Trigonometry

Trigonometry underlies many modern principles and engineering methodologies at a level so fundamental that it probably goes completely unnoticed in your life. Fourier series and transforms are used every time you listen a piece of recorded music or watch a video on your phone or computer. If you want to model any periodic activity, trigonometric functions work much better than polynomial functions.

1. Defining our Trigonometric Functions.

All of our standard trigonometric functions are defined in terms of a triangle and the standard cartesian plan. Recall from geometry that any pair of similar triangles has sides with a constant ratio and identical angles.

x. Graphing Trigonometric Functions

x +1. Trigonometric Identities

The Pythagorean identities will be the most commonly used identities. Recall the pythagorean theorem: $x^2 + y^2 = r^2$ where x and y are legs of a right triangle and r is the hypotenuse. Now, if we divide through by $r^2$ we have $\frac{x^2}{r^2} + \frac{y^2}{r^2} = 1$. Recall the definition of sine and cosine. $sin(\theta) = \frac{y}{r}$ and $cos(\theta) = \frac{x}{r}$. We use substitution and obtain:

$sin^2(\theta) + cos^2(\theta) = 1$

Similarly, we divide through the above identity by $sin^2(\theta)$ to obtain
$1 + cot^2(\theta) = csc^2(\theta)$
and by $cos(\theta)$ to obtain
$tan^2(\theta) + 1 = sec^2(\theta)$

These three identities will follow us throughout the course of our mathematical career so it is a good idea to commit at least one to memory.

Click Here for Worked Practice Problems

Graphing

1. Graph $1.5cos(2x + 2) – 1$ over two periods.

It is best to approach graphing starting with the argument and moving outwards. First, we will rewrite the argument by factoring:

$1.5cos(2x + 2) – 1 = 1.5cos(2(x + 1)) – 1$

Now, we can read off the most of the transformations directly. The period is $\frac{2\pi}{2} = \pi$ The function is shifted one unit to the left and one unit down. The function is scaled vertically by a factor of 1.5. A typical anchor I like to use for cosine graph’s is (0, 1). If examine what these transformations have done to that point, we can see that our point has undergone $(0 – 1, 1.5(1) -1) = (-1, .5)$. Now, the period has been reduced to $\pi$ so other important points would be $(-1 + \frac{\pi}{4}, -1)$ and $(-1 + \frac{\pi}{2}, -2.5)$. The full graph is given to the right.

2. Given the graph to the right, write a trigonometric equation that describes it. (Answers will vary.)

We have to make a decision about whether to describe this as a secant or cosecant function. It does not matter which you choose. Moreover, you will also get a slightly different answer should you decide on positive or negative leading coefficients. Again, both will work. Let’s make this a positive cosecant function. Now, we have

$A\cdot csc(b(x-h)) + k$

I’ve drawn a red box to capture a number of points about this graph. The horizontal distance is 2 (the graph is not to scale) implying the period is 2. Here, we obtain $\frac{2\pi}{b} = 2 \rightarrow b = \pi$. The point explictly labeled in black is the anchor which has been moved to the left 1 implying h = 1. The amplitude is half the height of the box (the vertical height is labeled in green). A = 3. The entire function is shifted down one. The centerline of the box is given by y = -1. If the function was not vertically shifted it’s center line would be y = 0.

Putting this information together, we obtain
$f(x) = 3csc(\pi(x-1)) – 1$

3. Graph $g(x) = 5tan(\pi x + \pi) + 1$ over two periods. Place at least 3 points on each graph using principal values and determine it’s period.

$g(x) = 5tan(\pi x + \pi) + 1$
$g(x) = 5tan(\pi (x + 1)) + 1$
So the period is $T = \frac{\pi}{\pi} = 1$. The period for tan and cot is $\pi$ not $2\pi$.
The graph is shifted one to the left and one up. Let’s look at this graph within the scope of transformations.

We see the graph of $y = tan(x)$ given to the left. We also see 3 points defined by principal values of the tan function. These points are $(\frac{-\pi}{4}, -1), (0, 0), (\frac{\pi}{4}, 1)$. Recall that it is salutary to think about an anchor point and what happens to that point. We’ll let (0, 0) be that anchor point here. It is shifted to (-1, 1). Note, that the other two points we selected are precisely $\frac{1}{4}$ of a period away from this anchor point. That distance is maintained. The rightmost point becomes $(-1 + \frac{1}{4}, 1 + 1(5)) = (-\frac{3}{4}, 6)$. It was one above our anchor point, so now it is 1 multiplied by the amplitude above our anchor point.

You can see that the red points on the lefthand graph became the green points on the righthand graph under our standard rules of functional transformations.

Identities

3. Verify $(1 + sin(x))(1 – sin(x)) = cos^2(x)$.

$(1 + sin(x))(1 – sin(x)) = $ We expand the righthand side by distributing

$1 + sin(x) – sin(x) – sin^2(x) =$ We cancel the terms with the same sign

$1 \cancel{+ sin(x) – sin(x)} – sin^2(x) =$

$1 – sin^2(x) =$ Finally, we apply substitution using the Pythagorean identity.

$cos^2(x)$

4. Verify $\frac{sin^2(x) – 1}{tan(x)sin(x) – tan(x)} = \frac{sin(x) + 1}{tan(x)}$.

$\frac{sin^2(x) – 1}{tan(x)sin(x) – tan(x)} =$ Here, I see the difference of squares in the numerator so I’m tempted to factor.

