Prvo rješenje.
Izlučivanjem zajedničkog faktora dobivamo
cos 2 x + cos 2 ( π 6 + x ) − 2 cos π 6 ⋅ cos x ⋅ cos ( π 6 + x ) = cos 2 x + cos ( π 6 + x ) ( cos ( π 6 + x ) − 2 cos π 6 cos x ) . \begin{aligned} &\cos^2x+\cos^2\left(\frac{\pi}{6}+x\right)-2\cos\frac{\pi}{6}\cdot\cos x\cdot\cos\left(\frac{\pi}{6}+x\right)\\ &=\cos^2x+\cos\left(\frac{\pi}{6}+x\right)\left(\cos\left(\frac{\pi}{6}+x\right)-2\cos\frac{\pi}{6}\cos x\right). \end{aligned} cos 2 x + cos 2 ( 6 π + x ) − 2 cos 6 π ⋅ cos x ⋅ cos ( 6 π + x ) = cos 2 x + cos ( 6 π + x ) ( cos ( 6 π + x ) − 2 cos 6 π cos x ) . Primjenom formule kosinusa zbroja slijedi
cos 2 x + ( cos π 6 cos x − sin π 6 sin x ) ⋅ ( cos π 6 cos x − sin π 6 sin x − 2 cos π 6 cos x ) = cos 2 x − ( cos π 6 cos x − sin π 6 sin x ) ⋅ ( cos π 6 cos x + sin π 6 sin x ) . \begin{aligned} &\cos^2x+\left(\cos\frac{\pi}{6}\cos x-\sin\frac{\pi}{6}\sin x\right)\\ &\quad\cdot\left(\cos\frac{\pi}{6}\cos x-\sin\frac{\pi}{6}\sin x-2\cos\frac{\pi}{6}\cos x\right)\\ &=\cos^2x-\left(\cos\frac{\pi}{6}\cos x-\sin\frac{\pi}{6}\sin x\right)\\ &\quad\cdot\left(\cos\frac{\pi}{6}\cos x+\sin\frac{\pi}{6}\sin x\right). \end{aligned} cos 2 x + ( cos 6 π cos x − sin 6 π sin x ) ⋅ ( cos 6 π cos x − sin 6 π sin x − 2 cos 6 π cos x ) = cos 2 x − ( cos 6 π cos x − sin 6 π sin x ) ⋅ ( cos 6 π cos x + sin 6 π sin x ) . Izraz sredimo
cos 2 x − ( cos 2 π 6 cos 2 x − sin 2 π 6 sin 2 x ) . \cos^2x-\left(\cos^2\frac{\pi}{6}\cos^2x-\sin^2\frac{\pi}{6}\sin^2x\right). cos 2 x − ( cos 2 6 π cos 2 x − sin 2 6 π sin 2 x ) . Konačno:
cos 2 x − 3 4 cos 2 x + 1 4 sin 2 x = 1 4 cos 2 x + 1 4 sin 2 x = 1 4 . \cos^2x-\frac34\cos^2x+\frac14\sin^2x=\frac14\cos^2x+\frac14\sin^2x=\frac14. cos 2 x − 4 3 cos 2 x + 4 1 sin 2 x = 4 1 cos 2 x + 4 1 sin 2 x = 4 1 . Drugo rješenje.
Izlučivanjem zajedničkog faktora dobivamo
cos 2 x + cos 2 ( π 6 + x ) − 2 cos π 6 ⋅ cos x ⋅ cos ( π 6 + x ) = cos 2 x + cos ( π 6 + x ) ( cos ( π 6 + x ) − 2 cos π 6 cos x ) . \begin{aligned} &\cos^2x+\cos^2\left(\frac{\pi}{6}+x\right)-2\cos\frac{\pi}{6}\cdot\cos x\cdot\cos\left(\frac{\pi}{6}+x\right)\\ &=\cos^2x+\cos\left(\frac{\pi}{6}+x\right)\left(\cos\left(\frac{\pi}{6}+x\right)-2\cos\frac{\pi}{6}\cos x\right). \end{aligned} cos 2 x + cos 2 ( 6 π + x ) − 2 cos 6 π ⋅ cos x ⋅ cos ( 6 π + x ) = cos 2 x + cos ( 6 π + x ) ( cos ( 6 π + x ) − 2 cos 6 π cos x ) . Primjenom formule umnoška kosinusa slijedi
cos 2 x + cos ( π 6 + x ) ( cos ( π 6 + x ) − 2 ⋅ 1 2 ( cos ( π 6 + x ) + cos ( π 6 − x ) ) ) = cos 2 x − cos ( π 6 + x ) cos ( π 6 − x ) . \begin{aligned} &\cos^2x+\cos\left(\frac{\pi}{6}+x\right)\left(\cos\left(\frac{\pi}{6}+x\right)-2\cdot\frac12\left(\cos\left(\frac{\pi}{6}+x\right)+\cos\left(\frac{\pi}{6}-x\right)\right)\right)\\ &=\cos^2x-\cos\left(\frac{\pi}{6}+x\right)\cos\left(\frac{\pi}{6}-x\right). \end{aligned} cos 2 x + cos ( 6 π + x ) ( cos ( 6 π + x ) − 2 ⋅ 2 1 ( cos ( 6 π + x ) + cos ( 6 π − x ) ) ) = cos 2 x − cos ( 6 π + x ) cos ( 6 π − x ) . Primjenom formule umnoška kosinusa dobivamo
cos 2 x − 1 2 cos π 3 − 1 2 cos 2 x . \cos^2x-\frac12\cos\frac{\pi}{3}-\frac12\cos2x. cos 2 x − 2 1 cos 3 π − 2 1 cos 2 x . Sada primijenimo kosinus dvostrukog broja i sredimo izraz
cos 2 x − 1 2 ⋅ 1 2 − cos 2 x + 1 2 = 1 4 . \cos^2x-\frac12\cdot\frac12-\cos^2x+\frac12=\frac14. cos 2 x − 2 1 ⋅ 2 1 − cos 2 x + 2 1 = 4 1 . Treće rješenje.
