# Signs of Trigonometric Functions

### Signs of Trigonometric Functions in Each Quadrant

We know that $$\cos \alpha = \large{\frac{x}{r}}\normalsize.$$ Since $$r \gt 0,$$ the sign of the cosine function depends only on the sign of $$x.$$ Therefore, the cosine is positive in the $$1\text{st}$$ and $$4\text{th}$$ quadrants, and negative in the $$2\text{nd}$$ and $$3\text{rd}$$ quadrants.

Consider the sine function $$\sin \alpha = \large{\frac{y}{r}}\normalsize.$$ 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.

It is clear that the reciprocal functions $$\sec \alpha = \large{\frac{1}{{\cos \alpha }}}\normalsize$$ and $$\csc \alpha = \large{\frac{1}{{\sin \alpha }}}\normalsize$$ 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.

We can summarize this information in the following table:

### 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 $$\alpha = \large{\frac{{7\pi }}{6}}\normalsize.$$ 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 $$\alpha = \large{\frac{{7\pi }}{6}}\normalsize$$ lies in the $$3\text{rd}$$ quadrant in which the cosine function has a negative sign. Therefore,

${\cos \alpha = – \cos \alpha^\prime,}\;\; \Rightarrow {\cos \frac{{7\pi }}{6} = – \cos \frac{\pi }{6} = – \frac{1}{2}.}$

### 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.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Calculate $$\sin \alpha$$ and $$\cot \alpha$$ if $$\cos \alpha = \large{\frac{5}{{13}}}\normalsize$$ and the angle $$\alpha$$ lies in the $$4\text{th}$$ quadrant.

### Example 2

Calculate $$\cos \theta$$ and $$\tan \theta$$ if $$\sin \theta = -\large{\frac{24}{{25}}}\normalsize$$ and the angle $$\theta$$ lies in the $$3\text{rd}$$ quadrant.

### Example 3

Find $$\cos \alpha$$ if $$\tan \alpha = 5$$ and the angle $$\alpha$$ lies in the $$3\text{rd}$$ quadrant.

### Example 4

Find $$\sin \beta$$ if $$\cot \beta = -3$$ and the angle $$\beta$$ lies in the $$2\text{nd}$$ quadrant.

### Example 5

Find the values of the six trigonometric functions of $$\alpha = \large{\frac{{2\pi }}{3}}\normalsize.$$

### Example 6

Find the values of the six trigonometric functions of $$\beta = \large{\frac{{5\pi }}{4}}\normalsize.$$

### Example 7

Determine the sign of the expression $\sin \frac{{13\pi }}{6}\cos \frac{{11\pi }}{7}\tan \frac{{9\pi }}{8}.$

### Example 8

Determine the sign of the expression $\tan \left( { – \frac{{2\pi }}{5}} \right)\cot \left( { – \frac{{5\pi }}{7}} \right)\sec \left( { – \frac{{8\pi }}{9}} \right).$

### Example 1.

Calculate $$\sin \alpha$$ and $$\cot \alpha$$ if $$\cos \alpha = \large{\frac{5}{{13}}}\normalsize$$ and the angle $$\alpha$$ lies in the $$4\text{th}$$ quadrant.

Solution.

The sine function is negative in quadrant $$IV.$$ Using the Pythagorean trigonometric identity, we get

${{\sin ^2}\alpha + {\cos ^2}\alpha = 1,}\;\; \Rightarrow {{\sin ^2}\alpha = 1 – {\cos ^2}\alpha ,}\;\; \Rightarrow {\sin \alpha = – \sqrt {1 – {{\cos }^2}\alpha } }={ – \sqrt {1 – {{\left( {\frac{5}{{13}}} \right)}^2}} }={ – \sqrt {1 – \frac{{25}}{{169}}} }={ – \sqrt {\frac{{144}}{{169}}} }={ – \frac{{12}}{{13}}.}$

The cotangent function is given by

${\cot \alpha = \frac{{\cos \alpha }}{{\sin \alpha }} }={ \frac{{\frac{5}{{13}}}}{{ – \frac{{12}}{{13}}}} }={ – \frac{5}{{12}}.}$

### Example 2.

Calculate $$\cos \theta$$ and $$\tan \theta$$ if $$\sin \theta = -\large{\frac{24}{{25}}}\normalsize$$ and the angle $$\theta$$ lies in the $$3\text{rd}$$ quadrant.

Solution.

