Question

in any triangle, prove that
c² cos ²B + b²cos²c + bc cos (B-C)
=1/2(a²+b²+c²)​

Answer 1

Answer:

The Pythagorean theorem helps find the lengths of the sides of a right triangle. It states that a²+b²=c², where a and b are the sides around the right angle and c is the hypotenuse.believe I have found a slightly neater way of doing this. Write cos2(B) and cos2(C) as 1−sin2(B) and 1−sin2(C) respectively. Then we are required to show

c2+b2−c2sin2(B)−b2sin2(C)+bccos(B−C)=12(a2+b2+c2)

which is equivalent to

bccos(B−C)−c2sin2(B)−b2sin2(C)=12(a2−b2−c2)

By the cosine rule, we have

a2−b2−c2=−2bccos(A)=2bccos(B+C)

since B+C=π−A . Then we need to show

bccos(B−C)−c2sin2(B)−b2sin2(C)=bccos(B+C)

Expand

cos(B±C)=cos(B)cos(C)∓sin(B)sin(C)

and subtract bccos(B)cos(C) from each side to show we need to prove

bcsin(B)sin(C)−c2sin2(B)−b2sin2(C)=−bcsin(B)sin(C)

Note that c2sin2(B)=b2sin2(C) from the sine rule to form the equivalent equation

bcsin(B)sin(C)=c2sin2(B)

or, equivalently:

bsin(C)=csin(B)

which we know to be true from the sine rule.