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How do you derive: 2cos(3-2x)?
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Don't you have a formula for Cos(A-B) ? the cosine of the difference of two angles...
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y = 2[cos1/2(f1 - f2)x][cos1/2(f1 + f2)x]
let f1 = 16 let f2 = 12 therefore y = 2[cos1/2(16 - 12)x][cos1/2(16 + 12)x] y =
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y = 2cos(2x)cos(14x) use the product rule: y' = 2(-sin(2x))(2)cos(14x) + 2cos(2x)(-sin(14x))(14) and
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cosA= 5/9 find cos1/2A
are you familiar with the half-angle formulas? the one I would use here is cos A = 2cos^2 (1/2)A - 1 5/9 +
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Yes, I am familiar with the half-angle formulas. In this case, we can use the half-angle formula for
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Determine the solution of the equetion: 2cos x-1=0? 2cos x=0? and 2cos x+1=0?
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2cosx - 1 = 0 2cosx = 1 cosx = 1/2 I know from my basic trig relations that cos 60° = 1/2 and by
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Define and then derive the expression for the MRTS. How do you derive this? I thought derive meant to receive or take something.
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MRTS is the marginal rate of technical substitution. In a two-input production function, it is
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how does 2cos(2x)-2cos(x)sin(x) becomes 2cos(2x)-sin(2x)? Try to explain each step simply.
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To understand how 2cos(2x) - 2cos(x)sin(x) becomes 2cos(2x) - sin(2x), let's break down the steps:
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the problem is
2cos^2x + sinx-1=0 the directions are to "use an identity to solve each equation on the interval [0,2pi). This is
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although, after looking at this problem i'm still not sure why i had to change the original problem?
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Multiply then use fundamental identities to simplify the expression below and determine which of the following is not equivalent
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recall that (a-b)(a+b) = a^2-b^2 so, 2^2 - (2cosx)^2 = 4 - 4cos^2x = 4(1-cos^2x) = 4sin^2x only B is
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What am I doing wrong?
Equation: sin2x = 2cos2x Answers: 90 and 270 .... My Work: 2sin(x)cos(x) = 2cos(2x) sin(x) cos(x) =
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wouldn't it be easier to solve for 2x then divide it by two? sin2x=2cos2x sin2x/cos2x=2 tan 2x=2
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Solve the equation on the interval 0 less than or equals theta less than 2 pi 0≤θ<2π. 2cos^(2)\theta +\sqrt(2cos\theta
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why do you write "the interval 0 less than or equals theta less than 2 pi" and then go right ahead
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Given that 2cos^2 x - 3sin x -3 = 0, show that 2sin^2 x + 3sin x +1 = 0?
Hence solve 2cos^2 x + 4sin x - 3 = 0, giving all
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2 cos^2 x - 3 sin x - 3 = 0 but cos^2 x = 1 - sin^2 x so 2 (1 -sin^2 x) - 3 sin 3 - 3 = 0 2 - 2
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A) -3sin(t)=15cos(t)sin(t) I have no clue... b) 8cos^2(t)=3-2cos(t) All i did was move
I'm trying to derive the formula
v^2 = v0^2 + 2a(x-x0) were zeros are subscripts my book tells me to derive it this way use the
I'm trying to derive the formula
v^2 = v0^2 + 2a(x-x0) were zeros are subscripts my book tells me to derive it this way use the
Derive absolute error in equations:
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The energy-separation curve for two atoms, a distance, r, apart is: U(r)=−Arm+Brn
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