Show that in any triangle ABC: (a+b)/(a-b)=(sinA+sinB)/(sinA-sinB) and solve the following for 0 ≤ x ≤ 2π: 2-sinx=2cos^(2)x

Answer 1

(SinA+SinB)/(SinA -SinB) = (a+b)/(a-b)

aSinA -aSinB + bSinA -bSinB = aSinA +aSinB -bSinA -bSinB

aSinA and -bSinB cancels

leaving

2aSinB = 2bSinA

divide by 2

aSinB =bSinA

divide both sides by ab

SinB/b = SinA/a which is always true by the Law of Sines

The sine of any of a triangles' angles divided by its opposite side is a constant for all 3 angles of the triangle

2-sinx = 2cos^2(x)

2-sinx =2(1-sin^2(x)

2-sinx = 2 -2sin^2(x) let y = sinx

2-y = 2- 2y^2

2y^2 -y = 0

y(y-1) = 0

y =0 or y =1

substitute back y=sinx

sinx = 0 or sinx =1

x = 0 or pi, or x=pi/2

x = 0, pi/2 or pi