We will start this lesson off with some definitions of basic trigonometric functions.
The Sine
The sine of an angle \(\theta\) is written as \(\sin(\theta)\) and is the distance from the intersection of the radius of the unit circle that has been rotated \(\theta\) counter clockwise about the origin from the x-axis and the diameter of the unit circle to the x-axis (as can be seen in the diagram above). Note that when this altitude to the x-axis is below the x-axis the sine of the angle is negative. When \(\theta\) is between \(0^{\circ}\) and \(90^{\circ}\) or \(0 rad\) and \(\frac{\pi}{2} rad\) (if you don't know what the radian is you should look it up because it is used very commonly), then \(\sin(\theta)\) is positive. In addition, when \(\theta\) is between \(0^{\circ}\) and \(90^{\circ}\), \(\sin(\theta)\) can be viewed in the context of a right triangle as the ratio of the length side opposite the angle to the length of the hypotenuse (think about how the radius of the unit circle is the hypotenuse of the triangle in the first definition and how from there we can scale it up for larger hypotenuses without changing the value of the sine).
The Cosine
The cosine of an angle \(\theta\) is written as \(\cos(\theta)\) and is the distance from the intersection of the radius of the unit circle that has been rotated \(\theta\) counter clockwise about the origin from the x-axis and the diameter of the unit circle to the x-axis (as can be seen in the diagram above). Similar to the sine, the cosine is negative when the point is to the left of the y axis (i.e. for \(90^{\circ}<\theta<180^{\circ}\)). In addition, for angles angles between \(0^{\circ}\) and \(90^{\circ}\), the cosine can be seen in the context of a right triangle as the ratio of the lengths of the side adjacent to the angle over the hypotenuse of the triangle (again, think about scaling up the unit circle).
The Tangent
The tangent of an angle \(\theta\) is written as \(\tan(\theta)\) and is the length of the line segment perpendlicar to the radius of the unit circle that has been rotated \(\theta\) counter clockwise about the origin from the x-axis with endpoints on the x-axis and the intersection of the aforementioned radius and the unit circle's diameter. The tangent is negative when exactly one of the sine cosine is negative. The tangent can also be seen as \(\frac{\sin{\theta}}{\cos{\theta}}\). Thinking about the right triangle definitions of sine and cosine, we can get that for angles between \(0^{\circ}\) and \(980^{\circ}\), the tangent in a right triangle is equal to the ratio of the side opposite the angle to the side adjacent to the angle.
There are a few other trigonometric functions and other ways to define the sine, cosine, and tangent, but these are far beyond the scope of the AMC 10.
There are a few important trigonometric identities and values that you need to know. This list will not include the sum and product identities because they are not needed on the AMC 10, but you should look them up if you are interested: they provide proofs for some trigonometric ratios that you should memorize. Also note that the square of a trigonometric value is placed before the parentheses. For example, we write \(\sin^2(\theta)\) and not \(\sin(\theta)^2\).
The sine of an angle \(\theta\) is written as \(\sin(\theta)\) and is the distance from the intersection of the radius of the unit circle that has been rotated \(\theta\) counter clockwise about the origin from the x-axis and the diameter of the unit circle to the x-axis (as can be seen in the diagram above). Note that when this altitude to the x-axis is below the x-axis the sine of the angle is negative. When \(\theta\) is between \(0^{\circ}\) and \(90^{\circ}\) or \(0 rad\) and \(\frac{\pi}{2} rad\) (if you don't know what the radian is you should look it up because it is used very commonly), then \(\sin(\theta)\) is positive. In addition, when \(\theta\) is between \(0^{\circ}\) and \(90^{\circ}\), \(\sin(\theta)\) can be viewed in the context of a right triangle as the ratio of the length side opposite the angle to the length of the hypotenuse (think about how the radius of the unit circle is the hypotenuse of the triangle in the first definition and how from there we can scale it up for larger hypotenuses without changing the value of the sine).
The Cosine
The cosine of an angle \(\theta\) is written as \(\cos(\theta)\) and is the distance from the intersection of the radius of the unit circle that has been rotated \(\theta\) counter clockwise about the origin from the x-axis and the diameter of the unit circle to the x-axis (as can be seen in the diagram above). Similar to the sine, the cosine is negative when the point is to the left of the y axis (i.e. for \(90^{\circ}<\theta<180^{\circ}\)). In addition, for angles angles between \(0^{\circ}\) and \(90^{\circ}\), the cosine can be seen in the context of a right triangle as the ratio of the lengths of the side adjacent to the angle over the hypotenuse of the triangle (again, think about scaling up the unit circle).
