Pertanyaan

6 cos 10 5 ∘ sin 7 5 ∘ = ....

$6 cos 10 5^{\circ} sin 7 5^{\circ} = ....$

N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

diperoleh 6 ⋅ cos 10 5 ∘ ⋅ sin 7 5 ∘ = − 2 3 .

diperoleh

6 \cdot cos 10 5^{\circ} \cdot sin 7 5^{\circ} = - \frac{3}{2}

Pembahasan

Ingat rumus jumlah dua sudut untuk fungsi sinus dan cosinus berikut. sin ( A + B ) cos ( A + B ) = = sin A cos B + sin B cos A cos A cos B − sin A sin B sehingga dapat ditentukan nilai dari: cos 10 5 ∘ sin 7 5 ∘ 6 ⋅ cos 10 5 ∘ ⋅ sin 7 5 ∘ = = = = = = = = = = = = = cos ( 6 0 ∘ + 4 5 ∘ ) cos 6 0 ∘ ⋅ cos 4 5 ∘ − sin 6 0 ∘ ⋅ sin 4 5 ∘ 2 1 ⋅ 2 1 2 − 2 1 3 ⋅ 2 1 2 4 1 2 − 4 1 6 sin ( 3 0 ∘ + 4 5 ∘ ) sin 3 0 ∘ ⋅ cos 4 5 ∘ + sin 4 5 ∘ ⋅ cos 3 0 ∘ 2 1 ⋅ 2 1 2 + 2 1 2 ⋅ 2 1 3 4 1 2 + 4 1 6 6 ⋅ ( 4 1 2 − 4 1 6 ) ⋅ ( 4 1 2 + 4 1 6 ) 6 ⋅ 4 1 ⋅ 4 1 ⋅ ( 2 − 6 ) ⋅ ( 2 + 6 ) 16 6 ⋅ ( 2 − 6 ) 8 3 ⋅ ( − 4 ) − 2 3 Dengan demikian, diperoleh 6 ⋅ cos 10 5 ∘ ⋅ sin 7 5 ∘ = − 2 3 .

Ingat rumus jumlah dua sudut untuk fungsi sinus dan cosinus berikut.

$sin (A + B) cos (A + B) = = sin A cos B + sin B cos A cos A cos B - sin A sin B$

sehingga dapat ditentukan nilai dari:

$cos 10 5^{\circ} sin 7 5^{\circ} 6 \cdot cos 10 5^{\circ} \cdot sin 7 5^{\circ} = = = = = = = = = = = = = cos (6 0^{\circ} + 4 5^{\circ}) cos 6 0^{\circ} \cdot cos 4 5^{\circ} - sin 6 0^{\circ} \cdot sin 4 5^{\circ} \frac{1}{2} \cdot \frac{1}{2} 2 - \frac{1}{2} 3 \cdot \frac{1}{2} 2 \frac{1}{4} 2 - \frac{1}{4} 6 sin (3 0^{\circ} + 4 5^{\circ}) sin 3 0^{\circ} \cdot cos 4 5^{\circ} + sin 4 5^{\circ} \cdot cos 3 0^{\circ} \frac{1}{2} \cdot \frac{1}{2} 2 + \frac{1}{2} 2 \cdot \frac{1}{2} 3 \frac{1}{4} 2 + \frac{1}{4} 6 6 \cdot (\frac{1}{4} 2 - \frac{1}{4} 6) \cdot (\frac{1}{4} 2 + \frac{1}{4} 6) 6 \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot (2 - 6) \cdot (2 + 6) \frac{6}{16} \cdot (2 - 6) \frac{3}{8} \cdot (- 4) - \frac{3}{2}$