Question

Find cos 36 with Trigonometry

Answer 1

or you can find cos 36 with a calculator :D

Answer 2

you have to do it from scratch or can we use the value of \(\sin 18^\circ \)?

Answer 3

do not use known values

Answer 4

http://mathforum.org/library/drmath/view/52139.html

Answer 5

let x = 36
5x = 180
3x = 180 - 2x
sin 3x = sin2x
3 sinx - 4sin^3 x = 2sinx cox
sinx =/= 0

3- 4 sin^2 x = 2cosx
3- 4 ( 1-cos^2 x) = 2 cos x
4 cos^2 x - 2cosx -1 =0
\[cosx = \frac{-(-2) \pm \sqrt{(-2)^2-4(1)(-1)}}{2(4)}\]
\[cos36 = \frac{2 \pm \sqrt{20}}{8} = \frac{1 \pm \sqrt{5}}{4} \]

Answer 6

assume that 36 is in degrees...

Answer 7

what is "sinx =/= 0" ?

Answer 8

How did 3x=180-2x suddenly turn into sin 3x = sin 2x? I'm sorry if I'm a bit slow :D

Answer 9

Taking sine of the both sides.

Answer 10

@im2bad
sinx≠0
since sinx = sin 36 ≠0

Answer 11

sin(pi-a)=sin a

Answer 12

sin (180 -2x) = sin 2x

Answer 13

nic work callisto

Answer 14

Right. Thank you.

Answer 15

sin 3x
= sin (2x+x)
= sin2x cosx + cos2x sinx
= (2sinx cosx) cosx + (1-2sin^2x)sinx
= 2sinx cos^2x + (1-2sin^2x)sinx
= 2sinx (1-sin^2 x) + (1-2sin^2x)sinx
= 2sinx - 2sin^3 x + sinx - 2sin^3 x
= 3sinx - 4sin^3 x

Answer 16

Ahh.. wait :|
One more step to go
since cos 36>0
\[cos36 =\frac{1+\sqrt{5}}{4}\]

Answer 17

how about this
\[\cos 36=x \ \ then \ \ \cos 72=2x^2-1 \\ \cos 36-\cos 72=2 \ \sin 54 \ \sin 18=2 \cos 36 \ \cos 72 \\ x-2x^2+1=2x(2x^2-1) \\ 4x^3+2x^2-3x-1=0 \\ (4x^2-2x-1)(x+1)=0 \\ \cos 36 \neq -1 \\ 4x^2-2x-1=0 \\ x=\frac{1+\sqrt{5}}{2}\]

Answer 18

err
\[x=\frac{1+\sqrt{5}}{4}\]

Answer 19

I learnt another way to solve it! Amazing!!!! ><

Answer 20

i forgot to medal u :)