# Unit Circles and Trigonometric Equations

The angles in a right-angled triangle are always acute, and basic trigonometry covers these acute triangles. However, angles than 90° are used in many situations, such as bearings. Negative angles are also used in areas such as engineering and science.

Drawing the graphs of trigonometric ratios can help us to see the change in signs as angles increase. We divide the domain 0° to 360° into 4 quadrants:

If we divide the circle into 4 quadrants, we notice that the x and y values have different signs in different quadrants. This is crucial to notice when looking at angles of any magnitude and explains the different signs you get when finding sin, cos and tan for angles greater than 90°.

Let (x, y) be a point on the unit circle.

Then $x=\cos{\theta}$  and $y=\sin{\theta}$.

Looking at the first quadrant, notice that x and y are both positive and that angle θ is turning anticlockwise from the x-axis. Point P(x, y) forms a triangle with sides 1, x and y, so we can find the trigonometric ratios for angle θ.

$\sin { \theta } =\cfrac { y }{ 1 } \\ \\ \cos { \theta } =\cfrac { x }{ 1 } \\ \\ \tan { \theta } =\cfrac { y }{ x }$

In the second quadrant, the angle around the circle is 180°-θ. We notice that all x values are negative and y values are positive. Hence, the point Q has the coordinates (-xy).

$\sin { (180^{ \circ }-\theta ) } =\sin { \theta } \\ \cos { (180^{ \circ }-\theta ) } =-\cos { \theta } \\ \tan { (180^{ \circ }-\theta ) } =-\tan { \theta }$

In the third quadrant, the angle around the circle is 180°+θ. We notice that the x and y values are both negative. Hence, the point R has the coordinates (-x, -y).

$\sin { (180^{ \circ }+\theta ) } =-\sin { \theta } \\ \cos { (180^{ \circ }+\theta ) } =-\cos { \theta } \\ \tan { (180^{ \circ }+\theta ) } =\tan { \theta }$

In the fourth quadrant, the angle around the circle is 360°-θ. We notice that the x value is positive, however the y value is negative. Therefore, the point S has the coordinates (x, -y).

$\sin { (360^{ \circ }-\theta ) } =-\sin { \theta } \\ \cos { (360^{ \circ }-\theta ) } =\cos { \theta } \\ \tan { (360^{ \circ }-\theta ) } =-\tan { \theta }$

To remember the four quadrants implications on  the positive and negative values, easily remember All Stations To Central (ASTC). The first quadrant's values of sin, cos and tan are all positive. While the second quadrant, only the sin values are positive, while the others are negative. In the third quadrant, the tan values are positive, while the others are negative. Lastly, the fourth quadrant is where the cos values are positive while the others are negative.

### Examples:

1. Find all the values of x between $0 < x < 2 \pi$ for $\sin { x } =\cfrac { 1 }{ \sqrt { 2 }}$

The idea is to find what are the values of x that satisfies the sin equation to obtain $\cfrac{ \pi }{ 4 }$.

We know by the simple trigonometric equation:

$x=\sin ^{ -1 }{ \cfrac { 1 }{ \sqrt { 2 } } }\\ \\ x=\cfrac { \pi }{ 4 }$

However, this only applies to angles between 0° to 90° Are there more values of x that gives $\cfrac { 1 }{ \sqrt { 2 }}$ between 90° to 360°?

Using the relation:

$\sin { (180^\circ-\theta )}=\sin { \theta } \\ \sin { (180^\circ-45^\circ) }=\sin { 45^\circ } \\ \sin { 135^\circ}= \sin { 45^\circ }$

Remember the Unit Circle, ASTC, and notice that all sin, cos and tan values are all positive in the first quadrant. Therefore all the angles in the first quadrant is acceptable. In addition, values of sin function in the second quadrant is also positive. The y-value is positive and therefore the value of sin function would also be positive.

Therefore, the possible values of $\sin { x } =\cfrac { 1 }{ \sqrt { 2 } }$ is:

$x=\cfrac { \pi }{ 4 } \ and \ x=\cfrac { 3 \pi }{ 4 }$

2. Find all the values of x between $0 < x < 2 \pi$ for

$\cfrac { 1+\sin { x } }{ \cos { x } } +\cfrac { \cos { x } }{ 1+\sin { x } } =4$

We can simplify the using identities:

Now that the equation is simplified:

$\cos { x } =\cfrac { 1 }{ 2 } \\ \\ x=\cos ^{ -1 }{ \cfrac { 1 }{ 2 } } \\ \\ x=60^\circ$

Again, since we are looking for the values of x between $0 < x < 2 \pi$, we refer back to ASTC, where the values of cosine is positive in the fourth quadrant.

We can also use the relation:

$\cos { (360^{ \circ }-\theta ) } =\cos { \theta } \\ \cos { (360^{ \circ }-60^{ \circ }) } =\cos { 60^{ \circ } } \\ \cos { 300^{ \circ } } =\cos { 60^{ \circ } } \\ \\ \therefore \quad the\quad values\quad of\quad x=60^{ \circ } \ and\quad 300^{ \circ }$

Watch this video to learn how to solve Trigonometric Equations: