We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
176
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+105618 
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+105618 
0

I am having trouble seeing why ...  ?

 

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

.
 Feb 6, 2019
 #3
avatar+7725 
+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

22 Online Users

avatar