Pertanyaan

tan 3 0 ∘ × cos 3 0 ∘ + sin 12 0 ∘ × cos 15 0 ∘ = ....

$tan 3 0^{\circ} \times cos 3 0^{\circ} + sin 12 0^{\circ} \times cos 15 0^{\circ}$ = ....

S. Nur

Master Teacher

Jawaban terverifikasi

Jawaban

diperoleh tan 3 0 ∘ × cos 3 0 ∘ + sin 12 0 ∘ × cos 15 0 ∘ = − 4 1 .

diperoleh

tan 3 0^{\circ} \times cos 3 0^{\circ} + sin 12 0^{\circ} \times cos 15 0^{\circ} = - \frac{1}{4}

Pembahasan

Ingat sudut berelasi serta nilai sudut istimewa berikut. sin ( 18 0 ∘ − α ) = sin α cos ( 18 0 ∘ − α ) = − cos α tan 3 0 ∘ = 3 1 3 cos 3 0 ∘ = 2 1 3 sin 6 0 ∘ = 2 1 3 Jadi nilai Dengan demikian diperoleh tan 3 0 ∘ × cos 3 0 ∘ + sin 12 0 ∘ × cos 15 0 ∘ = − 4 1 .

Ingat sudut berelasi serta nilai sudut istimewa berikut.

$sin (18 0^{\circ} - α) = sin α cos (18 0^{\circ} - α) = - cos α tan 3 0^{\circ} = \frac{1}{3} 3 cos 3 0^{\circ} = \frac{1}{2} 3 sin 6 0^{\circ} = \frac{1}{2} 3$

Jadi nilai

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell tan space 30 degree cross times cos space 30 degree plus sin space 120 degree cross times cos space 150 degree end cell row blank equals cell 1 third square root of 3 cross times 1 half square root of 3 plus sin space open parentheses 180 degree minus 60 degree close parentheses cross times cos space open parentheses 180 degree minus 30 degree close parentheses end cell row blank equals cell 1 half cross times 1 third cross times square root of 3 cross times square root of 3 plus sin space 60 degree cross times left parenthesis negative cos space 30 right parenthesis end cell row blank equals cell 1 over 6 cross times 3 plus 1 half square root of 3 cross times open parentheses negative 1 half square root of 3 close parentheses end cell row blank equals cell 3 over 6 minus 1 half cross times 1 half cross times square root of 3 cross times square root of 3 end cell row blank equals cell 3 over 6 minus 1 fourth cross times 3 end cell row blank equals cell 3 over 6 minus 3 over 4 end cell row blank equals cell 6 over 12 minus 9 over 12 end cell row blank equals cell fraction numerator negative 3 over denominator 12 end fraction end cell row blank equals cell negative 1 fourth end cell end table end style$

Dengan demikian diperoleh $tan 3 0^{\circ} \times cos 3 0^{\circ} + sin 12 0^{\circ} \times cos 15 0^{\circ} = - \frac{1}{4}$ .