Question

Diketahui sudut lancip alpha memenuhi sin a= 1/3 √3,maka nilai tan (phi/2-a)+ 3 cos a =...

Answer 1

Sin a = 1/3 √3 = √3/ 3 
cos a  = √( 1 -sin² a) = √(1 - 1/9)= √(2/3) = 1/3 √6
tan a = sin a / cos a = 1/2
cot a = 1/tan a = 2

tan (π/2 - a) + 3 cos a = cot a + 3 cos a = 2+ 3 (1/3 √6) = 2+√6

Answer 2

Sin a = 1/3 √3 = 1/√3
→ y = 1 ;  r = √3
→ x² = r² - y² = (√3)² - 1² = 3 - 1 = 2
⇒ x = √2

cos a = x/r = √2/√3 = 1/3 √6
cotan a = x/y = √2/1 = √2

tan (π/2 - a) + 3 cos a
= cot a + 3 cos a
= √2 + 3 (1/3 √6)
= √2 + √6

semoga membantu ^^