Pertanyaan

Diketahui sin A = 5 2 dan cos B = 13 − 2 . Sudut A sudut tumpul dan sudut B di kuadran III . Tentukan sin ( A + B ) !

Diketahui $sin A = \frac{2}{5}$ dan $cos B = \frac{- 2}{13}$ . Sudut $A$ sudut tumpul dan sudut $B$ di kuadran $III$ . Tentukan $sin (A + B)$ !

Ikuti Tryout SNBT & Menangkan E-Wallet 100rb

Habis dalam

Klaim

N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

sin ( A + B ) = 65 − 1 .

sin (A + B) = \frac{- 1}{65}

Pembahasan

Ingat bahwa, sin ( A + B ) = sinA ⋅ cosB + cosA ⋅ sinB mencari nilai cos A sin 2 A + cos 2 A ( 5 2 ) 2 + cos 2 A 5 4 + cos 2 A cos 2 A cos 2 A cos A cos A cos A = = = = = = = = 1 1 1 1 − 5 4 5 1 ± 5 1 ± 5 1 − 5 1 ( A sudut tumpul artinya di kuadran II ) mencari nilai sin B sin 2 B + cos 2 B = 1 sin 2 B + ( 13 − 2 ) 2 = 1 sin 2 B + 13 4 = 1 sin 2 B = 1 − 13 4 sin 2 B = 13 9 sin B = ± 13 9 sin B = ± 13 3 sin B = − 13 3 ( B di kuadran III ) mencari nilai sin (A+B) sin ( A + B ) = = = = sinA ⋅ cosB + cosA ⋅ sinB 5 2 ⋅ ( − 13 2 ) + 5 − 1 ⋅ ( − 13 3 ) 65 − 4 + 65 3 65 − 1 Dengan demikian sin ( A + B ) = 65 − 1 .

Ingat bahwa,

$sin (A + B) = sinA \cdot cosB + cosA \cdot sinB$

mencari nilai cos A

$sin^{2} A + cos^{2} A (\frac{2}{5})^{2} + cos^{2} A \frac{4}{5} + cos^{2} A cos^{2} A cos^{2} A cos A cos A cos A = = = = = = = = 111 1 - \frac{4}{5} \frac{1}{5} \pm \frac{1}{5} \pm \frac{1}{5} - \frac{1}{5} (A sudut tumpul artinya di kuadran II)$

mencari nilai sin B

$sin^{2} B + cos^{2} B = 1 sin^{2} B + (\frac{- 2}{13})^{2} = 1 sin^{2} B + \frac{4}{13} = 1 sin^{2} B = 1 - \frac{4}{13} sin^{2} B = \frac{9}{13} sin B = \pm \frac{9}{13} sin B = \pm \frac{3}{13} sin B = - \frac{3}{13} (B di kuadran III)$

mencari nilai sin (A+B)

$sin (A + B) = = = = sinA \cdot cosB + cosA \cdot sinB \frac{2}{5} \cdot (- \frac{2}{13}) + \frac{- 1}{5} \cdot (- \frac{3}{13}) \frac{- 4}{65} + \frac{3}{65} \frac{- 1}{65}$