Identifying quadrants from given conditions

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 $\sin\theta=\frac{1}{\sqrt{2}}$sinθ=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 $\cos\theta<0$cosθ<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.

Example 1

In which quadrant is the angle $\frac{7\pi}{6}$7π6​ and what are the values of $\sin\frac{7\pi}{6}$sin7π6​, $\cos\frac{7\pi}{6}$cos7π6​ and $\tan\frac{7\pi}{6}$tan7π6​ 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​

Example 2

If $\sec\theta=-\sqrt{2}$secθ=−√2 and $\tan\theta=-1$tanθ=−1, what is $\theta$θ?

We have $$. Hence, $\cos\theta=-\frac{1}{\sqrt{2}}$cosθ=−1√2​ and so, $\theta=\frac{3\pi}{4}$θ=3π4​ or $\theta=\frac{5\pi}{4}$θ=5π4​. The fact that $\tan\theta=-1$tanθ=−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​.

More Worked Examples

QUESTION 1

QUESTION 2

QUESTION 3