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

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

gus42298 Feb 4, 2019

#1**0 **

Have you looked at any of these?

Do you need it to specifically relate to your diagram?

Melody Feb 6, 2019

#2**0 **

I am having trouble seeing why ... ?

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

.Melody Feb 6, 2019

#3**+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.

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

.MaxWong Feb 7, 2019