The trigonometric functions are all related to each other. So, if we get the values of sine and cosine, we will be able to use them to calculate all the others.
First, let's put the point in a circle, at least approximately.
Since th x-value is negative and the y-value is positive, the point will be in the second quadrant:
The important information we get from that is that the sine value will be positive and the cosine value will be negative.
But, as we can see, we can use a right triangle with legs equal to 5 and 12 units to calculate it, and then we put the appropiate sign:
As we can see, this is a mirrored situation of what would be inthe first quadrant and the marked angle is the supplementary pair of θ.
Now, first we calculate the hypotenuse using the Pytagora's Theorem:
[tex]h=\sqrt[]{5^2+12^2}=\sqrt[]{25+144}=\sqrt[]{169}=13[/tex]
Thus, the sine of "a" is:
[tex]\sin a=\frac{opposite\, leg}{hypotenuse}=\frac{12}{13}[/tex]
And since we saw that this is in the second quadrant, so will be the sine of θ:
[tex]\sin \theta=\sin a=\frac{12}{13}[/tex]
And the cosine of "a" will be:
[tex]\cos a=\frac{adjacent\, leg}{hypotenuse}=\frac{5}{13}[/tex]
So, the cosine of θ, since it is in the second quadrant, will be the same but with the negative sign:
[tex]\cos \theta=-\cos a=-\frac{5}{13}[/tex]
With these values, we can find the others'.
Tangent of an angle is its sine divided by its cosine, so:
[tex]\tan \theta=\frac{\sin\theta}{\cos\theta}=\frac{\frac{12}{13}}{-\frac{5}{13}}=-\frac{12}{13}\cdot\frac{13}{5}=-\frac{12}{5}[/tex]
The cosecant of an angle is the inverse of its sine:
[tex]\csc \theta=\frac{1}{\sin \theta}=\frac{1}{\frac{12}{13}}=\frac{13}{12}[/tex]
The secant of an angle is the inverse of its cosine:
[tex]\sec \theta=\frac{1}{\cos \theta}=\frac{1}{-\frac{5}{13}}=-\frac{13}{5}[/tex]
And the cotangent of an angle is the inverse of its tangent:
[tex]\cot \theta=\frac{1}{\tan \theta}=\frac{1}{-\frac{12}{5}}=-\frac{5}{12}[/tex]
