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 ^_^