+22
How do i show as this but instead for cos(a+b)?
How do i show this for cos(a+b)?
+118677 Have you looked at any of these?
Do you need it to specifically relate to your diagram?
+118677 I am having trouble seeing why ... ?
\(\frac{DC}{AD}\cdot\frac{AD}{AB}=sina\cdot cos b\)
+9673 Make some changes to your diagram.
Now AF = 1, angle AFB = 90 degrees, and the angle at the bottom right corner is called G.
\(\quad\cos(a+b)\\ =\dfrac{\text{AC}}{\text{AB}}\\ =\dfrac{\text{AG} - \text{GC}}{\text{AB}}\\ =\dfrac{\text{AG}}{\frac{1}{\cos b}} - \dfrac{\text{GC}}{\frac{1}{\cos b}}\\ =\cos a \cos b - \tan b \sin a \cos b\\ = \cos a \cos b - \sin a \sin b\)
