3 years ago

Carly wants to buy 10 litres of

Mathematics

1 answer:

yuradex [85]3 years ago

4 0

Answer:

The correct answer is:

Home stores: £23.75

Paint Supplies: £27.30

The DIY Shop: £22.50

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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

kodGreya [7K]

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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