Trigo Plus-1

Geometry Level 3

( 1 cos 6 1 cos 1 ) ( 1 cos 6 2 cos 2 ) ( 1 cos 11 9 cos 5 9 ) \left(1-\frac{\cos 61^{\circ} }{\cos 1^{\circ}}\right) \left(1-\frac{\cos 62^{\circ}}{\cos 2^{\circ}} \right)\cdots \left (1-\frac{\cos 119^{\circ}}{\cos 59^{\circ}}\right )

Evaluate the expression above up to 3 decimal places.

This is part of the set Trigo Plus!


The answer is 1.00.

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1 solution

Ravi Dwivedi
Jul 12, 2015

1 cos ( 6 0 + k ) cos k = cos k cos ( 6 0 + k ) cos k = 2 sin 3 0 sin ( 3 0 + k ) cos k = cos ( 9 0 3 0 k ) cos k = cos ( 6 0 k ) cos k 1-\frac{\cos(60^{\circ}+k^{\circ})}{\cos k^{\circ}} = \frac{\cos k^{\circ}-\cos (60^{\circ}+k^{\circ})}{\cos k^{\circ}}\\ =\frac{2 \sin 30^{\circ} \sin(30^{\circ}+k^{\circ})}{\cos k^{\circ}}\\ =\frac{\cos (90^{\circ}-30^{\circ}-k^{\circ})}{\cos k^{\circ}}\\ =\frac{\cos (60^{\circ}-k^{\circ})}{\cos k^{\circ}}\\

So given expression becomes

( 1 cos 6 1 cos 1 ) ( 1 cos 6 2 cos 2 ) . . . ( 1 cos 11 9 cos 5 9 ) (1-\frac{\cos 61^{\circ}}{\cos 1^{\circ}}) (1-\frac{\cos 62^{\circ}}{\cos 2^{\circ}})... (1-\frac{\cos 119^{\circ}}{\cos 59^{\circ}})\\ = cos 1 cos 2 . . . cos 5 9 cos 1 cos 2 . . . cos 5 9 = 1 =\frac{\cos 1^{\circ} \cos2^{\circ} ... \cos59^{\circ}}{\cos 1^{\circ} \cos2^{\circ} ... \cos59^{\circ}} = \boxed{1}

Moderator note:

Oh that's nice. I wasn't aware of that trigo transformation. Thanks for sharing!

Deserves two up votes :)

Deepak Kumar - 5 years, 7 months ago

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