I Unit circle bug?

[aronclark1017]

 

[jedishrfu]

People dont use the unit circle for these kinds of problems instead they use trig identities to reduce it.

 

[PeroK]

View attachment 365996

Note that any point on the unit circle has coordinates $(\cos \theta, \sin \theta)$. This is perhaps the simplest way to define sine and cosine in the first place.

The negative y-axis is at an angle of $3\pi/2$. And the angle $3\pi/2 - x$ is an angle $x$ clockwise from the negative y-axis. From the geometry, the coordinates of that point are $(-\sin x, -cos x)$. So that:
$$\cos(3\pi/2 - x) = -\sin x, \ \ \sin(3\pi/2 - x) = - \cos x$$

 

[jedishrfu]

Thats a great insight @PeroK. I never that kind of derivation before.

 

[aronclark1017]

Note that any point on the unit circle has coordinates $(\cos \theta, \sin \theta)$. This is perhaps the simplest way to define sine and cosine in the first place.

The negative y-axis is at an angle of $3\pi/2$. And the angle $3\pi/2 - x$ is an angle $x$ clockwise from the negative y-axis. From the geometry, the coordinates of that point are $(-\sin x, -cos x)$. So that:
$$\cos(3\pi/2 - x) = -\sin x, \ \ \sin(3\pi/2 - x) = - \cos x$$

the following documentation is saying that sinx=cos(3pi/2-x). In all the former cases reflecting the triangle to come off of the Y axis as you see in 90-x derives the intended value but in this cases is not showing in the this documentation example. See..
Symmetric identities
If we draw a few copies of the triangle, we get:
$$\sin(x)=\cos(90-x)=-\cos(90+x)=\sin(180-x)=-\sin(180+x)=\cos(270-x)=-\cos(270+x)=-\sin(-x)$$
$$\cos(x)=\sin(90-x)=\sin(90+x)=-\cos(180-x)=-\cos(180+x)=\sin(270-x)=-\sin(270+x)=\cos(-x)$$
$$\tan(x)=\cot(90-x)=-\cot(90+x)=-\tan(180-x)=\tan(180+x)=\cot(270-x)=-\cot(270+x)=-\tan(-x)$$
The other three can be derived by taking the reciprocals of these three.

x is easier to type than theta

https://artofproblemsolving.com/wiki/index.php/Proofs_of_trig_identities

 

[PeroK]

the following documentation is saying that sinx=cos(3pi/2-x).

Try with $x = 30^\circ$ and you'll see this is wrong.

 

[aronclark1017]

yes it would appear to be missing the negative. But this documentation seems very creditable. I suppose it doesn't matter much as in this case of derivation of double angle formulas using unit circle.

 

[jedishrfu]

Sometimes credible things upon inspection are not so credible. This is the bane of self study students.