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

Question

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

Uniteasy Admin · Accepted Answer

According to the definition of conse
$$\cos 	heta = \dfrac{x}{\sqrt{x^2+9} } = \dfrac{\sqrt{10} }{10}x$$
$$\because x 
e 0 $$
$$ \dfrac{1}{\sqrt{x^2+9} } = \dfrac{\sqrt{10} }{10}$$
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 $$(1,3)$$
$$\sin 	heta = \dfrac{3}{\sqrt{10} }=\dfrac{3\sqrt{10} }{10} $$, $$	an 	heta  = 3$$
If $$ x = -1$$, the coordinate of $$P$$ is $$(-1,3)$$,
$$\sin 	heta = \dfrac{3}{\sqrt{10} }=\dfrac{3\sqrt{10} }{10} $$, $$	an 	heta  = -3$$

Question

Answer