If sinu= (-5/7) and pie<u< (3pie/2) find sin 2u
Calculate sin(2u) using the identity:
sin(2x)=2sin(x)cos(x) ....(1)
From "pie<u< (3pie/2)"
we conclude that u is in the third quadrant where sin(u)<0 and cos(u)<0.
From sin(u)=(-5/7), calculate cos(u) using cos²(u)+sin²(u)=1.
Make sure the value of both sin(u) and cos(u) are negative and substitute in equation (1) above.
To find the value of sin 2u, we can use the double-angle identity for sine:
sin 2u = 2sin u cos u
First, let's find the values of sin u and cos u. We are given that sin u = -5/7.
To find cos u, we can use the Pythagorean identity:
sin^2 u + cos^2 u = 1
Substituting the value of sin u, we get:
(-5/7)^2 + cos^2 u = 1
25/49 + cos^2 u = 1
cos^2 u = 1 - 25/49
cos^2 u = 24/49
cos u = ±√(24/49)
Since u is in the second quadrant (given that π < u < (3π/2)), the cosine value should be negative. Hence, cos u = -√(24/49).
Now, substitute the values of sin u and cos u into the double-angle identity:
sin 2u = 2sin u cos u
= 2(-5/7)(-√(24/49))
= 10√(24/49)/7
= (10/7)(√24/√49)
= (10/7)(√24/7)
= (10√24)/49
Therefore, sin 2u = (10√24)/49.