The double angle identities are
[tex]\sin(2x)=2\sin x\cos x[/tex]
[tex]\cos(2x)=\cos^2x-\sin^2x[/tex]
Then
[tex]2\sin\dfrac\pi6\cos\dfrac\pi6=\sin\left(\dfrac{2\pi}6\right)=\sin\dfrac\pi3[/tex]
[tex]\cos^2\dfrac\pi{10}-\sin^2\dfrac\pi{10}=\cos\left(\dfrac{2\pi}{10}\right)=\cos\dfrac\pi5[/tex]
The second identity together with the Pythagorean identity, [tex]\sin^2x+\cos^2x=1[/tex], gives us another equivalent expression:
[tex]\cos^2x-\sin^2x=\cos^2x-(1-\cos^2x)=2\cos^2x-1[/tex]
so
[tex]2\cos^2(0.5)-1=\cos(2\cdot0.5)=\cos1[/tex]
