Question

Could you explain how you get

cosb = cos^2b -1

Answer 1

using ( ) help

Answer 2

cos(b) = cos^2(b) - 1

or

cos(b) = cos^2(b-1)

Answer 3

cos(b) = cos^2(b) - 1

Answer 4

we could test it out first to see if that is true:

cos(pi) = -1

cos^2(pi) - 1 = 0

i dont see how they are equal

unless you need to find the points in which they intersect

Answer 5

first of all it is not true. take for example
\[b=\frac{\pi}{4}\] then
what amistre said

Answer 6

for which "b" values are they equal

Answer 7

perhaps "solve for b"?

Answer 8

then its a quadratic

Answer 9

c^2 -c - 1 = 0

Answer 10

whrere c = cos(b)

Answer 11

i think he wants 2 prove:cos(2b)=2cos^2(b)-1

Answer 12

yeah have fun solving that one. get
\[c=\frac{1\pm\sqrt{5}}{2}\] which is no big deal.
oh. never mind

Answer 13

i think he wants a strawberry milkshake, but that doesnt help the problem :)

Answer 14

well I got an answer before that cos(b) = 2cos^2 (b/2) -1 but I couldnt prove it myself ,
I got an problem how to get from :

sin(l) (1 + cos(b) ) to

4sin(l/2)cos(l/2)cos^2(b/2)

Answer 15

it not an identity ...

Answer 16

if it were an identity, the all values for b would equate, and they dont

Answer 17

I know , I just saw it in a book where he got from first sin(l) (1 + cos(b) ) to 4sin(l/2)cos(l/2)cos^2(b/2) but couldnt find how

Answer 18

@unniverzal
you cannot prove it because it is not true for all values of b. like saying
"prove
\[2x+1=x-4\] "

which is not true for all x. you can "solve for x" but you cannot prove that they are always equal.

Answer 19

i got no idea where:

sin(l) (1 + cos(b) ) to 4sin(l/2)cos(l/2)cos^2(b/2)

comes from .....

Answer 20

good luck with it tho :)

Answer 21

begin with

Answer 22

\[\sin ^{2}x +\cos ^{2}x =1\]

Answer 23

consider b=x