Easy Angles

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$30^{\circ }$, $45^{\circ }$, $60^{\circ }$, $90^{\circ }$

Sine and cosine are transcendental functions. They transcend Algebra. They are beyond our algebraic tools. That makes equations difficult when they involve sine and cosine.

This is true unless you work with angles that just happen to have nice values for sine and cosine. We have several: $30^{\circ }$, $45^{\circ }$, and $60^{\circ }$.

$\sin (30^{\circ }) = \frac {1}{2}$
$\cos (30^{\circ }) = \frac {\sqrt {3}}{2}$

And, since $30^{\circ }$ and $60^{\circ }$ make up a right triangle, we have

$\sin (60^{\circ }) = \frac {\sqrt {3}}{2}$
$\cos (60^{\circ }) = \frac {1}{2}$

$45^{\circ }$ cuts the quadrant in half making sine and cosine equal.

$\sin (45^{\circ }) = \frac {1}{\sqrt {2}}$
$\cos (45^{\circ }) = \frac {1}{\sqrt {2}}$

Add these to $0^{\circ }$, $90^{\circ }$, $180^{\circ }$, and $270^{\circ }$ and we can walk around the unit circle.

Analyze $f(t) = 3 \sin (2t - \pi ) - 2$

$\blacktriangleright $ One period

Start: $2t - \pi = 0$, which occurs at $t = \frac {\pi }{2}$where $f = 3 \cdot 0 - 2 = -2$

Finish: $2t - \pi = 2 \pi $, which occurs at $t = \frac {3 \pi }{2}$, where $f = 3 \cdot 0 - 2 = -2$

Period = $\frac {3 \pi }{2} - \frac {\pi }{2} = \pi $

Halfway between $\frac {\pi }{2}$ and $\frac {3\pi }{2}$ is $\pi $.

$\sin (2t - \pi )$ equals $0$ at $\pi $, where $f = 3 \cdot 0 - 2 = -2$.

A quarter of the way from Start to Finish is $\frac {3 \pi }{4}$, where $f = 3 \cdot 1 - 2 = 1$

The three-quarter mark is $\frac {5 \pi }{4}$, where $f = 3 \cdot -1 - 2 = -5$

$\blacktriangleright $ Then, extend periodically

We can calculate some additional exact values at the easy angles.

Our principle interval was $\left [ \frac {\pi }{2}, \frac {3\pi }{2} \right ]$, which has a length of $\pi $.

$\blacktriangleright $ $30^{\circ } = \frac {\pi }{6}$ is $\frac {1}{12}$ of the normal basic interval $[0, 2\pi ]$.

$\frac {1}{12}$ of our period here would be $\frac {1}{12} \cdot \pi = \frac {\pi }{12}$.

Therefore, if we move $\frac {\pi }{12}$ from $\frac {\pi }{2}$, then we should be at the angle where we normally get $\sin (\frac {\pi }{6}) = \frac {1}{2}$.

$\frac {\pi }{2} + \frac {\pi }{12} = \frac {7\pi }{12}$.

Let’s evaluate $f\left ( \frac {7\pi }{12} \right )$ and see if we get $\frac {\pi }{6}$ inside the sine function.

\[ f\left ( \frac {7\pi }{12} \right ) = 3 \sin \left (2 \left (\frac {7\pi }{12} \right ) - \pi \right ) - 2 = 3 \sin \left ( \frac {2\pi }{12} \right ) - 2 = 3 \sin \left ( \frac {\pi }{6} \right ) - 2 = 3 \cdot \frac {1}{2} - 2 = -\frac {1}{2} \]

$\blacktriangleright $ $60^{\circ } = \frac {\pi }{3}$ is $\frac {1}{6}$ of the normal basic interval $[0, 2\pi ]$.

$\frac {1}{6}$ of our period here would be $\frac {1}{6} \cdot \pi = \frac {\pi }{6}$.

Therefore, if we move $\frac {\pi }{6}$ from $\frac {\pi }{2}$, then we should be at the angle where we get $\sin (\frac {\pi }{3}) = \frac {\sqrt {3}}{2}$.

$\frac {\pi }{2} + \frac {\pi }{6} = \frac {4\pi }{6} = \frac {2\pi }{3}$.

Let’s evaluate $f\left ( \frac {2\pi }{3} \right )$ and see if we get $\frac {\pi }{3}$ inside the sine.

\[ f\left ( \frac {2\pi }{3} \right ) = 3 \sin \left (2 \left (\frac {2\pi }{3} \right ) - \pi \right ) - 2 = 3 \sin \left ( \frac {\pi }{3} \right ) - 2 = 3 \cdot \frac {\sqrt {3}}{2} - 2 = \frac {3\sqrt {3}-4}{2} \approx 0.5980762114 \]

Sine and Cosine

Sine and Cosine are basically shifts of each other. They follow the same periodic patterns.

They both oscillate between $-1$ and $1$, which makes their range $[-1,1]$.

They both have their maximum value of $1$ and their minimum value of $-1$ at the top, bottom, or sides of the unit circle.

When one has maximum or minimum, the other has a zero.

When we analyze sine and cosine functions, we usually start by locating when the inside is $0$, $\frac {\pi }{2}$, $\pi $, $\frac {3\pi }{2}$, or $2\pi $. This tells us when the function is experiencing a maximum, minimum, or zero.

Then we move to the easy angles ($\frac {\pi }{6}$, $\frac {\pi }{4}$, and $\frac {\pi }{3}$) to round out the shape.

$\blacktriangleright $ This is how we analyze compositions involving sine and cosine.

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more examples can be found by following this link
More Examples of Easy Angles

2025-05-17 22:59:12