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InterviewSolution
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sin 75° – sin 15° =_________________(a) \(\frac{\sqrt{3}}{\sqrt{2}}\)(b) \(\frac{\sqrt{2}}{\sqrt{3}}\)(c) \(\frac{\sqrt{3}}{2}\)(d) \(\frac{1}{\sqrt{2}}\)I have been asked this question in final exam.This is a very interesting question from Trigonometric Functions of Sum and Difference of Two Angles-2 topic in chapter Trigonometric Functions of Mathematics – Class 11 |
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Answer» RIGHT CHOICE is (d) \(\frac{1}{\sqrt{2}}\) To EXPLAIN: We KNOW, sin x – sin y = 2 cos (x+y)/2 sin (x-y)/2 sin 75° – sin 15° = 2 cos (75°+15°)/2SIN(75°-15°)/2 = 2 cos 45° sin 30° = 2*1/\(\sqrt{2}\)*1/2 = 1/\(\sqrt{2}\). |
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