# If trigonometric ratios are just ratios of length of sides, then why do they become negative?

This question might be basic, but it is confusing me:

\begin{align} \text{sine of \theta} &= \frac{\text{length of opposite}}{\text{length of hypotenuse}} \\[4pt] \text{cosine of \theta} &= \frac{\text{length of adjacent}}{\text{length of hypotenuse}} \\[4pt] \text{tangent of \theta} &= \frac{\text{length of opposite}}{\text{length of adjacent}} \end{align}

Thus, trigonometric ratios are just ratios of lengths of sides of a right-angled triangle.

So how come their sign changes in different quadrants? Lengths don't have signs.

Eg. $\sin 45^\circ = 0.7071\dots$, whereas $\sin(-45^\circ) = -0.7071\dots$. Why ?

• This answer might be helpful. – Blue Aug 27 '16 at 16:07
• The triangle definitions only apply (unmodified) for a narrow range of angles. – John Coleman Aug 27 '16 at 16:08
• Length have no sign but direction may – Sathasivam K Aug 27 '16 at 17:02
• The definitions that you quoted presuppose that theta is an angle in a right triangle (so that "opposite", "adjacent", and "hypotenuse" make sense). So these definitions apply only to theta in the range from $0$ to $\pi/2$ (or to 90 if you want to use degrees rather than radians). In that range, the trig functions are positive. To define the trig functions for theta outside that range, one needs more general definitions, and the accepted definitions lead to negative values in certain circumstances. – Andreas Blass Aug 27 '16 at 17:19

But when we see the second quadrant, the radius vector makes an OBTUSE ANGLE with positive direction of x-axis .But there is no right angle triangle which have one of three angle is obtuse or two angle as right angle. Hence we can't able to extend the definition based on the length of sides of right angle triangle to a triangle which contains an obtuse angle.when the angle is acute, all trigonometric ratio are positive as you said " $\sin\theta$=$\frac{opposite side}{hypotenuse}$ and so on.