Ako primijenimo formule za kosinus zbroja i činjenicu da je cos π 6 = 3 2 \cos\frac{\pi}{6}=\frac{\sqrt3}{2} cos 6 π = 2 3 sin π 6 = 1 2 \sin\frac{\pi}{6}=\frac12 sin 6 π = 2 1
cos 2 x + cos 2 ( π 6 + x ) − 2 cos π 6 ⋅ cos x ⋅ cos ( π 6 + x ) = cos 2 x + ( cos π 6 cos x − sin π 6 sin x ) 2 − 2 cos π 6 ⋅ cos x ( cos π 6 cos x − sin π 6 sin x ) . \begin{aligned} &\cos^2x+\cos^2\left(\frac{\pi}{6}+x\right)-2\cos\frac{\pi}{6}\cdot\cos x\cdot\cos\left(\frac{\pi}{6}+x\right)\\ &=\cos^2x+\left(\cos\frac{\pi}{6}\cos x-\sin\frac{\pi}{6}\sin x\right)^2\\ &\quad-2\cos\frac{\pi}{6}\cdot\cos x\left(\cos\frac{\pi}{6}\cos x-\sin\frac{\pi}{6}\sin x\right). \end{aligned} cos 2 x + cos 2 ( 6 π + x ) − 2 cos 6 π ⋅ cos x ⋅ cos ( 6 π + x ) = cos 2 x + ( cos 6 π cos x − sin 6 π sin x ) 2 − 2 cos 6 π ⋅ cos x ( cos 6 π cos x − sin 6 π sin x ) . cos 2 x + ( 3 2 cos x − 1 2 sin x ) 2 − 2 ⋅ 3 2 ⋅ cos x ⋅ ( 3 2 cos x − 1 2 sin x ) . \begin{aligned} &\cos^2x+\left(\frac{\sqrt3}{2}\cos x-\frac12\sin x\right)^2\\ &\quad-2\cdot\frac{\sqrt3}{2}\cdot\cos x\cdot\left(\frac{\sqrt3}{2}\cos x-\frac12\sin x\right). \end{aligned} cos 2 x + ( 2 3 cos x − 2 1 sin x ) 2 − 2 ⋅ 2 3 ⋅ cos x ⋅ ( 2 3 cos x − 2 1 sin x ) . cos 2 x + 3 4 cos 2 x − 3 2 cos x sin x + 1 4 sin 2 x − 3 2 cos 2 x + 3 2 cos x sin x . \begin{aligned} &\cos^2x+\frac34\cos^2x-\frac{\sqrt3}{2}\cos x\sin x+\frac14\sin^2x\\ &\quad-\frac32\cos^2x+\frac{\sqrt3}{2}\cos x\sin x. \end{aligned} cos 2 x + 4 3 cos 2 x − 2 3 cos x sin x + 4 1 sin 2 x − 2 3 cos 2 x + 2 3 cos x sin x . 1 4 cos 2 x + 1 4 sin 2 x = 1 4 ( cos 2 x + sin 2 x ) = 1 4 . \frac14\cos^2x+\frac14\sin^2x=\frac14\left(\cos^2x+\sin^2x\right)=\frac14. 4 1 cos 2 x + 4 1 sin 2 x = 4 1 ( cos 2 x + sin 2 x ) = 4 1 . 