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I have quite a few questions.

  1. First of all, for an angle $90^\circ\lt\theta\lt 180^\circ$, what would the sine/cosine/tangent of this angle be? What I'm saying is that a right angled triagle will always have all other angles acute. How is this possible?
  2. Why is only sine function positive in this quadrant while cosine and tangent are negative?

My teacher just told me to cram the values of trigonometric functions at different quadrants but I'm looking for a physical derivation for this. I'm also attaching a figure I drew to understand a right angled triangle with an obtuse angle.enter image description here

Eyy boss
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  • Have you learned about work and force in physics? You can come up with the right idea of sine and cosine of obtuse angle if you know about these definitions – Đào Minh Dũng Oct 04 '20 at 12:27
  • Yes I've learnt about them. Can you elaborate? – Eyy boss Oct 04 '20 at 12:33
  • The trigonometric functions may have convenient first definitions in relation to a right-angled triangle, but in fact these definitions are extended in natural ways which have huge mathematical significance. The sine, cosine and tangent of an obtuse angle can, for example, be easily constructed as physical distances in relation to a circle of unit radius with centre at the origin. – Mark Bennet Oct 04 '20 at 12:38
  • @Eyyboss Yes, if you have heard of the equation $W = |F|\cdot s \cdot \cos \alpha$, with $W$ as work, $|F|$ as force magnitude, $s$ as distance, and $\alpha$ is the angle between the force vector and the moving direction of the object. Now assuming that the object move in forward direction and $\alpha$ is greater than $90$, so the force is in backward direction. As we all see, the force hampers the effort of the object to move forward so the work must be negative, $W <0$. But $s, |F| > 0$, so $\cos \alpha$ must be negative. So the cosine of obtuse angle is negative – Đào Minh Dũng Oct 04 '20 at 12:39

2 Answers2

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First of all, I upvoted your query re very positive approach, re
"but I'm looking for a physical derivation for this".

To understand this, in the realm of trigonometry,
where sine and cosine are functions of angles,
you need to consider the sine and cosine functions
against the backdrop of the unit circle.

Imagine a unit circle centered at the origin, that hits the $x$ and $y$
axes at points (1,0), (0,1), (-1,0), and (0, -1).

Consider any point in the unit circle that is in the first (upper right) quadrant. The point will have coordinates $(x,y).$

Let $\theta$ denote the angle formed by (0,0) -- (1,0) with (0,0) -- (x,y). Since the radius of the circle is 1, $\cos \theta = x$ and
$\sin \theta = y.$

Now imagine traveling around the arc of the unit circle until you reach the point (0,1).

This point may be construed to represent $90^{\circ}$, just as one complete revolution around the circle can be construed to represent $360^{\circ}.$

It is easy to see that $\cos(90^{\circ}) = 0$ and $\sin(90^{\circ}) = 1.$

Now imagine traveling around the arc to any point on the unit circle that is in the 2nd (upper left) quadrant.

Here, the point $(x,y)$ in the 2nd quadrant will have $x < 0$ and $y > 0.$

Again, just as before, consider $\theta$ to be the angle formed by (0,0) -- (1,0) with (0,0) -- (x,y).

Here, by convention, $\cos \theta$ (again) $ = x$
and $\sin \theta$ (again) $ = y.$

Thus, it is easy to see that when $(x,y)$ is in the 2nd quadrant, and
$\theta$ is the angle formed by (0,0) -- (1,0) with (0,0) -- (x,y)
that $\cos \theta$ will by convention be $< 0$ and
$\sin \theta$ will by convention be $ > 0.$

A clear advantage of these conventions is that they facilitate the formulas shown at https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Angle_sum_and_difference_identities.

Another advantage of these conventions, which may not seem important in the realm of trigonometry, is that they facilitate the cosine and sine functions being continuous functions. This is a pandora's box that may not be worth exploring in the realm of trigonometry, but is still worth a very casual mention.

See https://www.mathopenref.com/triggraphsine.html.

user2661923
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  1. In obtuse angled triangle, you would draw a perpendicular. It may meet the third side externally.

  2. Sine is perpendicular upon hypotenuse. In second quadrant, perpendicular (i.e. y-axis coordinate) is positive. So, sine is positive. Here, base (i.e x-axis coordinate) is negative. So, cosine and tangent are negative.

Is this clear now?

aarbee
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  • I know that we're assuming coordinate system but isn't sine just ratio of two sides of a triangle? How can a side be negative? – Eyy boss Oct 04 '20 at 12:35
  • Draw the triangle on coordinate axis, with the angle in desired quadrant (the vertex for the desired angle being origin). – aarbee Oct 04 '20 at 12:40
  • @Eyy boss, indeed lengths of sides of triangle are positive. But on coordinate system, we take projection of hypotenuse on y-axis or x-axis. And by convention, projection on positive axis is considered positive and on negative axis, it is considered negative. – aarbee Oct 04 '20 at 12:51