Question

Let X=33/65 be the x-coordinate of the point P(x,y), where the terminal side of angle theta (in standard position) meets the unit circle. If P is in Quadrant 4, what is sin theta?

Answer 1

Answer:

The value of sinθ is $-\frac{56}{65}$.

Step-by-step explanation:

Let X=33/65 be the x-coordinate of the point P(x,y), where the terminal side of angle theta (in standard position) meets the unit circle.

Draw a diagram as shown below,

Use Pythagoras theorem in triangle OPQ,

$hypotenuse^2=base^2+perpendicular^2$

$(1)^2=(\frac{33}{65})^2+perpendicular^2$

$perpendicular=\sqrt{1-(\frac{33}{65})^2}$

$perpendicular=\sqrt{\frac{56}{65})^2}$

$perpendicular=\pm \frac{56}{65}$

In a right angled triangle,

$\sin\theta=\frac{perpendicular}{hypotenuse}$

$\sin\theta=\frac{\pm \frac{56}{65}}{1}$

$\sin\theta=\pm \frac{56}{65}$

It is given that point P is in Quadrant 4 and the value of sinθ is negative in fourth quadrant.

Therefore the value of sinθ is $-\frac{56}{65}$.

Answer 2

Sin ² ( theta ) = 1 - ( 33/65)²
sin² ( theta ) = 1 - 1089 / 4225
sin² ( theta ) = 3136/4225
sin ( theta ) = - √3136/4225 = - 56/65  ( because P is in Quadrant 4 ) 
Answer: sin (theta ) = - 56/65