An easy one this time

Geometry Level 1

Letting P = sin ⁡ A sin ⁡ B Q = sin ⁡ C cos ⁡ A R = sin ⁡ A cos ⁡ B S = cos ⁡ A cos ⁡ C P=\sin A \sin B \\ Q= \sin C \cos A \\ R = \sin A \cos B \\ S=\cos A \cos C

Find the value of 5 ( P 2 + Q 2 + R 2 + S 2 ) 5(P^2+Q^2+R^2+S^2) .


The answer is 5.

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Vaibhav Prasad by Vaibhav Prasad , 21, India

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2 solutions

Nihar Mahajan
Jul 18, 2015

5 ( P 2 + Q 2 + R 2 + S 2 ) = 5 ( sin ⁡ 2 A sin ⁡ 2 B + sin ⁡ 2 C cos ⁡ 2 A + sin ⁡ 2 A cos ⁡ 2 B + cos ⁡ 2 A cos ⁡ 2 C ) = 5 ( sin ⁡ 2 A sin ⁡ 2 B + sin ⁡ 2 A cos ⁡ 2 B + sin ⁡ 2 C cos ⁡ 2 A + cos ⁡ 2 A cos ⁡ 2 C ) = 5 [ sin ⁡ 2 A ( s i n 2 B + cos ⁡ 2 B ) + cos ⁡ 2 A ( s i n 2 C + cos ⁡ 2 C ) ] = 5 ( sin ⁡ 2 A + cos ⁡ 2 A ) = 5 5(P^2+Q^2+R^2+S^2) \\ =5(\sin^2 A \sin^2 B + \sin^2 C\cos^2 A + \sin^2 A \cos^2 B + \cos^2 A \cos^2 C) \\ = 5(\sin^2 A \sin^2 B + \sin^2 A\cos^2 B + \sin^2 C \cos^2 A + \cos^2 A \cos^2 C) \\ = 5[\sin^2 A(sin^2 B + \cos^2 B) + \cos^2 A(sin^2 C + \cos^2 C)] \\ = 5 (\sin^2 A + \cos^2 A) \\ = \boxed{5}

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I did the same, nice!

Mahdi Raza - 1 year, 4 months ago
Siddharth Singh
Jul 18, 2015

Substituting the values of P,Q,R and S.

5 ( s i n 2 A . sin ⁡ 2 B + sin ⁡ 2 C . cos ⁡ 2 A + sin ⁡ 2 A . cos ⁡ 2 B + cos ⁡ 2 A . cos ⁡ 2 C ) 5(sin^{ 2 }{ A }.\sin ^{ 2 }{ B } +\sin ^{ 2 }{ C } .\cos ^{ 2 }{ A } +\sin ^{ 2 }{ A } .\cos ^{ 2 }{ B } +\cos ^{ 2 }{ A } .\cos ^{ 2 }{ C } )

5 [ sin ⁡ 2 A ( sin ⁡ 2 B + cos ⁡ 2 B ) + cos ⁡ 2 A ( sin ⁡ 2 C + cos ⁡ 2 C ) ] 5[\sin ^{ 2 }{ A } (\sin ^{ 2 }{ B } +\cos ^{ 2 }{ B } )+\cos ^{ 2 }{ A } (\sin ^{ 2 }{ C } +\cos ^{ 2 }{ C } )]

5 [ 1 ] = 5 5[1]=\boxed{5}

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