﻿ In \triangle ABC, prove that \cos A+\cos B+ \cos C= 1+4\sin\dfrac{A}{2} \sin\dfrac{B}{2} \sin\dfrac{C}{2}

Question

In \triangle ABC, prove that

\cos A+\cos B+ \cos C= 1+4\sin\dfrac{A}{2} \sin\dfrac{B}{2} \sin\dfrac{C}{2}

Collected in the board: Trigonometry

Steven Zheng posted 2 years ago

The sum of internal angles of a triangle is equal to \pi

\because A+B+C = \pi

\therefore \cos C = -\cos(A+B)

\cos(\dfrac{\pi}{2}-\dfrac{C}{2})=\cos \dfrac{A+B}{2}

Applying Cofunction Identity

\sin \dfrac{C}{2} =\cos \dfrac{A+B}{2}

LHS = \cos A+\cos B - \cos(A+B)

Using sum identity and double angle identity,

RHS = 1+4\sin\dfrac{A}{2} \sin\dfrac{B}{2} \cos \dfrac{A+B}{2}

=1+4\sin\dfrac{A}{2} \sin\dfrac{B}{2}(\cos\dfrac{A}{2}\cos\dfrac{B}{2}-\sin\dfrac{A}{2}\sin\dfrac{B}{2} )

= 1+\dfrac{1}{4}\sin A \sin B - \sin^2\dfrac{A}{2}\sin^2\dfrac{B}{2}

=1+[\sin A \sin B - (1-\cos A)(1-\cos B)]

=1+[\sin A \sin B - (1-\cos A-\cos B+\cos A\cos B)]

=\cos A+\cos B - \cos (A+B)

=LHS

Steven Zheng posted 2 years ago

The sum of internal angles of a triangle is equal to \pi

\because A+B+C = \pi

\therefore \cos C = -\cos(A+B)

\cos(\dfrac{\pi}{2}-\dfrac{C}{2})=\cos \dfrac{A+B}{2}

Applying Cofunction Identity

\sin \dfrac{C}{2} =\cos \dfrac{A+B}{2}

Using Sum to Product identity and double angle identity

LHS = 2\cos\dfrac{A+B}{2}\cos\dfrac{A-B}{2}-\cos(A+B)

= 2\cos\dfrac{A+B}{2}\cos\dfrac{A-B}{2} - (2\cos^2\dfrac{A+B}{2}-1 )

=2\cos\dfrac{A+B} {2}(\cos\dfrac{A-B}{2}-\cos\dfrac{A+B}{2} )+1

=4\sin \dfrac{C}{2} \sin \dfrac{B}{2} \sin \dfrac{A}{2} +1

=RHS

Steven Zheng posted 2 years ago

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