1.) Prove 6sin4xcos4x = 3sin8x

2.) Derive sin( u + v) = sinucosv + cosusinv from the identity

cos(u+v) =cosucosv - sinusinv

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- September 4th 2006, 03:48 AM^_^Engineer_Adam^_^Trigo Help!!!
1.) Prove 6sin4xcos4x = 3sin8x

2.) Derive sin( u + v) = sinucosv + cosusinv from the identity

cos(u+v) =cosucosv - sinusinv - September 4th 2006, 04:07 AMCaptainBlackQuote:

Originally Posted by**^_^Engineer_Adam^_^**

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, 05:52 AMshubhadeepQuote:

Originally Posted by**^_^Engineer_Adam^_^**

Solve for sin(u+v) - September 4th 2006, 06:16 AMThePerfectHackerQuote:

Originally Posted by**^_^Engineer_Adam^_^**

Then, by co-funtions

Thus,

Thus,

Thus,

- September 4th 2006, 08:25 AMSoroban
Hello, ^_^Engineer_Adam^_^!

Quote:

1) Prove: .

You're expected to know the identity: .

We have: .

Therefore: .

- September 4th 2006, 02:22 PM^_^Engineer_Adam^_^
Thanks!!! :D

- September 6th 2006, 02:07 AMchanceyQuote:

Originally Posted by**shubhadeep**

Thanks - September 6th 2006, 07:31 AMCaptainBlackQuote:

Originally Posted by**chancey**

**Im**Perfect

Hacker, less algebra, less to go wrong.

RonL - September 6th 2006, 07:35 AMThePerfectHackerQuote:

Originally Posted by**CaptainBlank**

Because,

Is not true for all ! - September 6th 2006, 07:44 AMThePerfectHackerQuote:

Originally Posted by**ThePerfectHacker**

when it should be and,

when it should be.

But that is not ideal.

---

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. - September 6th 2006, 10:01 AMSoroban
Hello, ^_^Engineer_Adam^_^!

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

Quote:

2) Derive

from the identity:

Square the given identity:

. .

. .

Subtract from 1:

. .

Add and subtract and

. .**+ cosēv****+ sinēv****- cosēv****- sinēv**

Factor:

. .

Note that: . . . and we have:

. .

Factor: .

Therefore: .

I need a nap . . .

- September 9th 2006, 04:13 AM^_^Engineer_Adam^_^
just another question... what if the case would be deriving sin(u - v) to cos(u-v)?

Elow thanks people!!! - September 9th 2006, 04:29 AMCaptainBlackQuote:

Originally Posted by**^_^Engineer_Adam^_^**

now use the symmetry properties of and to get what you want:

but , and , so:

RonL