Question

Explain this trig identity for me?

Answer 1

I don't understand what they did for d...

Answer 2

For d?
They did some Half-Angle business.
Do you understand this?\[\large\rm \cos\left(\frac{\pi}{8}\right)=\cos\left(\frac{\pi/4}{2}\right)=\sqrt{\frac{1+\cos(\pi/4)}{2}}\]

Answer 3

\[\cos \left( x+y \right)=\cos x \cos y-\sin x \sin y\]
put y=x
\[\cos 2x=\cos ^2x-\sin ^2x=\cos ^2x-(1-\cos ^2x)=2\cos ^2x-1\]
\[2\cos ^2x=1+\cos 2x\]
\[\cos x=\pm \sqrt{\frac{ 1+\cos 2x }{ 2 }}\]
\[put~x=\frac{ \pi }{ 8 }\]
\[\cos \frac{ \pi }{ 8 }=\sqrt{\frac{ 1+\cos \frac{ \pi }{ 4 } }{ 2 }}\]
?

Answer 4

Zep, totally makes sense! Thanks!