## AravindG 3 years ago how to find value of sin 18

1. Argonx16

Take a calculator, and hit the "sin" button and type 18 in. You (may) need a scientific/graphing/google calculator. If you are asking how to do so by hand, I am not qualified to tell you.

2. AravindG

there must be a method using the formulas

3. AravindG

turing??

4. TuringTest

I am thinking....

5. SUROJ

(18*5)=90

6. AravindG

so?

7. TuringTest

all I got is the fact that we can write this as$\sin(\frac\pi{10})$thinking if we can go somewhere from there...

8. TuringTest

looking at some way to write this such that we can apply an identity to it...

9. SUROJ
10. AravindG

whoa!!

11. TuringTest

why is $\sin(3x)=\cos(2x)$?

12. SUROJ

Most of peoples are not smart....they are clever

13. campbell_st

use the power series expansion $\sin(x) = x/1! - x^3/3! + x^5/5! -x^7/7!.....$

14. AravindG

hw put valu 18>??

15. AravindG

bt it goes on

16. SUROJ

if x= 18, 2x+3x=90

17. AravindG

Another way to do this is to start with a regular pentagon where the length of each side is one unit. I'll outline the steps. (You can fill in the details.) Call the pentagon ABCDE. Draw diagonals AC and BE, which intersect at point F. Using the fact that each angle of the pentagon is 108 degrees, you can prove the triangles BFC and AFE are isosceles. Therefore the lengths of line segments CF and EF are also 1 unit. Also note the angles for the triangles are 72-72-36. Now show that ABF and CAB are similar triangles. Since they are similar, this means BE/AB = AB/BF. Since BE = EF + BF, substitute to get (EF + BF)/AB = AB/BF. Let's define x to be the length of BF. Then, since EF = 1, the equation becomes (1 + x)/1 = 1/x, or x^2 + x - 1 = 0. Use the quadratic equation to get x = (sqrt(5) - 1) / 2. Finally, let's create an 18-72-90 by drawing the altitude of triangle BFC from vertex C to side BF. Since BC = FC, the altitude hits BF at the midpoint of BF, which we will call point G. So, (a) triangle GBE is an 18-72-90 triangle, (b) BG = BF/2 = x/2. Using this 18-72-90 triangle, sin 18 degrees = BG / BC = x/2. Therefore sin 18 = (sqrt(5) - 1) / 4.

18. AravindG

i liked this

19. SUROJ

2x=90-3x

20. SUROJ

cos both side

21. AravindG

k

22. SUROJ

cos(2x)=sin(3x)

23. AravindG

thx evryone

24. SUROJ

catch it?

25. TuringTest

gotchya, thx

26. AravindG

campbell pls tell ur method