The "magnitude" of $\cos \theta, \sin \theta$ of an obtuse angles can be measure in terms of the acute angle its terminal arm forms with the $x$-axis, positive or negative. So an angle of, say, $\theta = \frac 34 \pi$ has its terminal arm pointing in the direction of Quadrant II, and the absolute values of $\cos\theta$, $\;\sin \theta$ are the same as the values of $\theta' = \pi/4$, the angle formed by the terminal arm of $\theta$ with respect to the negative $x$-axis: $\theta' = \pi - \frac{3\pi}{4}$.
The sign of that magnitude can be determined by the quadrant in which the angle's terminating ray is directed. (Recall that $\cos \theta$ corresponds with the $x$ coordinate, $\sin \theta$ with $y$):
Quadrant I: $\cos \theta, \sin \theta \gt 0$
Quadrant II: $\cos \theta \lt 0$, $\;\sin\theta > 0$
Quadrant III: $\cos \theta \lt 0,\;\;\sin\theta \lt 0$.
Quadrant IV: $\cos \theta \gt 0,\;\;\sin\theta \lt 0$.
Extending to other quadrants
If trigonometric identity formulas hold true so long as we are within the first quadrant, shouldn't we be able to extend our trig functions to all quadrants in this manner?
Mainly, you'd want the following identities.
$$\cos(\alpha+\theta)=\cos(\alpha)\cos(\theta)-\sin(\alpha)\sin(\theta)$$
$$\sin(\alpha+\theta)=\sin(\alpha)\cos(\theta)+\cos(\alpha)\sin(\theta)$$
Since we know $\sin\left(\frac\pi4\right)=\cos\left(\frac\pi4\right)=\frac{\sqrt2}2$, we can derive $\sin\left(\frac\pi2\right)=1$, $\cos\left(\frac\pi2\right)=0$, and so forth.
Similarly, we can derive what $\sin(-\theta)$ is by using $\sin(\theta-\theta),$ $\cos(\theta-\theta)$, and $\cos^2+\sin^2=1$, the Pythagorean identity. Two equations, and you can solve for $\cos(-\theta),\sin(-\theta)$ by substitution. Use Pythagorean identity for simplifying the answer.
It just happens to be that on the unit circle, the hypotenuse is by definition $1$, so...
$$\sin=\frac{\text{opp}}{\text{hyp}}=\text{opp}=y$$
$$\cos=\frac{\text{adj}}{\text{hyp}}=\text{adj}=x$$
And since both this definition and the one above derived by trig identities come out the same for $\theta>90\deg$ or $\pi$, then both are equally correct.
Best Answer
Let $P'(x',y')$ be a point on a circle centred at the origin $O,$ and $\theta$ be the angle swept out by $OP'$ as it rotates anticlockwise from the positive $x$-axis.
By definition, $$\cos\theta:=\frac {x'}{OP'}\,;\\\sin\theta:=\frac {y'}{OP'}\,;\\\tan\theta:=\frac {y'}{x'}\,.$$
When the circle is scaled into a unit circle with new general point $P(x,y),$ due to triangle similarity, these trigonometric ratios are preserved. In other words, the following set of definitions is equivalent to the above: $$\cos\theta:=x\,;\\\sin\theta:=y\,;\\\tan\theta:=\frac y x.$$
The unit-circle definition of the trigonometric functions, rather than the general-circle definition, is the standard/conventional presentation simply because it is simpler. But they are equivalent to each other.