Question

prove \[4\sin^236-1=4\cos^272\]

Answer 1

guys i got to do this question

Answer 2

can you help me

Answer 3

18*2=36
18*5=90
36*2=72

Answer 4

\[4\sin^236-1=-2(1-2\sin^236)+3\]

Answer 5

\[=-2(\cos 72)+3\]

Answer 6

guys original question is actually this
\[\frac{1}{2}\sqrt{4\sin^236-1}=\cos^272\]

Answer 7

i mean the original question is
\[\frac{1}{2}\sqrt{4\sin^236-1}=\cos 72\]

Answer 8

so maybe we shud prove
\[4\sin^236-4\cos^272=1\]

Answer 9

ie
\[4(\sin^236-\cos^272)\]

Answer 10

i think i need a good identity

Answer 11

The double angle identity for cosine, maybe?

Answer 12

\[\cos 2u = 1-2\sin^2 u\]
\[(\cos 2u)^2 = (1-2\sin^2 u)^2\]

Answer 13

try solving quadratic equation for sin(36) ... you might be able to reduce it into well known identity

Answer 14

and if you get the value of sin(36) then that's fine ...
http://www.wolframalpha.com/input/?i=sin%2836%29

Answer 15

okay so
\[\cos^2 72=(1-\sin^236)^2=1-2\sin^236+\sin^436\]not good

Answer 16

i'll be back ...thank you...

Answer 17

what's the problem, it's quadratic in sin(36)^2 i suppose you get it's value = (5 - sqrt(5))/8
the value of sin(36) is well known

Answer 18

\[1-2(\frac{5-\sqrt{5}}{5})+(\frac{5-\sqrt{5}}{5})^2\\=1-2(\frac{5-\sqrt{5}}{5})+1+\frac{2\sqrt{}5}{5}+\frac{1}{5}\]

Answer 19

=1/5