# Math Help - Trigo Help!!!

1. ## Trigo Help!!!

1.) Prove 6sin4xcos4x = 3sin8x

2.) Derive sin( u + v) = sinucosv + cosusinv from the identity
cos(u+v) =cosucosv - sinusinv

1.) Prove 6sin4xcos4x = 3sin8x
Use the identity:

sin(u + v) = sin(u) cos(v) + cos(u) sin(v)

with v=u=4x, then multiply the whole thing by 3.

RonL

2.) Derive sin( u + v) = sinucosv + cosusinv from the identity
cos(u+v) =cosucosv - sinusinv
$cos(u+v)=cos(u)cos(v)-sin(u)sin(v)$
$\sqrt{1-sin^{2}(u+v)}=cos(u)\sqrt{1-sin^{2}(v)}+sinu\sqrt{1-cos^{2}(v)}$
Solve for sin(u+v)

2.) Derive sin( u + v) = sinucosv + cosusinv from the identity
cos(u+v) =cosucosv - sinusinv
You have,
$\cos (u+v)=\cos u\cos v-\sin u\sin v$
Then, by co-funtions
$\sin (u+v)=\cos \left( \frac{\pi}{2}-(u+v) \right)$
Thus,
$\cos \left[ \left( \frac{\pi}{2}-u\right)-v\right]$
Thus,
$\cos \left( \frac{\pi}{2}-u\right)\cos u+\sin\left( \frac{\pi}{2}-u \right) \sin v$
Thus,
$\sin u\cos v+\cos u\sin v$

1) Prove: . $6\sin4x\cos4x \,= \,3\sin8x$

You're expected to know the identity: . $2\sin\theta\cos\theta \,=\,\sin2\theta$

We have: . $6\sin4x\cos4x \;=\:3(2\sin4x\cos4x) \:=\:3(\sin 8x)$

Therefore: . $6\sin4x\cos4x \:=\:3\sin8x$

6. Thanks!!!

$cos(u+v)=cos(u)cos(v)-sin(u)sin(v)$
$\sqrt{1-sin^{2}(u+v)}=cos(u)\sqrt{1-sin^{2}(v)}+sinu\sqrt{1-cos^{2}(v)}$
Solve for sin(u+v)
I see how you derived the irrationals, but can you please explain the next few steps?

Thanks

8. Originally Posted by chancey
I see how you derived the irrationals, but can you please explain the next few steps?

Thanks
Its much easier to do it using the method presented by ImPerfect
Hacker, less algebra, less to go wrong.

RonL

9. Originally Posted by CaptainBlank
Its much easier to do it using the method presented by ImPerfect
Hacker, less algebra, less to go wrong.

RonL
Also, I do not think it is entirely correct.
Because,
$\sqrt{1-\sin^2 x}=\cos x$
Is not true for all $x$!

10. Originally Posted by ThePerfectHacker
Also, I do not think it is entirely correct.
You can make it correct by definiton $\mbox{H4X}(x)$. It has the propery that,
$\mbox{H4X}(x)=\sqrt{1-\sin^2 x}$ when it should be and,
$\mbox{H4X}(x)=\sqrt{1-\sin^2 x}$ when it should be.
But that is not ideal.
---
Note: If you want to use the radical method you cannot write,
$\pm \sqrt{1-\sin^2 x}$. Because the expression would not be well-definied, i.e. not a function. My way through the H4X function is ye only way.

I found a way, but it's not worth the trouble . . .

2) Derive $\sin( u + v) \:= \:\sin u\cos v + \cos u\sin v$
from the identity: $\cos(u+v) \:=\:\cos u\cos v - \sin u\sin v$

Square the given identity:
. . $\cos^2(u + v)\:=\\cos u\cos v - \sin u\sinv)^2" alt="\cos^2(u + v)\:=\\cos u\cos v - \sin u\sinv)^2" />
. . $\cos^2(u + v)\:= \:\cos^2\!u\cos^2\!v - 2\sin u\sin v\cos u\cos v$ $+ \sin^2\!u\sin^2\!v$

Subtract from 1:
. . $1 - \cos^2(u + v) \:=\:1 - \cos^2\!u\cos^2\!v +$ $2\sin y\sin v\cos u\cos v - \sin^2\!u\sin^2\!v$

Add and subtract $\sin^2\!v$ and $\cos^2\!v$

. . $= \:1$ + cos²v $- \cos^2\!u\cos^2\!v + 2\sin u\sin v\cos u\cos v$ + sin²v $- \sin^2\!u\sin^2\!v$ - cos²v - sin²v

Factor:
. . $= \:1 + \cos^2\!v(1 - \cos^2\!u) + 2\sin u\sin v\cos u\cos v +$ $\sin^2\!v(1 - \sin^2\!u) - (\sin^2\!v + \cos^2\!v)$

Note that: $1 - (\sin^2\!v + \cos^2\!v) \:=\:0$ . . . and we have:
. . $\sin^2(u+v)\:=\:\sin^2\!u\cos^2\!v + 2\sin u\sin v\cos u\cos v$ $+ \sin^2\!v\s\cos^2\!u$

Factor: . $\sin^2(u+v) \:=\:\left(\sin u\cos v + \sin v\cos u)^2$

Therefore: . $\boxed{\sin(u + v)\:=\:\sin u\cos v + \sin v\cos u}$

I need a nap . . .

12. just another question... what if the case would be deriving sin(u - v) to cos(u-v)?

Elow thanks people!!!

just another question... what if the case would be deriving sin(u - v) to cos(u-v)?

Elow thanks people!!!
Just stick with what you have, you will have derived

$
\sin( u + v)=\sin (u)\cos (v) + \cos (u)\sin (v)
$

now use the symmetry properties of $\sin$ and $\cos$ to get what you want:

$
\sin(u-v)=\sin( u + (-v))=\sin (u)\cos (-v) + \cos (u)\sin (-v)
$

but $\sin(-x)=-\sin(x)$, and $\cos(-x)=\cos(x)$, so:

$
\sin(u-v)=\sin (u)\cos (v) - \cos (u)\sin (v)
$

RonL