$\frac{(sin(x) + 1)(sin(x) – 1)}{tan(x)sin(x) – tan(x)} =$ I can also factor a greatest common factor in the denominator

$\frac{(sin(x) + 1)(sin(x) – 1)}{tan(x)(sin(x) – 1)} =$ Now, I can cancel a factor that appears in the numerator and the denominator

$\frac{(sin(x) + 1)\cancel{(sin(x) – 1)}}{tan(x)\cancel{(sin(x) – 1)}} =$

$\frac{(sin(x) + 1)}{tan(x)}$

5. Verify $(tan^4(\theta)-sec^4(\theta))cos^2(\theta) = cos^2(\theta) – 2 $

Below we will apply two identities: the difference of perfect squares $x^2 – y^2 = (x – y)(x + y)$ and the Pythagorean identity

$1 + tan^2(\theta) = sec^2(\theta) \rightarrow$

$1 = sec^2(\theta) – tan^2(\theta) \rightarrow$

$-1 = tan^2(\theta) – sec^2(\theta)$

and

$sin^2(\theta) + cos^2(\theta) = 1 \rightarrow$

$sin^2(\theta) = 1 – cos^2(\theta)$

$(tan^4(\theta)-sec^4(\theta))cos^2(\theta) = $

$(tan^2(\theta)-sec^2(\theta))(tan^2(\theta)+sec^2(\theta))cos^2(\theta) = $

$(-1)(tan^2(\theta)+sec^2(\theta))cos^2(\theta) = $

$(-1)(sin^2(\theta)+1) = $

$(-1)(1 – cos^2(\theta)+1) = $

$(-1)(2 – cos^2(\theta)) = cos^2(\theta) – 2$

6. Verify $sin^8(\theta) – cos^8(\theta) = (sin^4(\theta) + cos^4(\theta))(sin^2(\theta) – cos^2(\theta))$

$sin^8(\theta) – cos^8(\theta) = $

$(sin^4(\theta) + cos^4(\theta))(sin^4(\theta) – cos^4(\theta)) = $

$(sin^4(\theta) + cos^4(\theta))(sin^2(\theta) – cos^2(\theta))(sin^2(\theta) + cos^2(\theta)) = $

$(sin^4(\theta) + cos^4(\theta))(sin^2(\theta) – cos^2(\theta))(1) = $

$(sin^4(\theta) + cos^4(\theta))(sin^2(\theta) – cos^2(\theta))$

7. Show that $sin^4(\theta) – cos^4(\theta) = – cos(2\theta)$.

$sin^4(\theta) – cos^4(\theta) = $

$(sin^2(\theta) – cos^2(\theta))(sin^2(\theta) + cos^2(\theta)) =$

$(sin^2(\theta) – cos^2(\theta))(1) = $

$(-1)(cos^2(\theta) – sin^2(\theta)) = $

$(-1)cos(2\theta) = -cos(2\theta)$

8. Show that $\frac{1 – sin(x)}{1 + sin(x)} = (sec(x) – tan(x))^2$

$\frac{1 – sin(x)}{1 + sin(x)} = $

$\frac{1 – sin(x)}{1 + sin(x)}\cdot\frac{1 – sin(x)}{1 – sin(x)} = $

$\frac{(1 – sin(x))^2}{1 – sin^2(x)} = $

$\frac{(1 – sin(x))^2}{cos^2(x)} = $

$\left(\frac{(1 – sin(x))}{cos(x)}\right)^2 = $

$\left( \frac{1}{cos(x)}-\frac{sin(x)}{cos(x)} \right)^2 = $

$(sec(x)-tan(x))^2 $

9. Show that $cos^4(\theta) = \frac{cos(4x)}{8} + \frac{cos(2x)}{2} + \frac{3}{8}$

$cos^4(\theta) = $

$cos^2(\theta)cos^2(\theta) = $

$\frac{cos(2x)+1}{2}\cdot\frac{cos(2x)+1}{2} = $

$\frac{cos^2(2x)+2cos(2x)+1}{4} = $

$\frac{cos^2(2x)}{4} + \frac{2cos(2x)+1}{4} = $

$\frac{\frac{cos(4x)+1}{2}}{4} + \frac{2cos(2x)+1}{4} = $

$\frac{cos(4x)+1}{8} + \frac{2cos(2x)+1}{4} = $

$\frac{cos(4x)}{8} + \frac{cos(2x)}{2} +\frac{3}{8}$

10. Show that $cos^2(x)sin^2(x) = \frac{cos(4x)}{8} + cos(2x) +\frac{7}{8}$.

$cos^2(x)sin^2(x) = $

$cos^2(x)(1-cos^2(x)) = $

$cos^2(x)-cos^4(x) = $

and here we’ll use the solution to problem 4 to facilitate a quicker answer

$\frac{cos(2x)+1}{2} – \frac{cos(4x)}{8} – \frac{cos(2x)}{2} -\frac{3}{8}$

$\frac{cos(4x)}{8} + cos(2x) +\frac{1}{8}$

11. Calculate $cos(15^0)$ exactly.

There are many ways to obtain a reasonable solution for this problem. I will apply the half angle identity.

$cos(15^0) = $

$\sqrt{\frac{cos(30^0) + 1}{2}} = $

$\sqrt{\frac{\frac{\sqrt{3}}{2} + 1}{2}} = $

$\frac{\sqrt{\sqrt{3}+2}}{2}$

12. Calculate $sin(7.5^0)$ exactly.

Here, we employ the sin half angle identity coupled with the result from the previous question!

$sin(7.5^0) = $

$\sqrt{\frac{1-cos(15^0)}{2}} = $

$\sqrt{\frac{1-\frac{\sqrt{\sqrt{3}+2}}{2}}{2}} = $

$\frac{\sqrt{2-\sqrt{\sqrt{3}+2}}}{2} = $