sin^2 (4pi/13) + cos^2 (4pi/13)

[jenn9580]

sin^2 (4pi/13) + cos^2 (4pi/13)

I have no idea how to get sin & cos from this. I can't draw a refernce triangle with a 55degree angle. We have only done isosceles triangles and 30-60 triangles. How do I even get started in the right direction?

 

[arthur ohlsten]

I am not sure what it is you need.
Have you been taught, sin^2@+cos^2@=1 for any alpha ?
then:
sin^2 [4pi/13] + cos^2 [4pi/13] =1

I suggest you sketch a circle of radius 1 about the origin of a x,y coordinate system. Then sketch a line from the origin to the circles circumference at a angle
of what you believe to be 55.5 degrees from the x axis.

This should satisfy your instructor
Arthur

 

[jenn9580]

Thank you. That makes sense now.

 

[arthur ohlsten]

this might help
On a xy coordinate system graph, sketch a line from the origin at a angle of @, [say about 30 degrees] for 1 unit long,[say a couple of inchs but called 1 unit]
make the line the hypoteneuse of a right triangle with the x axis the adjacent side , and a vertical line the opposite side.
Mark the adjacent side A , and the opposite side B and the hypotteneuse C
C^2=A^2+B^2
but A=cos@ and B= sin@ and C=1 OR
1=sin^2+cos^2

Arthur

 

[tkhunny]

sin^2 (4pi/13) + cos^2 (4pi/13)

I have no idea how to get sin & cos from this. I can't draw a refernce triangle with a 55degree angle. We have only done isosceles triangles and 30-60 triangles. How do I even get started in the right direction?

I seriously doubt this is a quiz on exact values of functions on 4pi/13. With very little doubt, I think it is a problem designed to get your attention and to see that you know a VERY IMPORTANT identity. Don't make things more complicated than they have to be.

\(\displaystyle \L\;sin^{2}(Gobble-de-gook)\;+\;cos^{2}(Gobble-de-gook)\;=\;1\)

...as long as Gobble-de-gook exists. Then move on to the next problem.

 

[jenn9580]

thank you. we went over it today, as to how to actually get to the =1. Thank you very much. i will have to brush up on my pythagorean stuff!