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What else can I help you with?
How do you proof that there is one and only one value of theta makes cos of theta equals 1 true?
You cannot prove it because it is not true! cos(0) = 1 cos(2*pi) = 1 cos(4*pi) = 1 ... cos(2*k*pi) = 1 for all integers k or, if you still work in degrees, cos(0) = 1 cos(360) = 1 cos(720) = 1 ... cos(k*360) = 1 for all integers k
Solve Cos x - 1 equals 0?
cos x - 1 = 0 cos(x) = 1 x = 0 +/- k*pi radians where k = 1,2,3,...
What does cos omak mean?
cos= vagina omak= mother's cos omak = Mother's vagina Basically its, f**k your mother in arabic.
Evaluate Arcsin cos 45 degrees?
It is 45 + 360*k deg or 135 + 360*k degrees where k is an integer.
What is the integral of -5.4 sin kt?
(5.4 / k) cos(kt)
What is the integral of sin3ycos5ydy?
Best way: Use angle addition. Sin(Ax)Cos(Bx) = (1/2) [sin[sum x] + sin[dif x]], where sum = A+B and dif = A-B To show this, Sin(Ax)Cos(Bx) = (1/2) [sin[(A+B) x] + sin[(A-B) x]] = (1/2) [(sin[Ax]Cos[Bx]+sin[Bx]cos[Ax]) + (sin[Ax]cos[-Bx]+sin[-Bx]cos[Ax])] Using the facts that cos[-k] = cos[k] and sin[-k] = -sin[k], we have: (1/2) [(sin[Ax]Cos[Bx]+sin[Bx]cos[Ax]) + (sin[Ax]cos[-Bx]+sin[-Bx]cos[Ax])] (1/2) [(sin[Ax]Cos[Bx]+sin[Bx]cos[Ax]) + (sin[Ax]cos[Bx]-sin[Bx]cos[Ax])] (1/2) 2sin[Ax]Cos[Bx] sin[Ax]Cos[Bx] So, Int[Sin(3y)Cos(5y)dy] = (1/2)Int[Sin(8y)-Sin(2y)dy] = (-1/16) Cos[8y] +1/4 Cos[2y] + C You would get the same result if you used integration by parts twice and played around with trig identities.
Which pair of angle measures satisfy the condition sin J cos K?
241
What is the solution to sec plus tan equals cos over 1 plus sin?
sec + tan = cos /(1 + sin) sec and tan are defined so cos is non-zero. 1/cos + sin/cos = cos/(1 + sin) (1 + sin)/cos = cos/(1 + sin) cross-multiplying, (1 + sin)2 = cos2 (1 + sin)2 = 1 - sin2 1 + 2sin + sin2 = 1 - sin2 2sin2 + 2sin = 0 sin2 + sin = 0 sin(sin + 1) = 0 so sin = 0 or sin = -1 But sin = -1 implies that cos = 0 and cos is non-zero. Therefore sin = 0 or the solutions are k*pi radians where k is an integer.
Integrate sin x cos 2x?
By Angle-Addition, cos(2x) = 2cos(x)^2-1 So, sin(x)cos(2x) = [2cos(x)^2-1]sin(x) = 2sin(x)cos(x)^2 - sin(x) Int[2sin(x)cos(x)^2 - sin(x)] = (-2/3)cos(x)^3 + cos(x) +K
What does x equal in this problem. cos(6x) sin(3x - 9)?
The solutions are:x = 11 + k*120x = 51 + k*120x = 87 + k*120where k is an integer.In the interval (0, 360) there are 9 solutions.
If you have Cos theta equals 0.92568 How do you find theta in radians?
cos(t) = 0.92568 therefore t = cos-1(0.92568) = 0.3880. If the answer comes out as 22.23, the calculator is set to degrees. Simply multiply that result by pi/180 to convert to radians (or reset the calculator to work in radians). Excel, for example, works in radians. From that primary value you get t = 0.3880 + 2*k*pi and t = 2*k*pi - 0.3880 for all integer values of k.
What kind of roles does the male lion have?
Bog roles. Do you get it? Cos of the defecation levels of male lions? No? K den.
