Jika cos(2x + 35°) = p dan cos(x + 25°) = q maka sin(3x + 60°) sin(x + 10°)?
(a) q^2 - p^2
(b) p^2 - q^2
(c) p^2 + q^2
(d) 1 - 2p^2q^2
(e) 2p^2q^2 - 1
(a) q^2 - p^2
(b) p^2 - q^2
(c) p^2 + q^2
(d) 1 - 2p^2q^2
(e) 2p^2q^2 - 1
Verified answer
TrigonometriKelas X
cos (2x + 35°) = p , sin (2x + 35°) = √(1 - p²)
cos (x + 25°) = q , sin (x + 25°) = √(1 - q²)
sin(3x + 60°) sin(x + 10°) = sin( (2x + 35°) + (x + 25°) ) sin ( (2x + 35°) - (x + 25°) )
= ( sin(2x + 35°) cos(x + 25°) + sin(x + 25°) cos(2x + 35°) )( sin(2x + 35°) cos (x + 25°) - sin(x + 25°) cos(2x + 35°) )
= (sin (2x + 35°) cos(x + 25°))² - (sin(x + 25°) cos(2x + 35°))²
= (q√(1 - p²))² - (p√(1 - q²))²
= q²(1 - p²) - p²(1 - q²)
= q² - p²