bkartch
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How to find exact value of tan 75°

Sagot :

kate200468
[tex]tan \alpha = \frac{sin \alpha }{cos \alpha } \\--------\\\\tan75^0= \frac{\big{sin75^0}}{\big{cos75^0}} \\\\sin75^0=sin(45^0+30^0)=sin45^0\cdot cos30^0+sin30^0\cdot cos45^0=\\\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{ \sqrt{2} }{2} \cdot \frac{ \sqrt{3} }{2} + \frac{1}{2} \cdot\frac{ \sqrt{2} }{2} = \frac{ \sqrt{6}+ \sqrt{2} }{4} [/tex]
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