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topjm [15]
3 years ago
14

If cos0=-3/5 in quadrant II, what is sin0

Mathematics
1 answer:
Dovator [93]3 years ago
4 0

Answer:

\displaystyle \sin \theta = \frac{4}{5} if \displaystyle \cos\theta = -\frac{3}{5} and \theta is in the second quadrant.

Step-by-step explanation:

By the Pythagorean Trigonometric Identity:

\left(\sin \theta\right)^2 + \left(\cos\theta)^2 = 1 for all real \theta values.

In this question:

\displaystyle \left(\cos\theta\right)^2 = \left(-\frac{3}{5}\right)^2 = \frac{9}{25}.

Therefore:

\begin{aligned} \left(\sin\theta\right)^2 &= 1 -\left(\cos\theta\right)^2 \\ &= 1 - \left(\frac{3}{5}\right)^2 = \frac{16}{25}\end{aligned}.

Note, that depending on \theta, the sign \sin \theta can either be positive or negative. The sine of any angles above the x axis should be positive. That region includes the first quadrant, the positive y-axis, and the second quadrant.

According to this question, the \theta here is in the second quadrant of the cartesian plane, which is indeed above the x-axis. As a result, the sine of this

It was already found (using the Pythagorean Trigonometric Identity) that:

\displaystyle \left(\sin\theta\right)^2 = \frac{16}{25}.

Take the positive square root of both sides to find the value of \sin \theta:

\displaystyle \sin\theta =\sqrt{\frac{16}{25}} = \frac{4}{5}.

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