+0

# ​ Trigonometric Identities - 2

0
80
9
+87

Details as follows:

OldTimer  Nov 17, 2018
#1
+94126
0

Hi Old Timer,

We are no ignoring your question (part 2 that is).  Chris and I just have not worked out how to do it yet.

Hopefully Heueka or Alan or Tiggsy or some other clever member or guest will show us all how it is done (hopefully in a nice easy to follow fashion.)

Edit:

I hope I haven't offended anyone by using names. Maybe Rom or ElectricPavlov or Omi or geno or  ...   We have lots of very knowledgable people here now days.

Melody  Nov 18, 2018
edited by Melody  Nov 18, 2018
#2
+87
+1

Hi Melody, Thanks for your update....you are very always helpful...I can vouch for that!

I have a partial solution which I will post later on tomorrow (loading is a bit tough)...but need to rework  as in desperation I worked back from the answer, but  still not perfect.

OldTimer  Nov 18, 2018
edited by OldTimer  Nov 18, 2018
#3
+87
+1

There must be some error in logic here:

OldTimer  Nov 18, 2018
#4
+94126
+1

So far I have not worked out how to derive those initial formulas that were given.

I have not looked at the rest of your logic yet, I'd really like to convince myself of the accuracy of the book extract first :/

Melody  Nov 18, 2018
#5
+20680
+9

​ Trigonometric Identities - 2

$$\text{Formula } 1: \boxed{\sin(A+B) = \dfrac{2\cdot \tan \left(\frac{A+B}{2} \right)} {1+\tan^2\left(\frac{A+B}{2} \right) } }\\ \text{Formula } 2: \boxed{\cos(A)-\cos(B) = -2\sin\left(\frac{A+B}{2} \right)\sin\left(\frac{A-B}{2} \right) } \\ \text{Formula } 3: \boxed{\sin(A)-\sin(B) = 2\cos\left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2} \right) }$$

$$\begin{array}{|rcll|} \hline \dfrac{p}{q} &=& \dfrac{\cos(A)-\cos(B) } {\sin(A)-\sin(B) } \\\\ &=& \dfrac{-2\sin\left(\frac{A+B}{2} \right)\sin\left(\frac{A-B}{2} \right) } { 2\cos\left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2} \right) } \\\\ &=& - \tan \left(\frac{A+B}{2} \right) \\ \mathbf{\tan \left(\frac{A+B}{2} \right)} &\mathbf{=}& \mathbf{- \dfrac{p}{q}} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \sin(A+B) &=& \dfrac{2\cdot \tan \left(\frac{A+B}{2} \right)} {1+\tan^2\left(\frac{A+B}{2} \right) } \\\\ &=& \dfrac{2\cdot \left( \dfrac{-p}{q} \right) } {1+\left( \dfrac{-p}{q} \right)^2 } \\\\ &=& -\dfrac{2\cdot p } {q\left(1+ \dfrac{p^2}{q^2} \right) } \\\\ &=& -\dfrac{2\cdot p } { q+ \dfrac{p^2}{q } } \\\\ \mathbf{\sin(A+B)} &\mathbf{=}& \mathbf{-\dfrac{2pq } { p^2+q^2 } } \\ \hline \end{array}$$

heureka  Nov 19, 2018
#6
+94126
+1

Thanks very much Heureka,

Those formulae  2 and 3 that you have used, do you have those memorised?

I mean can they be derived easily enough?

I do not remember seeing them before....

Melody  Nov 19, 2018
#7
+20680
+9

Hi Melody

For Example:

$$\begin{array}{|lrclrcl|} \hline (1) & \sin(x+y) &=& \sin(x)\cos(y) +\cos(x)\sin(y) \\ (2) & \sin(x-y) &=& \sin(x)\cos(y) -\cos(x)\sin(y) \\ \\ \hline \\ (1)-(2): & \sin(x+y)-\sin(x-y) &=& \sin(x)\cos(y) +\cos(x)\sin(y) \\ &&&-\Big(\sin(x)\cos(y) -\cos(x)\sin(y)\Big) \\ & \sin(x+y)-\sin(x-y) &=& \sin(x)\cos(y) +\cos(x)\sin(y) \\ &&& -\sin(x)\cos(y) +\cos(x)\sin(y) \\ & \sin(x+y)-\sin(x-y) &=& \cos(x)\sin(y) +\cos(x)\sin(y) \\ & \sin(x+y)-\sin(x-y) &=& 2\cos(x)\sin(y) \\ \\ \hline \\ (3): \mathbf{x+y = A} \\ (4): \mathbf{x-y = B} \\ \\ (3)+(4): & 2x = A+B \\ & \mathbf{x = \dfrac{A+B}{2}} \\\\ (3)-(4): & 2y = A-B \\ & \mathbf{ y = \dfrac{A-B}{2}} \\\\ \hline & \sin(x+y)-\sin(x-y) &=& 2\cos(x)\sin(y) \\ & \sin(A)-\sin(B) &=& 2\cos\left(\dfrac{A+B}{2}\right)\sin\left(\dfrac{A-B}{2} \right) \\ \hline \end{array}$$

heureka  Nov 19, 2018
#8
+94126
+1

Thanks very much Heureka :))

Melody  Nov 19, 2018
#9
+87
+1

Smooth as silk...Thanks a lot!

OldTimer  Nov 20, 2018