jika sin x = 1/4 nilai tan 2x

[tex] \sin(x) = \frac{1}{4} \\ \tan(x) = \frac{1}{ \sqrt{ {4}^{2} - {1}^{2} } } = \frac{1}{ \sqrt{15} } \\ \\ \tan(2x) = \frac{2 \tan(x) }{1 - { \tan }^{2} x} \\ = \frac{2 \times \frac{1}{ \sqrt{15} } }{1 - { ( \frac{1}{ \sqrt{15} } ) }^{2} } \\ = \frac{ \frac{2}{ \sqrt{15} } }{1 - \frac{1}{15} } \\ = \frac{ \frac{2}{ \sqrt{15} } }{ \frac{14}{15} } \\ = \frac{2}{ \sqrt{15} } \times \frac{15}{14} \\ = \frac{1}{7} \sqrt{15} [/tex]

Jika sin x = 1/4, maka cos x = √15/4

tan 2x = sin 2x/cos 2x
= 2sin x cos x/ (cos x)^2 - (sin x)^2
= (2)(1/4)(√15/4)/(√15/4)^2 - (1/4)^2
= (√15/8)/(15/16) - (1/16)
= (√15/8)/(14/16)
= (√15/8)(16/14)
= √15/7