Sin and cosines question?

2012-01-03 6:22 pm
Find sin x and cos x for x=0, 兀/2, 兀/6

I forgot how to do this, please explain.

回答 (6)

2012-01-03 6:56 pm
兀=180 degree cos(x)=cos 0=1
sin(x)=0 =sin o=0 cos(x)=cos 兀/2=90=0
sin(x)=兀/2=180/2=90=1 cos(x)=cos兀/6=30=√3/2
sin(x)=兀/6=180/6=30=1/2
2012-01-03 6:34 pm
{Sin[x], Cos[x]} for x = {0, pi/2, pi/6}


{{0, 1, 1/2}, {1, 0, Sqrt[3]/2}}
2012-01-03 6:33 pm
Use a calculator for now, and try to memorize them for future reference.

Tips:
cos(x) = sin(兀/2 - x), or in degrees, cos(x) = sin (90° - x). So you only have to memorize one set of answers.
The sin and cos of several commonly-used angles are of the form √n / 2.
2012-01-03 6:31 pm
For x=0, Sinx=0 & Cosx=1
For x=pi/2=90, Sinx=1 & Cosx=0
For x=pi/6=30, Sinx=(1/2) & Cosx=[sqrt(3)/2]
2012-01-03 6:27 pm
Anything in terms of π are in radians, where π = 180°.

You can simply set your scientific calculator into radian mode and just substitute the values for x in, or draw yourself a sine and cosine graph and read along, as the values for x you have given are pretty standard ones.

sin(0) = 0
sin(π/2) = 1
sin(π/6) = 0.5

cos(0) = 1
cos (π/2) = 0
cos (π/6) = √(3)/2

:) Hope this helps!
2012-01-03 6:27 pm
For this type of problem, a unit circle would be the most useful tool.

At x = 0, sin 0 is 0 and cos 0 is 1.

At x = pi/2, sin pi/2 is 1 and cos pi/2 is 0.

At x = pi/6, sin pi/6 is 1/2 and cos pi/6 is 3^(1/2) / 2.


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