Math help

[dghh70 5]

How do you do logarithms, Cos, Tan, Sin by hand?

[Easy Bakes 253]

Wont your math teacher help you?

[Alejandro24 575]

Logarithm base 10.

Cos, ln(cos)

ln(Cos) = ln(e^-ix)/2 + ln(e^ix)/2

or

ln (1/2 (e^-ix+e^ix))

Sen, ln(Sen)

ln(Sen) = ln(-i)(e^-ix)/2 - ln(e^ix)/2

or

ln (1/2 (-i)(e^-ix+e^ix))

Tan, ln(Tan)

ln(Tan)=(((-i)(e^-ix+e^ix))/((e^-ix+e^ix))

or

ln(((-i)(e^-ix+e^ix))/((e^-ix+e^ix))

Solve the bold expressions as common logarithms. The non bold italic expressions are the equivalent of the trigonometric functions. All the equivalents have imaginary parts, i (i, ln(i)), and real parts, carefull with that. Hope so this will guide you. Because is too difficult to write mathematics expressions here in ST. So ask to your teacher too, he or she must help you in any doubt.

  Edited October 27, 2011 by Alejandro24  

[Duke87 1,070]

Well, before the days when calculators could do this stuff, values were typically read from tables. There is no way to do trig by hand numerically, however one can estimate graphically with a compass, protractor, ruler, and pencil. Simple trick, really. Draw a circle with a radius of one and take any angle in it:

This is taking advantage of the definitions of the trig functions (sine is opposite over hypotenuse, cosine is adjacent over hypotenuse, etc.) by creating triangles with the denominator side with a length of 1. So, for instance, the sine is AC over OC, but OC (hypotenuse) is 1, so it's AC/1 or just AC. For tangent (opposite over adjacent) and secant (hypotenuse over adjacent), you have created a triangle with an adjacent side of length 1. For cotangent and cosecant you're taking advantage of complementary angles and the properties that cot(Θ) = tan(π/2-Θ), and csc(Θ)=sec(π/2-Θ).

For logarithms, I don't know of a method other than guess and check.

  Edited October 27, 2011 by Duke87  

[A Nonny Moose 8,261]

If you have access to a public library, any book on mensuration (measurements) should contain the appropriate methods. I seem to recall that there are algebraic series for most basic trigonometric and logarithmic expressions. It has been years since I did this. I always use a computer. Here is the wiki link for the common logarithm. And the same for Trig functions that gives the summation series.

You can have fun with these, but I really don't think other than curiosity it is worth doing. A $10 calculator has these expressions built in.

[Duke87 1,070]

I seem to recall that there are algebraic series for most basic trigonometric and logarithmic expressions.

That's called a Taylor series. Theory goes that any continuous differentiable function can be approximated by a polynomial.

And yeah, that would do it.

[A Nonny Moose 8,261]

I seem to recall that there are algebraic series for most basic trigonometric and logarithmic expressions.

That's called a Taylor series. Theory goes that any continuous differentiable function can be approximated by a polynomial.

And yeah, that would do it.

Takes a lot of patience to get the number of decimal places you want, and hand calculating those factorials has to be checked carefully. Then there is the horrible long divisions. All things considered, I am at a loss to understand why anyone would do that any more. Most calculators that cost more than two bucks have the built in chip for these functions.

When I was in high-school (1950-1955) many of the needed values were in my head from multiple use. for example, off the top of my head, Log₁₀(2) = 0.30103. In my day more problems worked out to have a result of either zero or 1. Calculators were not allowed in the class room and you were expected to use tables.

  Edited October 29, 2011 by A Nonny Moose