Using a sum/difference identity, help find the exact value of cos 465 degrees?
468102
cos(465) = cos(465-360) = cos(105)\n= cos(45+60)\nNow use the formula for cos(a+b)
1185389
There's ageneral formula for solving questions of this type\n \nRemeber \ncosx+cos2x+...+cosnx\n=cos(nx/2) sin((x+nx)/2)/(sin(x/2))\n\nNow just plug the values.\nhere x=1\nn=100\nSum= cos(100/2)sin((1+100)/2) / (sin1/2)\n=cos50sin(50.5)/sin0.5\n\nFormula is valid for all such series if angles increase in arithmetic series.\n\nTo see how the formula is derived\nJust form the series sinx+sin2x+..+sin nx =S\nAlso let cos x +cos 2x+...+cos nx=C   \nlet i=sqrt(-1)\nForm the sum C+iS=exp(ix)+exp(2ix)+...+exp(nix)\nThis is obtained by Euler's formula cos a+i sin a =exp(ia)\nSum the C+iS as geometric series and separate real and imaginary parts and you can see formula for both the series
925769
sin(θ) ≈ θ\nSo cos (θ) ≈ 1  because cos (0) = 1\nMore accurately cos (θ) ≈ 1 - θ²/2
601534
To convert from polar to cartesian co-ordinates, yousimply use:\n\nx = r cos (theta)\n\nand:\n\ny = r sin (theta)\n\nYou have radians - so make sure your calculator is set to radians.  This gives:\n\nx = 15 × cos (2.75) = -13.864...\ny = 15 × sin (2.75) = 5.7249...
40893
Remember this identity?\ncos A - cos B = - 2 sin ½(A + B) sin ½(A - B)\n\nIn this case, A is 5X and B is 5X\n3sin(3X)sin(2X) = -1.5(-2sin(½(5X+X))sin(½(5X-X)))\n\nUsing the identity I gave at first you get:\n-1.5(cos(5X)-cos(X)) = 1.5cos(X)-1.5cos(5X)\n\nYou could also write this using only additions as:\n1.5cos(X) + 1.5cos(5X+pi)
567245
simple x=r*cos(theta), y=r*sin(theta), r^2=x^2+y^2\nsubstitude, not hard
27656
I used the acronym sohcahtoa to remeber that sin=opp/hyp cos=adjacent/hyp and tan=opp/adj, so you have a angle you want to measure the degree of that angle, using the lengths of the sides of the triangle opposite, adjacent and hypotenuse and then using your calculater you divide the lengths and hit the button sin-1, cos-1 or tan-1 key, also cot=1/tan, hope that helps
1041436
I got  the first part. just get the cosx by itself. then you wind up with the cosx = -(the square root of 2/2)\nwhich is 225+360n and 315 +360n(in degrees)\nSorry i can't help you on the other one. Good luck, just keep trying new things with it, would be my advice.
994720
Well, I am going to assume that the function is: d=28+24cos[("pie"/5)(t-6)]. First of all, minus the 28, then divide by 24, then take the inverse cosine (arcosine), then divide by ("pie"/5) then add six, so it should result:\narccos [(d-28)/24] * (5/"pie") + 6= T
519602
Just draw the vectors\nMake sure you understand at what nagles they are\nr1 and r2 are inclined at 45 deg\nYou have asked only for magnitude\nJust apply the formula sqrt(4^2+2^2+2*4*2*cos 45)\nsqrt(16+4+16/sqrt(2))=5.595\nFor the difference part.Just reverse the sign for cos term\nyou get |r1-r2|=2.9472
94104
u can do it by trial and error. ok the possible digits for ones are just 1,3,5,7,9 since the statement says it is an odd number. ok then try to add 2 and equate it to 12. if it is = that's is the answer. 7+5= 12. so and 7 is greater than 5 by 2. the answer is 75 degrees.
797886
Have you learned trig identities yet?  I'd check to see if there was a trig identity you could substitute here.\n\nBTW, I am assuming you meant cos²x rather than cosx².\n\nTry this:  cos²x + sin²x = 1, so substitute 1-sin²x for cos²x:\n\n6cos²x + sinx - 4 = 0\n\n6(1-sin²x) + sinx - 4 = 0\n\n-sin²x + sinx + 2 = 0\n\n(multiply by -1 to make it a little easier)\n\nsin²x - sinx - 2 = 0\n\n(sinx-2)(sinx+1) = 0\n\nSo, sinx = 2 or sinx = -1\n\nSolving for x, this means x = sin^-1(2) or sin^-1(-1)
1029343
Unfortunately the above answer isn't right. The correct answer is:\n\n(-5/128)cos(4x) + (1/64)cos(8x) + (-1/384)cos(12x) +C\n\nThe trick with these problems is you need to use the power reducing trig identity to get the power of either \nsin(2x) or cos(2x) to be even. For example\n\nInt sin^4(2x)cos(2x) dx = (1/10) sin^5(2x) + C and\nInt cos^4(2x)sin(2x) dx = (-1/10) cos^5(2x) +C\n\nIf you have an odd power it screws up:\n\nInt sin^3(2x)cos(2x) = (1/64)(cos(8x) - 4cos(4x)) + C\n\nThis is because of the relationship sin^2(x)+cos^2(x)=1\n\nUse the trig identities:\n\nsin^2(2x) = (1-cos(4x))/2 or cos^2(2x) = (1+cos(4x))/2  \n\nThis will reduce the power by 1 and convert sine to cos or cos to sine. Since one can always be the derivative or the negative derivative of the other, for integration purposes you can do either. In your case use the sine squared reduction formula. Multiply the terms out and integrate termwise. \n\nHere's a link to useful trig identities:\n\nhttp://www.sosmath.com/trig/Trig5/trig5/trig5.html
949562
Use csx =1/sin and  1-cos^2=sin^2  and cot=cos/sin\n Ready...Go\ncscx - (cosx)(cotx) =  1/sin - cos^2/sin = (1-cos^2)/sin=sin^2/sin =sin\n       ole, ole , infree
1149867
We have x=rcos(θ), and y=rsin(θ).  \n\n\nTherefore x=4sin(θ)cos(θ) and y=4sin(θ)sin(θ).  \nAnd x^2+y^2= 16sin(θ)^2(cos(θ)^2+sin(θ)^2) =16sin(θ)=4y.  \nThus x^2+y^2-4y+4-4 =x^2+(y-2)^2-4=0 and x^2+(y-2)^2=4.\n\nAre you sure the second one is given by:\nρsin(θ)=2cos(θ) and not ρsin(φ)=cos(θ)?\n\nif it is ρsin(φ)=2cos(θ), then first notice that φ≠kπ for any integer k.\nThen we have ρ=2cos(θ)/sin(φ)\nNext we have x=ρcos(θ)sin(φ)=2cos(θ)^2, and y=ρsin(θ)sin(φ)=2cos(θ)sin(θ).\n\nThus x^2+y^2= 4cos(θ)^2(cos(θ)^2+sin(θ)^2) =2x,\nand x^2-2x+1-1+y^2= (x-1)^2-1+y^2=0,\nso (x-1)^2+y^2=1
151637
sin 140cos 50 + cos 140sin50\n= sin (140 + 50)\n= sin 190\nIf the purpose is to express the answer as a single trigo function, then that is the answer.\nHowever, if the value is to be solved, then a scientific calculator has to be used. Also,\n\nsin 190 = sin (180-190) = sin (-10) = -sin 10
500080
Add a short diagonal line to one of the plus signs to make it a 4:\n5 + 545 = 550