QUIZ MATEMATIKA Buktikan sin⁸75° - cos⁸75° = [tex] \frac{7}{16} \sqrt{3} [/tex]

Topik: Trigonometri

Lakukan pemfaktoran atas bentuk tersebut:
[tex]$\begin{align}\sin^875-\cos^875&=(\sin^475+\sin^475)(\sin^475-\sin^475) \\ &=(\sin^475+\sin^475)(\sin^275+\sin^275) \\ &(\sin^275-\cos^275) \\ &=(\sin^475+\cos^475)\times 1\times(\sin^275-\cos^275) \\ &=(\sin^475+\cos^475)(\sin^275-\cos^275)\end{align}[/tex]

Lakukan sedikit manipulasi identitas pada faktor pertama memberikan:
[tex]$\begin{align}\sin^875-\cos^875&=\left((\sin^275+\cos^275)^2-2\sin^275\cos^275\right) \\ &(\sin^275-\cos^275) \\ &=-(1-2(\sin75\cos75)^2)(\cos^275-\sin^275) \\ &=\left(\frac12\times(2\sin75\cos75)^2-1\right)(\cos^275-\sin^275)\end{align}[/tex]

Gunakan sifat sudut ganda, dengan:
[tex]\boxed{\begin{array}{rcl}\sin2x&=&2\sin x\cos x \\ \cos2x&=&\cos^2x-\sin^2x\end{array}}[/tex]

Diperoleh:
[tex]$\begin{align}\sin^875-\cos^875&=\left(\frac12\times(\sin(2\times75))^2-1\right)\times\cos(2\times75) \\ &=\left(\frac12\sin^2150-1\right)\times\cos150 \\ &=\left(\frac12\times\left(\frac12\right)^2-1\right)\times\left(-\frac12\sqrt3\right) \\ &=\left(\frac18-1\right)\times\left(-\frac12\sqrt3\right) \\ &=-\frac78\times\left(-\frac12\sqrt3\right) \\ &=\frac7{16}\sqrt3\end{align}[/tex]