainations and trig identities need a lot more than that!
730722
put sinx=t\ncosxdx=dt\nI=fcos(pi t)dt with limits -1,+1\nI=(sin(pi t))/pi with limits +1,-1\nI=(sin(pi)-sin(-pi))/pi=0\nthanks & good luck
449531
Knowing trig makes learning calculus easier.