Find values of sin 671-2 - and cos 671-2 -

sin 6712^∘=sin(90^∘ 2212^∘)=cos 22 12^∘=√(1+cos 2 (221/2^∘)2)=√(1+cos 45^∘2)=√(1+1/√(2)2)=12√(2+√(2)) cos 6712^∘=cos(90^∘ 2212^∘)=sin 22 12^∘=√(1 cos ...

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Standard XII

Mathematics

Domain and Range of Basic Inverse Trigonometric Functions

Find values o...

Question

Find values of $sin 67 {\frac{1}{2}}^{\circ}$ and $cos 67 {\frac{1}{2}}^{\circ}$

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Solution

$sin 67 {\frac{1}{2}}^{\circ} = sin (90^{\circ} - 22 {\frac{1}{2}}^{\circ}) = cos 22 {\frac{1}{2}}^{\circ} = \sqrt{\frac{1 + cos 2 (22 {\frac{1}{2}}^{\circ})}{2}} = \sqrt{\frac{1 + cos 45^{\circ}}{2}} = \sqrt{\frac{1 + 1 / \sqrt{2}}{2}} = \frac{1}{2} \sqrt{2 + \sqrt{2}}$
$cos 67 {\frac{1}{2}}^{\circ} = cos (90^{\circ} - 22 {\frac{1}{2}}^{\circ}) = sin 22 {\frac{1}{2}}^{\circ} = \sqrt{\frac{1 - cos 2 (22 {\frac{1}{2}}^{\circ})}{2}} = \sqrt{\frac{1 - cos 45^{\circ}}{2}} = \sqrt{\frac{1 - 1 / \sqrt{2}}{2}} = \frac{1}{2} \sqrt{2 - \sqrt{2}}$

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Find values of sin6712∘ and cos6712∘

sin6712∘=sin(90∘−2212∘)=cos2212∘=√1+cos2(2212∘)2=√1+cos45∘2=√1+1/√22=12√2+√2cos6712∘=cos(90∘−2212∘)=sin2212∘=√1−cos2(2212∘)2=√1−cos45∘2=√1−1/√22=12√2−√2

Find values of $sin 67 {\frac{1}{2}}^{\circ}$ and $cos 67 {\frac{1}{2}}^{\circ}$