How to find the exact value of tan 11¼° using the value of cos 45°?
Solution:
For all values of the angle A we know that, 2 sin\(^{2}\) \(\frac{A}{2}\) = 1 - cos A
Again, for all values of the angle A we know that, 2 sin \(\frac{A}{2}\) cos \(\frac{A}{2}\) = sin A
Now tan 11¼°
= \(\frac{sin 11¼°}{cos 11¼°}\)
= \(\frac{sin 11¼°}{cos 11¼°}\) × \(\frac{2 sin 11¼°}{2 sin 11¼°}\)
= \(\frac{2 sin^{2} 11¼°}{2 sin 11¼° cos 11¼°}\)
= \(\frac{1 - cos 22½°}{sin 22½°}\)
= \(\frac{1 - \sqrt{\frac{1 + cos 45°}{2}}}{\sqrt{\frac{1 - cos 45°}{2}}}\)
= \(\frac{\sqrt{2} - \sqrt{1 + cos 45°}}{\sqrt{1 - cos 45°}}\)
= \(\frac{\sqrt{2} - \sqrt{1 + \frac{1}{\sqrt{2}}}}{\sqrt{1 - \frac{1}{\sqrt{2}}}}\)
= \(\frac{\sqrt{2} - \sqrt{\frac{\sqrt{2} + 1}{\sqrt{2}}}}{\sqrt{\frac{\sqrt{2} - 1}{\sqrt{2}}}}\)
= \(\frac{\sqrt{2\sqrt{2}} - \sqrt{\sqrt{2} + 1}}{\sqrt{\sqrt{2} - 1}}\)
= \(\frac{\sqrt{2\sqrt{2}} - \sqrt{\sqrt{2} + 1}}{\sqrt{\sqrt{2} - 1}}\) × \(\frac{\sqrt{\sqrt{2} + 1}}{\sqrt{\sqrt{2} + 1}}\)
= \(\frac{\sqrt{2\sqrt{2}}\cdot \sqrt{\sqrt{2} + 1} - \sqrt{(\sqrt{2} + 1)^{2}}}{\sqrt{(\sqrt{2} + 1)(\sqrt{2} - 1)}}\)
= \(\frac{\sqrt{2\sqrt{2}{(\sqrt{2} + 1})}-(\sqrt{2} + 1)}{{\sqrt{2 - 1}}}\)
= \(\sqrt{4 + 2\sqrt{2}} - (\sqrt{2} + 1)\)
11 and 12 Grade Math
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