# Trigo Help!!!

• September 4th 2006, 04:48 AM
Trigo Help!!!
1.) Prove 6sin4xcos4x = 3sin8x

2.) Derive sin( u + v) = sinucosv + cosusinv from the identity
cos(u+v) =cosucosv - sinusinv
• September 4th 2006, 05:07 AM
CaptainBlack
Quote:

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
• September 4th 2006, 06:52 AM
Quote:

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)
• September 4th 2006, 07:16 AM
ThePerfectHacker
Quote:

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,
$\sin u\cos v+\cos u\sin v$
• September 4th 2006, 09:25 AM
Soroban

Quote:

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

• September 4th 2006, 03:22 PM
Thanks!!! :D
• September 6th 2006, 03:07 AM
chancey
Quote:

$cos(u+v)=cos(u)cos(v)-sin(u)sin(v)$
$\sqrt{1-sin^{2}(u+v)}=cos(u)\sqrt{1-sin^{2}(v)}+sinu\sqrt{1-cos^{2}(v)}$
Solve for sin(u+v)

I see how you derived the irrationals, but can you please explain the next few steps?

Thanks
• September 6th 2006, 08:31 AM
CaptainBlack
Quote:

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
• September 6th 2006, 08:35 AM
ThePerfectHacker
Quote:

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 $x$!
• September 6th 2006, 08:44 AM
ThePerfectHacker
Quote:

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.
• September 6th 2006, 11:01 AM
Soroban

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

Quote:

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

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 . . .

• September 9th 2006, 05:13 AM
just another question... what if the case would be deriving sin(u - v) to cos(u-v)?

Elow thanks people!!!
• September 9th 2006, 05:29 AM
CaptainBlack
Quote:

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