Does (sin^2)(x) mean [sin x]^2 or sin(x)^2?  Is there a difference?
400641
There's a difference.  It refers to (sin x)^2.  Plug in a few numbers and you can easily tell the difference.
180190
let f(x) = (x^2) sin(x+2)\nthe derivitive of f(x) [ f'(x) = 2x sin(x+2) + (x^2) cos(x+2)]\n\nThis is done by taking the product rule {g(x)h(x)}' = g'(x)h(x) + g(x)h'(x)( the derivitive of the first times the second plus the derivitive of the second times the first) \nwhere g(x) = x^2 and h(x) = sin(x+2)
1148883
I assume what looks 'funny' to you is the wild oscillations near x=0.  The reason is that for x near zero, the quantity 1/x gets really large, and as x gets closer to zero the value of 1/x changes a LOT for a small change in x.  What this means for sin(1/x) is that the sine function goes through many oscillations over just a small range of x (near zero) because its argument changes so much.  Does that make sense?\n\nSee if you can make an argument for what behavior to expect for sin (1/x) when x is far from zero instead (positive or negative).
843808
If sin A = x and sinB = -x  then \nA = .... , -4π - B, -3π + B, -2π - B, -π + B,.-B, π + B, 2π - B, 3π + B, 4π - B, .... \n= nπ + (-1)^nB (n is any integer)\n\nSo arcsin(x) - arcsin(-x) = nπ + (-1)^nB - B\n= nπ + ((-1)^(n) -1)B\n\nNo this looks too clumsy ....new strategy!!\n\nsinA + sinB = 0\nSo 2 sin(½(A + B))cos(½(A - B)) = 0\n\nThus sin(½(A + B)) = 0 or cos(½(A - B)) = 0\nie ½(A + B) = 0, ±π, ±2π, ±3π, .... , ±nπ, \nie A + B = ±2nπ\nie arcsinx + arcsin (-x) = ±2nπ\n\nOr ½(A - B) =  ±π/2, ±3π/2, ±5π/2, ... , ±(2n - 1)π/2\nSo A - B = ±(2n - 1)π\nSo arcsinx - arcsin(-x) = ±(2n - 1)π
876211
2 cos^2 x + sin^2 x = 1\n\ncos^2 x + cos^2 x + sin^2 x = 1\n\ncos^2 x + (cos^2 x + sin^2 x) = 1\n\ncos^2 x + 1 = 1\n\ncos^2 x = 0\n\ncos x = 0\n\nx = pi/2 or 3 pi/2
880711
sin(2A)=sin(A+A)=sinAcosA + sinAcosA=2sinAcosA
1154919
<1>...u know lim xsin(1/x)=lim (sin(1/x))/(1/x),and when x goes to infinity ,(1/x) goes to <0>....and we know that (sina)/a when <a> goes to <0> is 1...and here <a> is <1/x>...so it's limit is <1>
23861
juancho needed the square outside the parenthesis.  jota needed one more parenthesis before the square.  you need to square the whole [sin(x^2+2x+1)] thing.\nozlem is correct.\nhussein f is wrong.
162293
On both equations, you have x^ but don't say what x was raised to. I'm assuming that it is x^2. Is this correct?
1015891
Isn't it sinx divided by 6?\nSimple just cancel 'n' from the numerator and the denominator.\nU will be left with six=6
638739
This is generally used for answering or solving problems.\nFor example if you do something like:\n4sin^2(theta)-9cos^2(theta)=(2sin(theta)+3cos(theta))(2sin(theta)-3cos(theta)) \nYou may write "Using x^2-y^2=(x+y)(x-y)"\nHope this helps...
200647
sin^2(x)=0.5*(1-cos(2x)).this can be integrated
753754
I presume you mean\n\n[(6x^2) / (4x^2y-12xy)] * [(x^2+x-12) / (3x^2+12x)]\n\nThere seems to be something wrong. Are you sure about those "y" terms?
72362
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.
579155
The first one isn't, because it's undefined when x=4 (assuming the (x-4) is the denominator, not just the x)\n\nThe second one is; it's a horizontal line with no breaks or holes.
947972
(sec x)^2 + (tan x)^2 = 1 is not an identity.\nHere's how we arrive at that.\nStart with the identity (sin x)^2 + (cos x)^2 = 1\nDivide everything by (cos x)^2\n(sin x)^2 / (cos x)^2 + (cos x)^2 / (cos x)^2 = 1 / (cos x)^2\nYou know that (sin x) / (cos x) = tan x, \nso (sin x)^2 / (cos x)^2 = (tan x)^2\n(cos x)^2 / (cos x)^2 = 1\nand 1 / (cos x)^2 = (sec x)^2\nWe get (tan x)^2 + 1 = (sec x)^2\nWe can rearrange this to get\n(sec x)^2 - (tan x)^2 = 1\nAs you can see, this is not the same as (sec x)^2 + (tan x)^2 = 1
925769
sin(θ) ≈ θ\nSo cos (θ) ≈ 1  because cos (0) = 1\nMore accurately cos (θ) ≈ 1 - θ²/2