Question

Jika tan 3/4 dengan sudut x lancip maka nilai dari cos 2x adalah

Answer 1

Tan a = y/x = 3/4
r = akar (3^2 + 4^2) =5
cos 2a
= (cos a) ^2 - (sin a)^2
= (x/r)^2 - (y/r)^2
= (4/5)^2 - (3/5)^2
= 16/25 - 9/25
= 7/25

Answer 2

Jawab:

cos 2x = \frac{7}{25}

Penjelasan dengan langkah-langkah:

1. menemukan nilai cos² x

tan x = \sqrt{sec^{2} x - 1} \\\frac{3}{4} = \sqrt{sec^{2} x - 1}\\\frac{9}{16} = sec^{2} x - 1\\\frac{9}{16} + 1 = \frac{9}{16} + \frac{16}{16} = \frac{9+16}{16} = \frac{25}{16} = sec^{2} x\\sec^{2} x = \frac{1}{cos^{2}x}  = \frac{25}{16}\\cos^{2}x = \frac{16}{25}

2. menemukan sin² x

sin^{2} x= 1 - cos^{2} x\\sin^{2} x= 1 - \frac{16}{25} = \frac{25}{25} - \frac{16}{25} = \frac{25-16}{25} = \frac{9}{25}

3. final cos 2x

cos 2x = cos^{2}x-sin^{2}x = 1 - 2 \times sin^{2}x\\cos 2x = \frac{16}{25} - \frac{9}{25} = 1 - 2 \times \frac{9}{25}\\cos 2x = \frac{7}{25} = \frac{25}{25} - \frac{18}{25}\\cos 2x = \frac{7}{25} = \frac{7}{25}