Question

Cos(−θ)=−√2/5 , sinθ>0 What is the value of sinθ ?

Answer 1

Answer:

sin theta = +(√23)/5

Step-by-step explanation:

Did you know that the cosine is an even function, so that cos (-theta) = cos (+theta)?

Thus we have:

cos(θ)= −√2/5 , sinθ>0

If the cosine of theta is negative, that means that the terminal side of theta is in either Quadrant II or III.  Since the sine of theta is positive, we can deduce that theta is in Quadrant II.

Given cos(θ)= −√2/5, we square both sides, obtaining (cos theta)^2 = 2/25.  Using the formula (sin theta)^2 + (cos theta)^2 = 1, we arrive at:

(sin theta)^2 = 1 - 2/25, or 23/25.

Then sin theta = +(√23)/5.

Answer 2

first off, let's recall the symmetry trig identity that cos(-θ) = cos(θ).

now, the hypotenuse is just a radius unit, and therefore never negative, so in that cosine equation, the negative must be the numerator.

sin(θ) > 0, is another way to say, the sine is positive, and since the hypotenuse is always positive, then the opposite side in the angle is also positive.

$\bf cos(-\theta )=-\cfrac{\sqrt{2}}{5}\implies \stackrel{\textit{symmetry identity}}{cos(\theta )=-\cfrac{\sqrt{2}}{5}}\implies cos(\theta )=\cfrac{\stackrel{adjacent}{-\sqrt{2}}}{\stackrel{hypotenuse}{5}} \\\\\\ \textit{let's find the \underline{opposite side}} \\\\\\ \stackrel{\textit{using the pythagorean theorem}}{c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}}$

$\bf \pm\sqrt{5^2-(-\sqrt{2})^2}=b\implies \pm\sqrt{25-2}=b \\\\\\ \pm\sqrt{23}=b\implies \stackrel{\textit{recall that }sin(\theta )>0}{+\sqrt{23}=b} \\\\\\ sin(\theta )=\cfrac{\stackrel{opposite}{\sqrt{23}}}{\stackrel{hypotenuse}{5}}$