Jika sudut lancip α memenuhi sin⁡ α=1/3 √3. Nilai 2 cos⁡ α+cos⁡ (π/2+α)+sin⁡(π-α)=​

Α lancip -> semua fungsi bernilai positif

sin α = 1/3 √3 = (√3)/(3) = y/r

x=√(r²-y²) –> x= √6

cot α = x/y = √6/ (√3) =√2

cos α = x/r = √6/(3)

tan (π/2 - α) + 3 cos α = cot α + 3 cos α

= √2 + 3(√6)/ 3)

=√2 + √6

SEMOGA BERMANFAAT

JANGAN LUPA, LIKE, COMENT & FOLLOW YA

sin a = ⅓√3 = 1/√3

y = 1, r = √3

x = √((√3)² – 1²) = √(3-1)= √2

maka

2 cos a + cos (π/2 + a) + sin(π – a)

= 2 (√2/√3) + cos (90 + a) + sin (180 – a)

= ⅔√6 – sin a + sin a

= ⅔ √6

^_^ semoga membantu ^_^