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This is so much work that it is not worthwhile to do in practice, although the formulae themselves are actually quite simple. The basic method is to use a so-called "infinite series". The angle must be expressed in radians. If the angle is in degrees, multiply it by (pi/180), to get the equivalent angle in radians. Then, use the formula:

sin(x) = x - x3/3! + x5/5! - x7/7! + x9/9!...

The individual terms become smaller and smaller, quite quickly, so the idea is to continue adding more terms until you see that the terms become so small that you can ignore them (depending on the desired degree of accuracy). An expression like 5!, read "five factorial" or "the factorial of five", means to multiply all natural numbers up to five: 5! = 1 x 2 x 3 x 4 x 5.

Similarly,

cos(x) = 1 - x2/2! + x4/4! - x6/6! + x8/8!...

There is a more complicated formula for tan(x), or simply calculate as follows:

tan(x) = sin(x) / cos(x)

The formulae for sin(x) and cos(x) are derived from the Taylor expansion, explained in basic calculus books.

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Continue Learning about Trigonometry

Can you transform sine functions into cosine functions?

If you know the measure of one angle, and the length of one side of a triangle, you can find the measures of the other sides and angles. From there, you can find the values of the other trig functions. cos (x) = sin (90-x) in degrees there are other identities such as cos^2+sin^2=1, so cos^2=1-sin^2


How tan9-tan27-tan63 tan81 equals 4?

tan(9) + tan(81) = sin(9)/cos(9) + sin(81)/cos(81)= {sin(9)*cos(81) + sin(81)*cos(9)} / {cos(9)*cos(81)} = 1/2*{sin(-72) + sin(90)} + 1/2*{sin(72) + sin(90)} / 1/2*{cos(-72) + cos(90)} = 1/2*{sin(-72) + 1 + sin(72) + 1} / 1/2*{cos(-72) + 0} = 2/cos(72) since sin(-72) = -sin(72), and cos(-72) = cos(72) . . . . . (A) Also tan(27) + tan(63) = sin(27)/cos(27) + sin(63)/cos(63) = {sin(27)*cos(63) + sin(63)*cos(27)} / {cos(27)*cos(63)} = 1/2*{sin(-36) + sin(90)} + 1/2*{sin(72) + sin(36)} / 1/2*{cos(-36) + cos(90)} = 1/2*{sin(-36) + 1 + sin(36) + 1} / 1/2*{cos(-36) + 0} = 2/cos(36) since sin(-36) = -sin(36), and cos(-36) = cos(36) . . . . . (B) Therefore, by (A) and (B), tan(9) - tan(27) - tan(63) + tan(81) = tan(9) + tan(81) - tan(27) - tan(63) = 2/cos(72) – 2/cos(36) = 2*{cos(36) – cos(72)} / {cos(72)*cos(36)} = 2*2*sin(54)*sin(18)/{cos(72)*cos(36)} . . . . . . . (C) But cos(72) = sin(90-72) = sin(18) so that sin(18)/cos(72) = 1 and cos(36) = sin(90-36) = sin(54) so that sin(54)/cos(36) = 1 and therefore from C, tan(9) – tan(27) – tan(63) + tan(81) = 2*2*1*1 = 4


What is cos2 A?

Cos(2A) = Cos(A + A) Double Angle Indentity Cos(A+A) = Cos(A)Cos(A) - Sin(A)Sin(A) => Cos^(2)[A] - SIn^(2)[A] => Cos^(2)[A] - (1 - Cos^(2)[A] => 2Cos^(2)[A] - 1


How do you prove that 2 sin 3x divided by sin x plus 2 cos 3x divided by cos x equals 8 cos 2x?

You need to know the trigonometric formulae for sin and cos of compound angles. sin(x+y) = sin(x)*cos(y)+cos(x)*sin(y) and cos(x+y) = cos(x)*cos(y) - sin(x)*sin(y) Using these, y = x implies that sin(2x) = sin(x+x) = 2*sin(x)cos(x) and cos(2x) = cos(x+x) = cos^2(x) - sin^2(x) Next, the triple angle formulae are: sin(3x) = sin(2x + x) = 3*sin(x) - 4*sin^3(x) and cos(3x) = 4*cos^3(x) - 3*cos(x) Then the left hand side = 2*[3*sin(x) - 4*sin^3(x)]/sin(x) + 2*[4*cos^3(x) - 3*cos(x)]/cos(x) = 6 - 8*sin^2(x) + 8cos^2(x) - 6 = 8*[cos^2(x) - sin^2(x)] = 8*cos(2x) = right hand side.


What is this expression as the cosine of an angle cos30cos55 plus sin30sin55?

cos(30)cos(55)+sin(30)sin(55)=cos(30-55) = cos(-25)=cos(25) Note: cos(a)=cos(-a) for any angle 'a'. cos(a)cos(b)+sin(a)sin(b)=cos(a-b) for any 'a' and 'b'.