Both the angle $\theta$ and the shaded triangle share the same adjacent and hypotenuse $-3/5$ since $\cos = \text{adjacent/hypotenuse}$. But, since we know $\cos(135^\circ)$ is not equal to $\cos(45^\circ)$, how should I interpret the shaded triangle angle?

enter image description here


Both the angle $\theta$ and the shaded triangle share the same adjacent and hypotenuse $-3/5$

This uses the definition of $\cos$ for the entire range $\theta \in [0, 2\pi)$ based on the unit circle, where the sides are considered as signed segments and, indeed, $\cos(\theta) = -3/5$.

since $\cos = \text{adjacent/hypotenuse}$.

This, on the other hand, uses the geometric definition of $\cos\,$ for acute angles, where the sides of the right triangle are considered as positive magnitudes. In this case, $\cos(\pi-\theta)=3/5\,$, indeed.

Quoting from wikipedia's Trigonometric functions - Right-angled triangle definitions:

In ordinary Euclidean geometry, according to the triangle postulate, the inside angles of every triangle total $180^\circ$ ($\pi$ radians). Therefore, in a right-angled triangle, the two non-right angles total $90^\circ$ ($\pi / 2$ radians), so each of these angles must be in the range of $(0, \pi/2)$ as expressed in interval notation. The following definitions apply to angles in this $0 – \pi/2$ range. They can be extended to the full set of real arguments by using the unit circle, or $\;\dots$

  • $\begingroup$ In other words, both acute and obtuse angles are right to use the definition of adj/hyp for cosine. The only key difference comes along if we apply it with degree. i.e. cos(45) and cos(135) ? $\endgroup$ – ilovetolearn Jul 16 '17 at 7:37
  • 2
    $\begingroup$ @youcanlearnanything Key difference is that the right-triangle definition uses the positive magnitudes of the sides and is only valid for acute angles, while the unit circle definition uses signed lenghts for the sides and is valid for any angle. This has nothing to do with degrees vs. radians or other angle measures, but with how adj and hyp are interpreted in each definition. $\endgroup$ – dxiv Jul 16 '17 at 7:41
  • 1
    $\begingroup$ @dvix now I got it. thank you! $\endgroup$ – ilovetolearn Jul 16 '17 at 8:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.