Question

Suppose there exists a point P(x,3) \, (x \ne 0) on the terminal side of an angle \theta with its vertex at the origin, and \cos \theta = \dfrac{\sqrt{10} }{10}x . Show the value of \sin \theta and \tan \theta

Collected in the board: Trigonometry

Steven Zheng posted 2 months ago


Answer

According to the definition of conse


\cos \theta = \dfrac{x}{\sqrt{x^2+9} } = \dfrac{\sqrt{10} }{10}x

\because x \ne 0

\dfrac{1}{\sqrt{x^2+9} } = \dfrac{\sqrt{10} }{10}


Square both sides and simplify,

\dfrac{1}{x^2+9}=\dfrac{1}{10}

Solve the equation

x=1 or x= -1

If x = 1, the coordinate of P is (1,3)

\sin \theta = \dfrac{3}{\sqrt{10} }=\dfrac{3\sqrt{10} }{10} , \tan \theta = 3

If x = -1, the coordinate of P is (-1,3),

\sin \theta = \dfrac{3}{\sqrt{10} }=\dfrac{3\sqrt{10} }{10} , \tan \theta = -3

Steven Zheng posted 2 months ago

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