the terminal point of 0 is (3/2, 1/2). what is sin 0 ?

[tex] \bigg( \frac{3}{2}, \: \: \frac{1}{2} \bigg) = (x, \: y) \\ \\ \implies \: x = \frac{3}{2} \: \: y = \frac{1}{2} \\ \\ r = \sqrt{ {x}^{2} + {y}^{2} } \\ = \sqrt{ { \bigg( \frac{3}{2} } \bigg)^{2} + { \bigg( \frac{1}{2} } \bigg)^{2}} \\ = \sqrt{ \frac{9}{4} + \frac{1}{4} } \\ = \sqrt{ \frac{10}{4} } \\ r = \frac{ \sqrt{10} }{2} \\ \\ \sin \theta = \frac{y}{r} \\ \\ \sin \theta = \frac{ \frac{1}{2} }{ \frac{ \sqrt{10} }{2} } \\ \\ \sin \theta = \frac{1}{ \sqrt{10} } [/tex]

sakshigupta4990 sakshigupta4990 · Answer 2 · 2022-05-24T12:22:30+02:00

Answer:

The answer is,[tex]\frac{1}{\sqrt{10} }[/tex]

Step-by-step explanation:

Definition of terminal point:

The Unit Circle's Terminal Points Begin at the unit circle's point (1,0). Walk for 't' units in a counterclockwise direction. The "terminal point" is where you finally arrive (x,y).

According, to the question:

[tex](\frac{3}{2} ,\frac{1}{2} =(x,y)\\x=\frac{3}{2},\frac{1}{2} \\r=\sqrt{x^{2} +y^{2} } \\\sqrt{\frac{9}{4}+\frac{1}{4} } \\\sqrt{\frac{10}{4} } \\r=\sqrt{10} /2[/tex]

[tex]sin[/tex]θ=[tex]\frac{y}{x}[/tex]

[tex]sin[/tex]θ= [tex]\frac{1}{2} /\sqrt{10}/2\\[/tex]

[tex]sin[/tex]θ=[tex]\frac{1}{\sqrt{10} }[/tex]

the terminal point are [tex]\frac{1}{\sqrt{10} }[/tex]

To know more about terminal point here, https://brainly.com/question/24085885

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