Trigonometric Identities

Question

If ABC is a triangle,where cos(theta)=(a-b)/c is given (a,b,c are standard lengths of the triangle ABC & angle theta is a real value)

(a)Show that cos[(A-B)/2]=[(a+b)*sin(theta)]/2*(ab)^1/2

(b)show cos[(A+B)/2]=[c*sin(theta)]/2*(ab)^1/2

Answer 1

Ashley,

with a lot of algebra and a couple of trig tools you can solve this, I did.

put down 2*cos((A+B)/2)*2*cos((A-B)/2)=2*(cos(A)+cosB) . that is cosA +cosB=2*cos((A+B/2)*cos((A-B/2)

guess what in triangle , 2*(cos(A)+cos(B))=[(c²+b²-a²)/(b*c) + ((a²+c²-b²)/(a*c)]

from what you are given 2*cos((A+B)/2)*2*cos((A-B)/2) should equal [(c*(a+b))*sin²(theta)]/(a*b)

with cos(theta)=(a-b)/c.

for sin²(theta) use sin²(theta)=1-cos^2(theta), that is =1-[(a-b)/c)]²

show [(c²+b²-a²)/(b*c) + ((a²+c²-b²)/(a*c)]= [(c*(a+b)/a*b)-((c*(a-+b)/a*b)*((a-b)²/c²)

first thing, make denominators on both sides equal to a*b*c, allowing you to combine numerators, and allowing you to multiply out the denominator.

then whalla, both sides are equal so yes when theta has property cos(theta)=(a-b)c a,b,c triangle sides you can say that cos((A+B)/2)=(a+b)*sin(theta)/((2*(a*b)^(1/2)))and cos((A-B)/2)=c*sin(theta)/((2*(a*b)^(1/2)))