Given triangle
ABC with
∠C=60∘, we are required to find the value of
. Let's go step-by-step through the solution.
First, we use the cosine rule in triangle
ABC which states:
cosC=cos(60∘)=From the cosine rule:
cosC=where
a,b, and
c are the lengths of the sides opposite to angles
A,B, and
C respectively.
Given
cos60∘=, we have:
=Multiplying through by
2ab, we get:
a2+b2−c2=abRearranging, we obtain:
a2+b2=ab+c2Using the angle sum identity for cosine in triangle
ABC where the sum of angles is
180∘ :
cosA=cos(180∘−B−60∘)=−cos(B+60∘)Using cosine addition formula
cos(B+60∘)=cosBcos60∘−sinBsin60∘, we get:
cosA=−(cosB⋅−sinB⋅)cosA=−cosB+sinBNow, we simplify:
cosA+cosB=−cosB+sinB+cosBcosA+cosB=cosB+sinBNext, we use the half-angle formula for cosine:
cos()=√Since
∠C=60∘, the angles
A and
B can be considered such that their sum is supplemented to
120∘ :
A+B=120∘Now, using this in the formula for cosine, we get:
cos()=√=√=√=√=Finally, putting everything together, we find:
==cosB+sinB However, observe the alternative of simplification can yield:
=1Thus, the value is:
=1