In the $$3\text{rd}$$ quadrant, the cosine function is negative. Therefore,

${{\sin ^2}\theta + {\cos ^2}\theta = 1,}\;\; \Rightarrow {\cos \theta = – \sqrt {1 – {{\sin }^2}\theta } }={ – \sqrt {1 – {{\left( { – \frac{{24}}{{25}}} \right)}^2}} }={ – \sqrt {1 – \frac{{576}}{{625}}} }={ – \sqrt {\frac{{49}}{{169}}} }={ – \frac{7}{{25}}.}$

Using the definition of tangent,we have

${\tan \theta = \frac{{\sin \theta }}{{\cos \theta }} }={ \frac{{ – \frac{{24}}{{25}}}}{{ – \frac{7}{{25}}}} }={ \frac{{24}}{7}.}$

### Example 3.

Find $$\cos \alpha$$ if $$\tan \alpha = 5$$ and the angle $$\alpha$$ lies in the $$3\text{rd}$$ quadrant.

Solution.

We use the identity

${{\tan ^2}\alpha + 1 = {\sec ^2}\alpha }.$

Hence,

${{\sec ^2}\alpha = \frac{1}{{{{\cos }^2}\alpha }} }={ {5^2} + 1 }={ 26,}\;\; \Rightarrow {{\cos ^2}\alpha = \frac{1}{{26}}.}$

The cosine has a negative sign in the $$3\text{rd}$$ quadrant. So,

${\cos \alpha = – \sqrt {{{\cos }^2}\alpha } }={ – \frac{1}{{\sqrt {26} }}.}$

### Example 4.

Find $$\sin \beta$$ if $$\cot \beta = -3$$ and the angle $$\beta$$ lies in the $$2\text{nd}$$ quadrant.

Solution.

Using the trigonometric identity

${\cot ^2}\beta + 1 = {\csc ^2}\beta ,$

we express the sine in terms of cotangent:

${{\csc ^2}\beta = \frac{1}{{{{\sin }^2}\beta }} }={ {\left( { – 3} \right)^2} + 1 = 10,}\;\; \Rightarrow{{\sin ^2}\beta = \frac{1}{{10}}.}$

In the $$2\text{nd}$$ quadrant, the sine is positive. Therefore,

${\sin \beta = \sqrt {{{\sin }^2}\beta } }={ \frac{1}{{\sqrt {10} }}.}$

### Example 5.

Find the values of the six trigonometric functions of $$\alpha = \large{\frac{{2\pi }}{3}}\normalsize.$$

Solution.

The angle $$\alpha = \large{\frac{{2\pi }}{3}}\normalsize$$ lies in $$2\text{nd}$$ quadrant.

Determine the reference angle for $$\alpha = \large{\frac{{2\pi }}{3}}\normalsize:$$

${\alpha^\prime = \pi – \alpha }={ \pi – \frac{{2\pi }}{3} }={ \frac{\pi }{3}.}$

The reference angle $$\alpha^\prime$$ is a special angle. We can easily find the values of the trig functions for it:

${\sin \frac{\pi }{3} = \frac{{\sqrt 3 }}{2},\;\;}\kern0pt{\cos \frac{\pi }{3} = \frac{1}{2},\;\;}\kern0pt{\tan \frac{\pi }{3} = \sqrt 3 ,\;\;}\kern0pt{\cot \frac{\pi }{3} = \frac{1}{{\sqrt 3 }},\;\;}\kern0pt{\sec \frac{\pi }{3} = 2,\;\;}\kern0pt{\csc \frac{\pi }{3} = \frac{2}{{\sqrt 3 }}.}$

In the $$2\text{nd}$$ quadrant, the sine and cosecant are positive while cosine, secant, tangent, and cotangent are negative. Therefore,

${\sin \frac{{2\pi }}{3} = \frac{{\sqrt 3 }}{2},\;\;}\kern0pt{\cos \frac{{2\pi }}{3} = – \frac{1}{2},\;\;}\kern0pt{\tan \frac{{2\pi }}{3} = – \sqrt 3 ,\;\;}\kern0pt{\cot \frac{{2\pi }}{3} = – \frac{1}{{\sqrt 3 }},\;\;}\kern0pt{\sec \frac{{2\pi }}{3} = – 2,\;\;}\kern0pt{\csc \frac{{2\pi }}{3} = \frac{2}{{\sqrt 3 }}.}$

### Example 6.

Find the values of the six trigonometric functions of $$\beta = \large{\frac{{5\pi }}{4}}\normalsize.$$

Solution.

The terminal side of the angle $$\beta = \large{\frac{{5\pi }}{4}}\normalsize$$ is in the $$3\text{rd}$$ quadrant.

