0
Calculating trigonometric functions, such as sin, cos, tan, requires some fairly involved calculations. If you don't have a calculator, you best use tables.
Such functions are calculated with Taylor series; for example, if you want to calculate the sine of an angle, and the angle is specified in degrees, multiply by (pi/180) to convert to radians. Then, having the angle "x" in radians, you can use the formula:
sin x = x - x3/3! + x5/5! - x7/7! ...
Similarly:
cos x = 1 - x2/2! + x4/4! - x6/6! ...
Note that, although these are infinite series, they converge pretty quickly, especially for small angles. That means that the individual terms quickly get smaller and smaller.
Still curious? Ask our experts.
Chat with our AI personalities
Professor
Blake
CoachAdd your answer:
Earn +20 pts
What is the numerical value of cos 35 x sin 24?
In degrees? cos(35˚) = .81915, sin(24˚) = .40673;cos(35˚) * sin(24˚) = .33318In radians? cos(35) = -.90367, sin(24) = -.90558;cos(35) * sin(24) = .81836A calculator will achieve these results faster than wiki.answers. 9 times out of 10, at least.:-)
Solution for tan x plus cot x divided by sec x csc x?
(tan x + cot x)/sec x . csc x The key to solve this question is to turn tan x, cot x, sec x, csc x into the simpler form. Remember that tan x = sin x / cos x, cot x = 1/tan x, sec x = 1/cos x, csc x = 1/sin x The solution is: [(sin x / cos x)+(cos x / sin x)] / (1/cos x . 1/sin x) [(sin x . sin x + cos x . cos x) / (sin x . cos x)] (1/sin x cos x) [(sin x . sin x + cos x . cos x) / (sin x . cos x)] (sin x . cos x) then sin x. sin x + cos x . cos x sin2x+cos2x =1 The answer is 1.
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.
Find the derivatives of y equals 2 sin 3x and show the solution?
y = 2 sin 3x y' = 2(sin 3x)'(3x)' y' = 2(cos 3x)(3) y' = 6 cos 3x
How can you prove that 1-2 cosine squared over sine times cosine is equal to tangent minus cotangent?
sin2 + cos2 = 1 So, (1 - 2*cos2)/(sin*cos) = (sin2 + cos2 - 2*cos2)/(sin*cos) = (sin2 - cos2)/(sin*cos) = sin2/(sin*cos) - cos2/(sin*cos) = sin/cos - cos-sin = tan - cot
