please show workings thank you
r^2 [sin2 θ] = 6
r^2 (2sin θ cos θ) = 6 divide through by 2
r^2 (sin θ cos θ) = 3 [ sin θ = y/r , cos θ= x / r ]
So....
r^2 ( y/r)(x/r) = 3
r^(2) ( xy) / r^2 = 3
xy = 3
y = 3 / x
how does sin20 become 2sin0 cos0?
sorry im terrible at this topic
This is a trig identity
sin (2 θ) = 2sin θ cos θ
ok thank you very much
x^4 = x^2 + y^2 [x^2 + y^2 = r^2, x = r cos θ, y = r sin θ ]
(r cos θ)^4 = r^2
r^4 (cos θ)^4 = r^2
