In a ΔABC, if cos C = sin A/2 sin B, prove that the triangle is isosceles.

Question

In a ΔABC, if cos C = \(\cfrac{sin \,A}{2sin\,B}\), prove that the triangle  is isosceles.

Answer 1

The key point to solve the problem:

To prove a triangle isosceles our task is to show either any two angles equal or two sides equal.

Idea of cosine formula -  cos C = \(\cfrac{b^2+a^2-c^2}{2ab}\)

The idea of sine Formula:

Given, 

cos C = \(\cfrac{sin \,A}{2sin\,B}\)

As it has sin terms involved so that sine formula can work, and cos C is also there so we might need cosine formula too.

Let’s apply sine formula keeping a target to prove any two sides equal.

Using sine formula we have –

∴ sin A = ak and sin B = bk

If we apply cosine formula, we will get an equation in terms of sides only that may give us any two sides equal.

⇒ b² + a² – c² = a² ⇒ b² = c²

⇒ b = c

Hence 2 sides are equal.

∴ Δ ABC is isosceles. ….proved