exact value using double angles - please help (1 Viewer)

[boogaaaaa]

find the exact value of sin 67.5 degrees using sin 2 x. I'm having trouble with this one, because after a certain point i have no idea what to do..

my attempt:

sin 2(67.5)=2sin(67.5)cos(67.5)
sin 135 = 2sin(67.5)cos(67.5)
1/√2 = 2sin(67.5)cos(67.5)

and i have no idea what to do afterwards. any help will be greatly appreciated.

 

[omgmynameislong]

sin(22.5+45)= sin(22.5)cos(45) + cos(22.5)sin(45)

cos45= 1-2sin^2(22.5) (cos2x=1-2sin^2(x))

work out what exact value of sin 22.5 is

cos(22.5+22.5)=2cos^2(22.5) - 1 (cos2x=2cos^2(x)-1)

work out exact value for cos 22.5

then just put the values into sin(22.5+45)= sin(22.5)cos(45) + cos(22.5)sin(45)

 

[cutemouse]

or more briefly,

 

[GUSSSSSSSSSSSSS]

find the exact value of sin 67.5 degrees using sin 2 x. I'm having trouble with this one, because after a certain point i have no idea what to do..

my attempt:

sin 2(67.5)=2sin(67.5)cos(67.5)
sin 135 = 2sin(67.5)cos(67.5)
1/√2 = 2sin(67.5)cos(67.5)

and i have no idea what to do afterwards. any help will be greatly appreciated.

or more briefly,

ummmmm you shud maybe read the question ....
oh well, these things happen =S lol

 

[addikaye03]

find the exact value of sin 67.5 degrees using sin 2 x. I'm having trouble with this one, because after a certain point i have no idea what to do..

my attempt:

sin 2(67.5)=2sin(67.5)cos(67.5)
sin 135 = 2sin(67.5)cos(67.5)
1/√2 = 2sin(67.5)cos(67.5)

and i have no idea what to do afterwards. any help will be greatly appreciated.

sin2(67.5)=2sin(67.5)cos(67.5)

1/2rt2=sin(67.5)sin(22.5)

[sin(45+22.5)]sin(22.5)=1/2rt2

[sin(45)cos(22.5)+cos(45)sin(22.5)]sin(22.5)=1/2rt2

sin45[sin(22.5)cos(22.5)]+cos45[sin^2(22.5)]=1/2rt2

sin45[1/2(sin45)]+cos45[sin^2(22.5)]=1/2rt2

[1/2(sin45)]+[sin^2(22.5)]=1/2

2sin^2(22.5)=1-(1/rt2)

sin^2(22.5)=1/2-1/2rt2

Since sin^2@>0, sin@>0

sin(22.5)=rt ( 1/2-1/2rt2)

Which i suppose could be made to look neater, i believe that's the answer though, since it used sin2x=2sinxcosx