+0  
 
0
53
3
avatar+22 

 

 

How do i show as this but instead for cos(a+b)?

 

 

 

 

How do i show this for cos(a+b)?

 Feb 4, 2019
 #1
avatar+99309 
0

Have you looked at any of these?

 

https://goo.gl/HqnAmb

 

Do you need it to specifically relate to your diagram?

 Feb 6, 2019
 #2
avatar+99309 
0

I am having trouble seeing why ...  ?

 

\(\frac{DC}{AD}\cdot\frac{AD}{AB}=sina\cdot cos b\)

.
 Feb 6, 2019
 #3
avatar+7372 
+2

Make some changes to your diagram.

Now AF = 1, angle AFB = 90 degrees, and the angle at the bottom right corner is called G.

\(\quad\cos(a+b)\\ =\dfrac{\text{AC}}{\text{AB}}\\ =\dfrac{\text{AG} - \text{GC}}{\text{AB}}\\ =\dfrac{\text{AG}}{\frac{1}{\cos b}} - \dfrac{\text{GC}}{\frac{1}{\cos b}}\\ =\cos a \cos b - \tan b \sin a \cos b\\ = \cos a \cos b - \sin a \sin b\)

.
 Feb 7, 2019

16 Online Users

avatar
avatar
avatar
avatar
avatar