Question

Find the value of \sin50\degree(1+ \sqrt{3}\tan10\degree )

Collected in the board: Trigonometry

Steven Zheng posted 4 hours ago

Answer

\sin50\degree(1+ \sqrt{3}\tan10\degree )

=\sin50\degree \cdot \dfrac{\cos10\degree+\sqrt{3}\sin10\degree }{\cos10\degree}

=2\sin50\degree \cdot \dfrac{\dfrac{1}{2} \cos10\degree+\dfrac{\sqrt{3}}{2}\sin10\degree }{\cos10\degree}

=2\sin50\degree \cdot \dfrac{\cos60\degree \cos10\degree+\sin60\degree \sin10\degree }{\cos10\degree}

=2\sin50\degree \dfrac{\cos50\degree }{\cos10\degree}

=\dfrac{\sin100\degree }{\cos10\degree}=\dfrac{\cos10\degree }{\cos10\degree} = 1


Steven Zheng posted 4 hours ago

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