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In lower classes, students are told that $\cos\theta=\dfrac{\text{base}}{\text{hypotenuse}}$. Then when they come to higher class, they are told that $\cos\theta=\dfrac{\text{base}}{\text{hypotenuse}}$ is only valid for right angle triangle.

So for giving the general definition, we draw a circle of radius $r$. A point $P(x,y)$ is taken on circle and radius vector $OP$ is drawn where $O$ is origin. Angle which radius vector makes with positive $X$ axis is considered to be $\theta$, then $\cos\theta=\dfrac{\text{ $x$ coordinates of point }}{\text{radius of circle}}$.

But this definition would only be valid when we have a coordinate plane, what if there is no coordinate plane. Suppose only a triangle is given to you and its angles and then you are asked calculate $\cos\theta,\sin\theta$ etc.

Also I don't understand the significance of definition of $\cos\theta$ in coordinate plane i.e $\cos\theta=\dfrac{\text{ $x$ coordinates of point }}{\text{radius of circle}}$. Please help in understanding this?

user3290550
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3 Answers3

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It depends on context. Acute triangles are positive in nature cause they're basic and we can use them as the standard. In the coordinate plane that corresponds with what is taught to be the first quadrant.

Therefore, for higher angles, they have some difference as compared to acute angles.

For example, how do you use this formula for area of a triangle $\frac{1}{2}ab\sin \theta$ if $\theta$ is obtuse? How do you calculate the height of the triangle then?

In that case, we can adapt our definition of trigonometric ratios to suit these differences. With the first quadrant gaining all the positivity of coordinates, the axiom set is for positive rotations to be anticlockwise.

Here's how the trigonometric ratios were defined for us officially:

$\sin \theta$ is the vertical displacement caused by a rotation of $\theta$ relative to a corresponding point on the unit circle.

Basically if the rotation $\theta$ brings you from the official point of rest at $(1,0)$ to a point on the unit circle, then the sine is encoded in the vertical displacement caused.

Same for the cosine.

$\cos \theta$ is the horizontal displacement caused by a rotation of $\theta$ relative to a corresponding point on the unit circle.

Check out the most upvoted answer here too.

Visually stunning math concepts which are easy to explain

Nεo Pλατo
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Here is an intuitive description of the trigonometric functions of a real number:

  • First consider the unit circle centred at the origin of axes in a cartesian coordinates system, and choose as the origin on this circle the point $A=(1,0)$. We wind up (a copy of) the real axis around the circle, making its origin $0$ coincide with $A$, counterclockwidse for the positive part, clockwise for the negative part of the axis.
  • A real number $a$ then coincides with a point $M$ on the unit circle. By definition, the abscissa of its projection $H$ on the $x$-axis is $\sin a$, the ordinate of its projection on the $y$-axis is $\cos a$.

  • These definitions coincide with the old definitions: indeed, in the triangle $OHM$, the radian measure of angle $\widehat{HOM}$ is $a$, and its cosine in the old sense is $\dfrac{\overline{OH}}{OM}=\overline{OH}$. Similarly, its sine in the old sense is $\dfrac{\overline{HM}}{OM}=\overline{HM}=\overline{OK}$.

  • There are similar definitions for the tangent and cotangent.

I hope this will help.

Bernard
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Two powerful tools you can use in a general triangle are sine and cosine law. In facticular:

A) In a triangle $ABC$, we have: $$\frac{a}{\sin(\alpha)}=\frac{b}{\sin(\beta)}=\frac{c}{\sin(\gamma)}$$ B) In a triangle $ABC$, we have: $$\cos(\alpha)=\frac{\overline{AB}^2+\overline{AC}^2-\overline{BC}^2}{2\overline{AB}\cdot\overline{AC}}$$

Matteo
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