1
$\begingroup$

I'm learning Trigonometry right now with myself and at current about how to find trigonometry ratios of angles greater than $90^\circ$. I came to know that for finding trigonometric ratio of these angles we can use reference angle because they both have the same absolute numeric value. Can anybody tell me why they both have same trigonometric ratio. Is there any graphical or or any kind of proof to show this.

Please help. Thankyou in advance.

$\endgroup$
0
$\begingroup$

Perhaps the two pictures below will help. The sine is positive when the height measured is to a point above the horizontal axis, and negative if below. The tangent is similarly positive or negative if the height is to a point respectively above or below the axis, but here we must clarify what happens if the point whose tangent we are taking is in the left half of the circle. You draw a line through that point and the center, and it still intersects the vertical tangent line to the right of the circle, and we go by that point. The secant is positive if the point on the circle is between the center and the intersection with that vertical tangent line; thus it is positive in the right half of the circle. In the left half of the circle, the center is between the point on the circle and the point of intersection with the vertical tangent line on the right; thus the secant is negative.

The tangent or the secant of $90^\circ$ is $\infty$, and this $\infty$ is neither $+\infty$ nor $-\infty$ but rather a single $\infty$ that is approached by going in either the positive or the negative direction along the line.

$\endgroup$
  • $\begingroup$ thank you for your answer. Sorry I didn't get the Figure. $\endgroup$ – Yogesh Tripathi Jan 27 '16 at 18:04
  • $\begingroup$ Do you mean you can't see the images, or that you don't understand them? Or something else? $\qquad$ $\endgroup$ – Michael Hardy Jan 27 '16 at 19:29
  • $\begingroup$ I don't understand them. $\endgroup$ – Yogesh Tripathi Jan 27 '16 at 21:48

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.