1.) Prove 6sin4xcos4x = 3sin8x
2.) Derive sin( u + v) = sinucosv + cosusinv from the identity
cos(u+v) =cosucosv - sinusinv
You can make it correct by definiton . It has the propery that,Originally Posted by ThePerfectHacker
when it should be and,
when it should be.
But that is not ideal.
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Note: If you want to use the radical method you cannot write,
. Because the expression would not be well-definied, i.e. not a function. My way through the H4X function is ye only way.
Hello, ^_^Engineer_Adam^_^!
I found a way, but it's not worth the trouble . . .
2) Derive
from the identity:
Square the given identity:
. . \cos u\cos v - \sin u\sinv)^2" alt="\cos^2(u + v)\:=\\cos u\cos v - \sin u\sinv)^2" />
. .
Subtract from 1:
. .
Add and subtract and
. . + cosēv + sinēv - cosēv - sinēv
Factor:
. .
Note that: . . . and we have:
. .
Factor: .
Therefore: .
I need a nap . . .