Trigonometric Functions

NZ Level 8 (NZC) Level 3 (NCEA) [In development]

Identifying quadrants from given conditions

Lesson

Consistent with the periodic nature of the trigonometric functions, we find that as the domain variable rotates from $0$0 to $2\pi,$2π, each function takes on positive values in two quadrants and negative values in two quadrants. This is explained more fully in the chapter on the Unit Circle definitions of sine, cosine and tangent.

Hence, if we are given a function value only, it is not possible to know which of two possible domain values led to that range value unless further information is available. For example, if $\sin\theta=\frac{1}{\sqrt{2}}$`s``i``n``θ`=1√2, it is not possible to tell whether $\theta=\frac{\pi}{4}$`θ`=π4 or $\theta=\frac{3\pi}{4}$`θ`=3π4. However, if we are given the additional piece of information that $\cos\theta<0$`c``o``s``θ`<0, then the angle must be in the second quadrant, since this is the only one in which sine is positive and cosine is negative.

In which quadrant is the angle $\frac{7\pi}{6}$7π6 and what are the values of $\sin\frac{7\pi}{6}$`s``i``n`7π6, $\cos\frac{7\pi}{6}$`c``o``s`7π6 and $\tan\frac{7\pi}{6}$`t``a``n`7π6 in terms of a first quadrant argument?

Since $\pi<\frac{7\pi}{6}<\frac{3\pi}{2}$π<7π6<3π2, the angle is in the third quadrant.

Hence,

$\sin\frac{7\pi}{6}$sin7π6 |
$=$= | $-\sin\frac{\pi}{6}$−sinπ6 |

$\cos\frac{7\pi}{6}$cos7π6 |
$=$= | $-\cos\frac{\pi}{6}$−cosπ6 |

$\tan\frac{7\pi}{6}$tan7π6 |
$=$= | $\tan\frac{\pi}{6}$tanπ6 |

If $\sec\theta=-\sqrt{2}$`s``e``c``θ`=−√2 and $\tan\theta=-1$`t``a``n``θ`=−1, what is $\theta$`θ`?

We have $$. Hence, $\cos\theta=-\frac{1}{\sqrt{2}}$`c``o``s``θ`=−1√2 and so, $\theta=\frac{3\pi}{4}$`θ`=3π4 or $\theta=\frac{5\pi}{4}$`θ`=5π4. The fact that $\tan\theta=-1$`t``a``n``θ`=−1 implies that $\theta=\frac{3\pi}{4}$`θ`=3π4 or $\theta=\frac{7\pi}{4}$`θ`=7π4. On comparing these results we see that the solution is $\theta=\frac{3\pi}{4}$`θ`=3π4.

If $\theta$`θ` is an angle such that $\sin\theta$`s``i``n``θ`$>$>$0$0 and $\cos\theta$`c``o``s``θ`$<$<$0$0, which quadrant(s) does it lie in?

quadrant $I$

`I`Aquadrant $II$

`I``I`Bquadrant $IV$

`I``V`Cquadrant $III$

`I``I``I`Dquadrant $I$

`I`Aquadrant $II$

`I``I`Bquadrant $IV$

`I``V`Cquadrant $III$

`I``I``I`D

If $\theta$`θ` is an angle such that $\tan\theta$`t``a``n``θ`$>$>$0$0 and $\cot\left(\theta\right)$`c``o``t`(`θ`)$>$>$0$0, which quadrant(s) does it lie in?

quadrant $IV$

`I``V`Aquadrant $III$

`I``I``I`Bquadrant $II$

`I``I`Cquadrant $I$

`I`Dquadrant $IV$

`I``V`Aquadrant $III$

`I``I``I`Bquadrant $II$

`I``I`Cquadrant $I$

`I`D

If $P$`P`$\left(x,y\right)$(`x`,`y`) corresponds to a point on the unit circle at a rotation of $\left(\theta\right)$(`θ`) such that $\sin\theta=\frac{3}{5}$`s``i``n``θ`=35 and $\cos\theta=-\frac{4}{5}$`c``o``s``θ`=−45, which quadrant is $\left(x,y\right)$(`x`,`y`) located in?

quadrant $II$

`I``I`Aquadrant $IV$

`I``V`Bquadrant $III$

`I``I``I`Cquadrant $I$

`I`Dquadrant $II$

`I``I`Aquadrant $IV$

`I``V`Bquadrant $III$

`I``I``I`Cquadrant $I$

`I`D

Form and use trigonometric, polynomial, and other non-linear equations

Apply trigonometric methods in solving problems