The correct option is A sin 75^∘ nbsp;= sin(90^∘ nbsp; 15^∘) = cos 15^∘ nbsp;= cos(45^∘ nbsp; 30^∘) = nbsp; √(3) + 1/2√(2). ...

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Range of Trigonometric Ratios from 0 to 90 Degrees

sin 750=[MNR ...

Question

sin $75^{0}$ =

[MNR 1979]

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Solution

The correct option is B

$s i n 75^{\circ} = s i n (90^{\circ} - 15^{\circ}) = c o s 15^{\circ} = c o s (45^{\circ} - 30^{\circ})$

= $\frac{\sqrt{3} + 1}{2 \sqrt{2}}$ .

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