The Tangent
The tangent of an angle \(\theta\) is written as \(\tan(\theta)\) and is the length of the line segment perpendlicar to the radius of the unit circle that has been rotated \(\theta\) counter clockwise about the origin from the x-axis with endpoints on the x-axis and the intersection of the aforementioned radius and the unit circle's diameter. The tangent is negative when exactly one of the sine cosine is negative. The tangent can also be seen as \(\frac{\sin{\theta}}{\cos{\theta}}\). Thinking about the right triangle definitions of sine and cosine, we can get that for angles between \(0^{\circ}\) and \(980^{\circ}\), the tangent in a right triangle is equal to the ratio of the side opposite the angle to the side adjacent to the angle.
There are a few other trigonometric functions and other ways to define the sine, cosine, and tangent, but these are far beyond the scope of the AMC 10.
There are a few important trigonometric identities and values that you need to know. This list will not include the sum and product identities because they are not needed on the AMC 10, but you should look them up if you are interested: they provide proofs for some trigonometric ratios that you should memorize. Also note that the square of a trigonometric value is placed before the parentheses. For example, we write \(\sin^2(\theta)\) and not \(\sin(\theta)^2\).
- \(\sin(60^\circ)=\frac{\sqrt{3}}{2}\)
- \(\sin(30^\circ)=\frac{1}{2}\)
- \(\sin(30^\circ)=\frac{1}{2}\)
- \(\sin(45^\circ)=\frac{1}{\sqrt{2}}\)
- \(\sin(15^\circ)=\frac{\sqrt{6}-\sqrt{2}}{4}\)
- \(\sin(75^\circ)=\frac{\sqrt{6}+\sqrt{2}}{4}\)
- \(\sin(\theta)=\sin(180^\circ-\theta)=-\sin(-\theta)\) (should be apparent if you think of the unit circle)
- \(\cos(\theta)=-\cos(180^\circ-\theta)=\cos(-\theta)\) (\(\cos(-\theta)=\cos(360^\circ -\theta), \text{this identity is apparent if you think of the unit circle}\))
- \(\sin(\theta)=\cos(90^{\circ}-\theta)\) (\(\cos(-\theta)=\cos(360^\circ -\theta)\))
- \(\sin^2(\theta)+\cos^2(\theta)=1\)
- \(\sin(180^\circ + \nu)=-\sin(\nu)\)
- \(\cos(180^\circ +\kappa)=-\cos(\kappa)\)
Sample Problem 1. Prove that \(\sin(\theta)=\cos(90^{\circ}-\theta)\) .
For this problem, when we seen that one angle is \(90^\circ\) minus another, we think of the angles of a right triangle, especially because right triangles have so much to do with trigonometry. By looking at a right triangle and seeing how two two non-right angles add to \(90^\circ\), we can easily confirm this identity to be true for \(0^\circ<\theta<90^\circ\) (\(\cos(\gamma)=\frac{CA}{CB}=\sin(\beta)\)). We will proceed by casework on the quadrant that angle \(\theta\) is in. If it is in the second quadrant, then we can say \(\theta = 90^\circ +\chi\) for \(0^\circ<\chi<90^\circ\). Thus, \(\cos{90^\circ-(90^\circ +\chi)}=\cos(-\chi)=-\cos(\chi)=-\sin(90^\circ -\chi)=-(-\sin(180^\circ - (90^\circ - \chi))=\sin(90^\circ+\chi))\). For the third and fourth quadrants, note that \(\cos(90^\circ - (180^\circ + \theta))=\cos(-90^\circ -\theta)=\cos(90^\circ + \theta)=-\cos(180^\circ - (90^\circ + \theta))=-\cos(90^\circ -\theta)=-\sin(\theta)=\sin(180^\circ + \theta)\). Thus, by building on each previous result, we are done.
Now, you may be thinking that it this is nice and all and will help find lengths in right triangles, but how often do we really use right triangles? Well, we do use right triangles pretty often, but we can extend this beyond right triangles. We can use trigonometry to help us in any triangle.
Sample Problem 2. In triangle ABC let BC=a, AC=b, AB=C, \(m\measuredangle BAC=A\), \(m\measuredangle ABC=B\), \(m\measuredangle ACB=C\), and \(R\) be the radius of the circumscribed circle. Prove that \(\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}=2R\).
