Question

Without using trigonometric tables, evaluate the follow
${cos1}^0{\cos 2}^0{\cos 3}^0.......{\cos 180}^0$

Answer 1

$cosx=-cos(180-x)$

so this gives

$cos1=-cos179$

$cos2=-cos178$

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and so on till cos $90$we will not find pair for that it will be left alone so after putting these values expression becomes,

$cos90+cos180=0–1=-1$