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Strike441 [17]
3 years ago
12

find the value of the trigonometric function sin (t) if sec t = -4/3 and the terminal side of angle t lies in quadrant II

Mathematics
1 answer:
kodGreya [7K]3 years ago
8 0

Answer:

sin(t) =\frac{\sqrt{7}}{4}

Step-by-step explanation:

By definition we know that

sec(t) = \frac{1}{cos(t)}

and

cos ^ 2(t) = 1-sin ^ 2(t)

As sec(t) = -\frac{4}{3}

Then

sec(t) = -\frac{4}{3}\\\\\frac{1}{cos(t)} =-\frac{4}{3}\\\\cos(t) = -\frac{3}{4}

Now square both sides of the equation:

cos^2(t) = (-\frac{3}{4})^2

cos^2(t) = \frac{9}{16}\\\\

1-sin^2(t) =\frac{9}{16}\\\\sin^2(t) =1-\frac{9}{16}\\\\sin^2(t) =\frac{7}{16}\\\\sin(t) =\±\sqrt{\frac{7}{16}}

In the second quadrant sin (t) is positive. Then we take the positive root

sin(t) =\sqrt{\frac{7}{16}}

sin(t) =\frac{\sqrt{7}}{4}

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