1.) Prove 6sin4xcos4x = 3sin8x
2.) Derive sin( u + v) = sinucosv + cosusinv from the identity
cos(u+v) =cosucosv - sinusinv
You have,Originally Posted by ^_^Engineer_Adam^_^
$\displaystyle \cos (u+v)=\cos u\cos v-\sin u\sin v$
Then, by co-funtions
$\displaystyle \sin (u+v)=\cos \left( \frac{\pi}{2}-(u+v) \right)$
Thus,
$\displaystyle \cos \left[ \left( \frac{\pi}{2}-u\right)-v\right]$
Thus,
$\displaystyle \cos \left( \frac{\pi}{2}-u\right)\cos u+\sin\left( \frac{\pi}{2}-u \right) \sin v$
Thus,
$\displaystyle \sin u\cos v+\cos u\sin v$
Hello, ^_^Engineer_Adam^_^!
1) Prove: .$\displaystyle 6\sin4x\cos4x \,= \,3\sin8x$
You're expected to know the identity: .$\displaystyle 2\sin\theta\cos\theta \,=\,\sin2\theta$
We have: .$\displaystyle 6\sin4x\cos4x \;=\:3(2\sin4x\cos4x) \:=\:3(\sin 8x)$
Therefore: .$\displaystyle 6\sin4x\cos4x \:=\:3\sin8x$
You can make it correct by definiton $\displaystyle \mbox{H4X}(x)$. It has the propery that,Originally Posted by ThePerfectHacker
$\displaystyle \mbox{H4X}(x)=\sqrt{1-\sin^2 x}$ when it should be and,
$\displaystyle \mbox{H4X}(x)=\sqrt{1-\sin^2 x}$ when it should be.
But that is not ideal.
---
Note: If you want to use the radical method you cannot write,
$\displaystyle \pm \sqrt{1-\sin^2 x}$. 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 $\displaystyle \sin( u + v) \:= \:\sin u\cos v + \cos u\sin v$
from the identity: $\displaystyle \cos(u+v) \:=\:\cos u\cos v - \sin u\sin v$
Square the given identity:
. . $\displaystyle \cos^2(u + v)\:=\\cos u\cos v - \sin u\sinv)^2$
. . $\displaystyle \cos^2(u + v)\:= \:\cos^2\!u\cos^2\!v - 2\sin u\sin v\cos u\cos v$$\displaystyle + \sin^2\!u\sin^2\!v$
Subtract from 1:
. . $\displaystyle 1 - \cos^2(u + v) \:=\:1 - \cos^2\!u\cos^2\!v +$$\displaystyle 2\sin y\sin v\cos u\cos v - \sin^2\!u\sin^2\!v $
Add and subtract $\displaystyle \sin^2\!v$ and $\displaystyle \cos^2\!v$
. . $\displaystyle = \:1 $ + cosēv $\displaystyle - \cos^2\!u\cos^2\!v + 2\sin u\sin v\cos u\cos v$ + sinēv $\displaystyle - \sin^2\!u\sin^2\!v$ - cosēv - sinēv
Factor:
. . $\displaystyle = \:1 + \cos^2\!v(1 - \cos^2\!u) + 2\sin u\sin v\cos u\cos v +$ $\displaystyle \sin^2\!v(1 - \sin^2\!u) - (\sin^2\!v + \cos^2\!v)$
Note that: $\displaystyle 1 - (\sin^2\!v + \cos^2\!v) \:=\:0$ . . . and we have:
. . $\displaystyle \sin^2(u+v)\:=\:\sin^2\!u\cos^2\!v + 2\sin u\sin v\cos u\cos v$$\displaystyle + \sin^2\!v\s\cos^2\!u$
Factor: .$\displaystyle \sin^2(u+v) \:=\:\left(\sin u\cos v + \sin v\cos u)^2$
Therefore: .$\displaystyle \boxed{\sin(u + v)\:=\:\sin u\cos v + \sin v\cos u}$
I need a nap . . .
Just stick with what you have, you will have derivedOriginally Posted by ^_^Engineer_Adam^_^
$\displaystyle
\sin( u + v)=\sin (u)\cos (v) + \cos (u)\sin (v)
$
now use the symmetry properties of $\displaystyle \sin$ and $\displaystyle \cos$ to get what you want:
$\displaystyle
\sin(u-v)=\sin( u + (-v))=\sin (u)\cos (-v) + \cos (u)\sin (-v)
$
but $\displaystyle \sin(-x)=-\sin(x)$, and $\displaystyle \cos(-x)=\cos(x)$, so:
$\displaystyle
\sin(u-v)=\sin (u)\cos (v) - \cos (u)\sin (v)
$
RonL