(Assuming 100 is 100 degrees)
cos 100 degrees is equal to - sin 10 degrees.
In radians, this is - sin (pi/18).
Approximating pi as 22/7, this is - sin (11/63)
Using four terms of a Taylor series, this is approximately:
- (11/63) + (11/63)^3 /6 - (11/63)^5 / 120 + (11/63)^7 / 5040.
- (11/63) + (11^3/63^3) /6 - (11^5/63^5) / 120 + (11^7/63^7) / 5040.
Rewriting the fractions with the common denominator 63^7:
- (11*63^6/63^7) + (11^3*63^4/63^7) /6 - (11^5*63^2/63^7) / 120 + (11^7/63^7) / 5040.
Rewriting again with common denominators of 5040:
- 5040*(11*63^6/63^7)/5040 + 840*(11^3*63^4/63^7)/5040 - 42*(11^5*63^2/63^7) / 5040 + (11^7/63^7) / 5040.
Now add it all up:
[- 5040*11*63^6 + 840*11^3*63^4 - 42*11^5*63^2 + 11^7] / (5040*63^7)
Now, some seriously long multiplication gives:
(- 3466302962466960 + 17612440516440 - 26846879598 + 19487171) / 19852462421401680
Some really easy addition:
-3448717349342947 / 19852462421401680
And finally, do the long division:
-0.17371735939 ... and so on.
The actual value is -0.17364817766693034885171662676931
So, I got 3 decimal places right. Maybe use a better approximation of pi (like 355/113) and more terms of the Taylor series. But you get the picture.
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This can be done on a graphing calculator by making sure you have your calculator in degrees mode, and then tentering the cos(23). You get an answer of 0.9205048535.
Cos is short for Cosine ( Complementary Sine) Similrly Sin is short for Sine Tan is short for Tangent.
You can use the Pythagorean identity to solve this:(sin theta) squared + (cos theta) squared = 1.
To solve the equation 2cos(x) + cos(x) - 1 = 0, we first combine like terms to get 3cos(x) - 1 = 0. Then, we isolate the cosine term by adding 1 to both sides to get 3cos(x) = 1. Finally, we divide by 3 to solve for cos(x), which gives cos(x) = 1/3. Therefore, x = arccos(1/3) or approximately 70.53 degrees.
There are two ways to solve for the double angle formulas in trigonometry. The first is to use the angle addition formulas for sine and cosine. * sin(a + b) = sin(a)cos(b) + cos(a)sin(b) * cos(a + b) = cos(a)cos(b) - sin(a)sin(b) if a = b, then * sin(2a) = sin(a)cos(a) + cos(a)sin(a) = 2sin(a)cos(a) * cos(2a) = cos2(a) - sin2(b) The cooler way to solve for the double angle formulas is to use Euler's identity. eix = cos(x) + i*sin(x). Yes, that is "i" as in imaginary number. we we put 2x in for x, we get * e2ix = cos(2x) + i*sin(2x) This is the same as * (eix)2 = cos(2x) + i*sin(2x) We can substitute our original equation back in for eix. * (cos(x) + i*sin(x))2 = cos(2x) + i*sin(2x) We can distribute the squared term. * cos2(x) + i*sin(x)cos(x) + i*sin(x)cos(x) + (i*sin(x))2 = cos(2x) + i*sin(2x) And simplify. Because i is SQRT(-1), the i squared term becomes negative. * cos2(x) + 2i*sin(x)cos(x) - sin2(x) = cos(2x) + i*sin(2x) * cos2(x) - sin2(x) + 2i*sin(x)cos(x) = cos(2x) + i*sin(2x) Now you can plainly see both formulas in the equation arranged quite nicely. I don't yet know how to get rid of the i, but I'm working on it.