1. ## Addition Formulae for Trigonometry

I need someone to check if part a) is correct and need help on part b)
Thanks

2. Originally Posted by r_maths

I need someone to check if part a) is correct and need help on part b)
Thanks
Here's (a). your logic is correct, however, you can't call both bases adj, since they represend different lengths, call one adj1 and the other adj2, i used different names though

3. Originally Posted by r_maths

I need someone to check if part a) is correct and need help on part b)
Thanks
Here's (b)

Darn it, I forgot to say "this is my 12th post" last time!

4. Thanks for the speedy response Jhevon!
Congrats on 1200
I'm privileged.

5. Hello, r_maths!

You're doing fine!
Code:
              C
*
*| *
* |   *
*  |     *
*   |h      *
*    |         *
*     |           *
* α    |           β *
A * * * * * * * * * * * * * B
D

In right triangle CDA: .tanα = h/AD . . AD = h/tanα

In right triangle CDB: .tanβ = h/DB . . DB = h/tanβ

. . . . . . . . . . . . . . . . . . . . h . . . . h
Then: .AB .= .AD + DB .= .------ + ------
. . . . . . . . . . . . . . . . . . . tanα . . tanβ

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

For part (b), we have:

. . . . . . . . h . . . . .h . . . . h搾osβ . . h搾osα
. . AB .= .------ + ------ .= .-------- + --------
. . . . . . . tanβ . . tanα . . . .sinβ . . . . sinα

. . . . . . .h新inα搾osβ + h新inβ搾osα . . . .h(sinα搾osβ + sinB搾osα)
. . . . = .-------------------------------- .= .------------------------------
. . . . . . . . . . . . sinα新inβ . . . . . . . . . . . . . . sinα新inβ

. . . . . . .h新in(α + β)
. . . . = .---------------
. . . . . . . .sinα新inβ

Edit:Too slow . . again

6. Originally Posted by Soroban
Hello, r_maths!

You're doing fine!
Code:
              C
*
*| *
* |   *
*  |     *
*   |h      *
*    |         *
*     |           *
* α    |           β *
A * * * * * * * * * * * * * B
D

In right triangle CDA: .tanα = h/AD . . AD = h/tanα

In right triangle CDB: .tanβ = h/DB . . DB = h/tanβ

. . . . . . . . . . . . . . . . . . . . h . . . . h
Then: .AB .= .AD + DB .= .------ + ------
. . . . . . . . . . . . . . . . . . . tanα . . tanβ

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

For part (b), we have:

. . . . . . . . h . . . . .h . . . . h搾osβ . . h搾osα
. . AB .= .------ + ------ .= .-------- + --------
. . . . . . . tanβ . . tanα . . . .sinβ . . . . sinα

. . . . . . .h新inα搾osβ + h新inβ搾osα . . . .h(sinα搾osβ + sinB搾osα)
. . . . = .-------------------------------- .= .------------------------------
. . . . . . . . . . . . sinα新inβ . . . . . . . . . . . . . . sinα新inβ

. . . . . . .h新in(α + β)
. . . . = .---------------
. . . . . . . .sinα新inβ

i like your part (b), it's much more elegant and shorter than mine. i saw that i could do mine shorter afterwards.

our part (a)'s are identical, except i called C what you called D, and D what you called C