Knowing that we are dealing with right triangles and that we are dealing with a circumradius, we will want to draw the circumcircle of our triangle (remember that because any angle is half the arc it intersects, any angle that intersects the endpoints of a diameter will be right). There doesn't appear to be any reason to to draw the diameter through any vertex of the triangle as opposed to another one, so we try drawing the diameter through the middle vertex. We connect the point the point on the opposite endpoint of the diameter we just drew to one of the vertices because we then get an angle intersecting the same arc as one angle of our triangle does. From here we can see that \(m\measuredangle C = m\measuredangle AGB\). We can also see that \(m\measuredangle GBA=90^\circ\) because the angle intersects the diameter of our circle. Thus, we can get \(\sin(C)=\sin(\measuredangle AGB)=\frac{c}{GA}=\frac{c}{2R}\implies \frac{c}{\sin(C)}=2R\).
For any acute triangle, we would simply repeat this process for each angle to get the desired result. However, because we have an obtuse triangle, we need to do something different for the obtuse angle, \(\angle A\). What d=we did worked well, so we want to do something similar. We want to have an angle that is at least intersecting the arc opposite the arc \(\angle A\) intersects (using the fact that an inscribed angle is half the measure of the arc intersects we get that the angles add to \(18-^\circ\), and we know that \(\sin(\theta)=\sin(180^\circ -\theta)\)). Similar to how we did last time, we connect the point on hte opposite endpoint of the endpoint we just drew to point \(C\). We again have that angle BCG is a right angle. From here we get that \(\sin(A)=\sin(G)=\frac{a}{2R} \implies \frac{a}{\sin(A)}=2R\). This theorem is trivial to prove for right triangles. Thus, we are done. This relationship we just derived is called the Extended Law of Sines.
For any acute triangle, we would simply repeat this process for each angle to get the desired result. However, because we have an obtuse triangle, we need to do something different for the obtuse angle, \(\angle A\). What d=we did worked well, so we want to do something similar. We want to have an angle that is at least intersecting the arc opposite the arc \(\angle A\) intersects (using the fact that an inscribed angle is half the measure of the arc intersects we get that the angles add to \(18-^\circ\), and we know that \(\sin(\theta)=\sin(180^\circ -\theta)\)). Similar to how we did last time, we connect the point on hte opposite endpoint of the endpoint we just drew to point \(C\). We again have that angle BCG is a right angle. From here we get that \(\sin(A)=\sin(G)=\frac{a}{2R} \implies \frac{a}{\sin(A)}=2R\). This theorem is trivial to prove for right triangles. Thus, we are done. This relationship we just derived is called the Extended Law of Sines.
Another relationship we can get using trigonometry is that, using the same labeling system as the previous problem, \(c^2=a^2+b^2-2ab\cos(C)\). This is called the Law of Cosines. We will not prove this theorem, but we encourage you to try to prove this on your own (Hint: What theorem does this look like? What do we always often use in trigonometry?) or to look up a proof.
Exercises
- Prove that \(\sin^2(\zeta)+\cos^2(\zeta)=1\).
- Prove the values given for the sine of \(60^\circ\) and \(30^\circ\).
Solution 1
First of all, we don't need to worry about the signs of the sines because we are squaring them. Next, this theorem reminds us of the Pythagorean Theorem. In addition, we saw in the first diagram that the sine and cosine represent lengths in a right triangle in the unit circle. However, through this observation we are done! The hypotenuse of the right triangle seen is the first diagram is the radius of the unit circle, and thus it has length of 1. Thus, for any angle, we will get the desired equality simply by using the Pythagorean Theorem and the unit circle definitions for sine and cosine.
Solution 2
See the values that we saw earlier and angles that are related to \(60^\circ\), we think equilateral triangles. We will draw the equilateral triangle with side length 2. However, this doesn't get us a \(30^\circ\) angle or an altitude. Thus, as we know that the altitude from a vertex of an equilateral triangle is also an angle bisector, we draw an altitude. From there, we have both of the desired angles, so we just need to find CD. We can do this easily using the Pythagorean Theorem to get \(CD^+1=4\implies CD=\sqrt{3}\). Thus, we get that \(sin(60^\circ = \frac{\sqrt{3}}{2})\) and \(sin(30^\circ = \frac{1}{2})\).