Question

Sederhanakan ekspresi identitas trigonometri bentuk akar berikut
[tex]\displaystyle \sqrt{2+\sqrt{2+\sqrt{2+2\cos 8x}}}[/tex]

Answer 1

Verified answer

Jawab:
= 2cos(x)

Penjelasan dengan langkah-langkah:
Sederhanakan

[tex]\displaystyle\sqrt{2+\sqrt{2+\sqrt{2+2\cos(8x)}}}\\\\=\sqrt{2+\sqrt{2+\sqrt{2(1+\cos(8x))}}}\\\\=\sqrt{2+\sqrt{2+\sqrt{2(1+\cos(8x))}}}[/tex]
Identitas trigonometri
[tex]\cos(2x)=\cos^2(x)-\sin^2(x)\\\cos(2x)=\cos^2(x)-(1-\cos^2(x))\\\cos(2x)=2\cos^2(x)-1[/tex]
Maka
[tex]\because\:\:1+\cos(2x)=2\cos^2(x)\:\:\therefore[/tex]
[tex]\displaystyle\sqrt{2+\sqrt{2+\sqrt{2(1+\cos(8x))}}}\\\\=\sqrt{2+\sqrt{2+\sqrt{2(1+\cos(2(4x)))}}}\\\\=\sqrt{2+\sqrt{2+\sqrt{2(2\cos^2(4x))}}}\\\\=\sqrt{2+\sqrt{2+\sqrt{4\cos^2(4x)}}}\\\\=\sqrt{2+\sqrt{2+\sqrt{4}\not\sqrt{\cos^{\not2}(4x)}}}\\\\=\sqrt{2+\sqrt{2+2\cos(4x)}}[/tex]
Difaktorin lagi..
[tex]\displaystyle=\sqrt{2+\sqrt{2(1+\cos(4x))}}\\\\=\sqrt{2+\sqrt{2(1+\cos(2(2x)))}}\\\\=\sqrt{2+\sqrt{2(2\cos^2(2x))}}\\\\=\sqrt{2+\sqrt{4\cos^2(2x)}}\\\\=\sqrt{2+\sqrt{4}\not\sqrt{\cos^{\not2}(2x)}}\\\\=\sqrt{2+2\cos(2x)}}[/tex]
Difaktorin lagi
[tex]\displaystyle =\sqrt{2(1+\cos(2x))}}\\\\=\sqrt{2(2\cos^2(x))}}\\\\=\sqrt{4\cos^2(x)}}\\\\=\sqrt{4}\not\sqrt{\cos^{\not2}(x)}}[/tex]
= 2cos(x)
(xcvi)