﻿ In \triangle ABC, if \cos^2A+\cos^2 B+\cos^2C=1 , prove that the triangle is a right triangle.

#### Question

In \triangle ABC, if \cos^2A+\cos^2 B+\cos^2C=1 , prove that the triangle is a right triangle.

Collected in the board: Trigonometry

Steven Zheng posted 2 years ago

Using the Pythagorean identity to reduce the power of the trigonometric expression

\cos^2\alpha =\dfrac{\cos 2\alpha +1 }{2}

Then,

\cos^2A+\cos^2 B+\cos^2C

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

\dfrac{\cos 2A+\cos 2B}{2} +\cos^2C=0

Using sum to product identity,

\cos \alpha + \cos \beta = 2\cos\dfrac{\alpha +\beta }{2} \cos\dfrac{\alpha -\beta }{2}

Then,

\cos(A+B)\cos(A-B)+\cos^2C=0

\because A+B+C=\pi

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

\cos^2C-\cos C\cos(A-B)=0

\cos C[\cos C-\cos(A-B)]=0

Using the difference to product identity

\cos \alpha - \cos \beta = -2\sin\dfrac{\alpha +\beta }{2} \sin\dfrac{\alpha -\beta }{2}

Then,

\cos C\sin\dfrac{C+A-B}{2}\sin\dfrac{A-B-C}{2} =0

\because A+B+C = \pi

\cos C\sin(\dfrac{\pi}{2}-B)\sin(A-\dfrac{\pi}{2})=0

\cos C\cos B \cos A = 0

\therefore A=\dfrac{\pi}{2} or B=\dfrac{\pi}{2} or C=\dfrac{\pi}{2}

\triangle ABC is a right triangle

Steven Zheng posted 2 years ago

\because \cos^2A+\cos^2 B+\cos^2C=1

\cos^2A+\cos^2 B = 1-\cos^2C =\sin^2C

\cos^2A+\cos^2 B - \sin^2C = 0

Using the Pythagorean identities

\cos^2A+\cos^2 B - \sin^2C

= \cos^2A+\dfrac{\cos 2B+1}{2} -\dfrac{1-\cos 2C}{2} =0

\cos^2A+\dfrac{1}{2}(\cos 2B+\cos 2C ) = 0

Using sum to product identity

\cos \alpha + \cos \beta = 2\cos\dfrac{\alpha +\beta }{2} \cos\dfrac{\alpha -\beta }{2}

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

= \cos^2A+ \cos(B+C)\cos(B-C) = 0

\because A+B+C = \pi

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

\cos A\cdotp [-\cos(B+C)] - \cos A \cos(B-C) = 0

\cos A[\cos(B+C)+ \cos(B-C) ] = 0

Using sum to product identity again

\cos A\cos B \cos C = 0

\therefore A=\dfrac{\pi}{2} or B=\dfrac{\pi}{2} or C=\dfrac{\pi}{2}

\triangle ABC is a right triangle

Steven Zheng posted 2 years ago

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