Signs of Trigonometric Functions

Signs of Trigonometric Functions in Each Quadrant

We know that cos α = x/r. Since r > 0, the sign of the cosine function depends only on the sign of x. Therefore, the cosine is positive in the 1st and 4th quadrants, and negative in the 2nd and 3rd quadrants.

Signs of cosine and secant functions in each quadrant. — Figure 1.

Consider the sine function \(\sin \alpha = \frac{y}{r}.\) Its sign is determined by the sign of \(y.\) Hence, the sine is positive in the \(1\text{st}\) and \(2\text{nd}\) quadrants, and negative in the \(3\text{rd}\) and \(4\text{th}\) quadrants.

Signs of sine and cosecant functions in each quadrant. — Figure 2.

It is clear that the reciprocal functions \(\sec \alpha = \frac{1}{{\cos \alpha }}\) and \(\csc \alpha = \frac{1}{{\sin \alpha }}\) have the same signs as \(\cos \alpha\) and \(\sin \alpha,\) respectively.

The signs of tangent and cotangent depend on the signs of sine and cosine. The tangent and cotangent are positive when \(x\) and \(y\) are both positive or both negative. This occurs in the \(1\text{st}\) and \(3\text{rd}\) quadrants. In the \(2\text{nd}\) and \(4\text{th}\) quadrants, these functions are negative.

Signs of tangent and cotangent functions in each quadrant. — Figure 3.

We can summarize this information in the following table:

Signs of trigonometric functions — Figure 4.

Evaluating Trigonometric Functions

To find a trigonometric function of any angle, it is convenient to use the concept of reference angle. This involves the following steps:

Determine the reference angle for the given angle in standard position. The reference angle is always acute.
Calculate the trigonometric function value for the reference angle.
Determine the sign of the trigonometric function depending on the quadrant in which the terminal side of the given angle lies.

Example

Calculate the value of \(\cos \frac{{7\pi }}{6}.\)

Solution.

We denote \(\alpha = \frac{{7\pi }}{6}.\) The reference angle of \(\alpha\) is equal to

\[\alpha^\prime = \alpha - \pi = \frac{{7\pi }}{6} - \pi = \frac{\pi }{6}.\]

The cosine of the reference angle can be found from the table:

\[\cos \alpha^\prime = \cos \frac{\pi }{6} = \frac{1}{2}.\]

The initial angle \(\alpha = \frac{{7\pi }}{6}\) lies in the \(3\text{rd}\) quadrant in which the cosine function has a negative sign. Therefore,

\[\cos \alpha = - \cos \alpha^\prime,\;\; \Rightarrow \cos \frac{{7\pi }}{6} = - \cos \frac{\pi }{6} = - \frac{\sqrt{3}}{2}.\]

Pythagorean Trigonometric Identities for Any Angle

We already know that, for any acute angle, the following identities are valid:

\[{\sin ^2}\alpha + {\cos ^2}\alpha = 1,\]

\[{\tan ^2}\alpha + 1 = {\sec ^2}\alpha,\]

\[{\cot ^2}\alpha + 1 = {\csc ^2}\alpha.\]

It turns out that these identities remain valid for any angle \(\alpha.\)

Consider, for example, the first identity. Let \(M\left( {x,y} \right)\) be a point on the terminal side of the angle \(\alpha.\) Then no matter which quadrant the angle belongs to, we can always build a right triangle with the legs \(\left| x \right|,\) \(\left| y \right|\) and hypotenuse \(r = OM.\) Hence,

\[{\left| y \right|^2} + {\left| x \right|^2} = {r^2},\;\; \Rightarrow {y^2} + {x^2} = {r^2}.\]

It follows from here that

\[{\left( {\frac{y}{r}} \right)^2} + {\left( {\frac{x}{r}} \right)^2} = 1,\]

which means

\[{\sin ^2}\alpha + {\cos ^2}\alpha = 1.\]

If we know one of the trigonometric functions of an angle and the quadrant in which the angle lies, we can determine all other trigonometric functions of this angle. This can be done using the identities listed above and definitions of trigonometric functions. Note that if a trigonometric expression contains a quotient, it is valid for only those angles at which the denominator is not zero.

See solved problems on Page 2.

Precalculus

Trigonometry

Signs of Trigonometric Functions