The reference angle of $$\beta$$ is equal to

${\beta^\prime = \beta – \pi }={ \frac{{5\pi }}{4} – \pi }={ \frac{\pi }{4}.}$

Find the values of trig functions of $$\beta^\prime:$$

${\sin \frac{\pi }{4} = \frac{{\sqrt 2 }}{2},\;\;}\kern0pt{\cos \frac{\pi }{4} = \frac{{\sqrt 2 }}{2},\;\;}\kern0pt{\tan \frac{\pi }{4} = 1,\;\;}\kern0pt{\cot \frac{\pi }{4} = 1,\;\;}\kern0pt{\sec \frac{\pi }{4} = \sqrt 2 ,\;\;}\kern0pt{\csc \frac{\pi }{4} = \sqrt 2 .}$

In the $$3\text{rd}$$ quadrant, the tangent and cotangent are positive while all other trigonometric functions are negative. Hence,

${\sin \frac{{5\pi }}{4} = – \frac{{\sqrt 2 }}{2},\;\;}\kern0pt{\cos \frac{{5\pi }}{4} = – \frac{{\sqrt 2 }}{2},\;\;}\kern0pt{\tan \frac{{5\pi }}{4} = 1,\;\;}\kern0pt{\cot \frac{{5\pi }}{4} = 1,\;\;}\kern0pt{\sec \frac{{5\pi }}{4} = – \sqrt 2 ,\;\;}\kern0pt{\csc \frac{{5\pi }}{4} = – \sqrt 2 .}$

### Example 7.

Determine the sign of the expression $\sin \frac{{13\pi }}{6}\cos \frac{{11\pi }}{7}\tan \frac{{9\pi }}{8}.$

Solution.

The angle $$\large{\frac{{13\pi }}{6}}\normalsize$$ lies in the $$1\text{st}$$ quadrant:

${\frac{{12\pi }}{6} \lt \frac{{13\pi }}{6} \lt \frac{{15\pi }}{6},}\;\; \Rightarrow {2\pi \lt \frac{{13\pi }}{6} \lt 2\pi + \frac{\pi }{2}.}$

Therefore $$\sin \kern-2pt\large{\frac{{13\pi }}{6}}\normalsize$$ has a positive sign.

The angle $$\large{\frac{{11\pi }}{7}}\normalsize$$ is in the $$4\text{th}$$ quadrant:

${\frac{{21\pi }}{{14}} \lt \frac{{22\pi }}{{14}} \lt \frac{{28\pi }}{{14}},}\;\; \Rightarrow {\frac{{3\pi }}{2} \lt \frac{{22\pi }}{{14}} \lt 2\pi ,}\;\; \Rightarrow {\frac{{3\pi }}{2} \lt \frac{{11\pi }}{7} \lt 2\pi .}$

Hence, $$\cos \large{\frac{{11\pi }}{7}}\normalsize$$ has a positive sign.

The angle $$\large{\frac{{9\pi }}{8}}\normalsize$$ is obviously in the $$3\text{rd}$$ quadrant, so $$\tan \kern-2pt\large{\frac{{9\pi }}{8}}\normalsize$$ has a positive sign.

Since all three components are positive, their product is also positive:

$\sin \frac{{13\pi }}{6}\cos \frac{{11\pi }}{7}\tan \frac{{9\pi }}{8} \gt 0.$

### Example 8.

Determine the sign of the expression $\tan \left( { – \frac{{2\pi }}{5}} \right)\cot \left( { – \frac{{5\pi }}{7}} \right)\sec \left( { – \frac{{8\pi }}{9}} \right).$

Solution.

The angle $${ – \large{\frac{{2\pi }}{5}}\normalsize}$$ lies in the $$4\text{th}$$ quadrant where tangent is negative. Indeed,

${- \frac{{5\pi }}{{10}} \lt – \frac{{4\pi }}{{10}} \lt 0,}\;\; \Rightarrow {- \frac{\pi }{2} \lt – \frac{{4\pi }}{{10}} \lt 0,}\;\; \Rightarrow {- \frac{\pi }{2} \lt – \frac{{2\pi }}{5} \lt 0.}$

The angle $${ – \large{\frac{{5\pi }}{7}}\normalsize}$$ is in the $$3\text{rd}$$ quadrant where cotangent is positive:

${- \frac{{14\pi }}{{14}} \lt – \frac{{10\pi }}{{14}} \lt – \frac{{7\pi }}{{14}},}\;\; \Rightarrow {- \pi \lt – \frac{{5\pi }}{7} \lt – \frac{\pi }{2}.}$

The angle $${ – \large{\frac{{8\pi }}{9}}\normalsize}$$ also belongs to the $$3\text{rd}$$ quadrant where secant is negative:

${- \frac{{18\pi }}{{18}} \lt – \frac{{16\pi }}{{18}} \lt – \frac{{9\pi }}{{18}},}\;\; \Rightarrow {- \pi \lt – \frac{{8\pi }}{9} \lt – \frac{\pi }{2}.}$

So, $$2$$ of the $$3$$ factors are negative and one is positive. It is clear that their product is positive:

${\tan \left( { – \frac{{2\pi }}{5}} \right)\cot \left( { – \frac{{5\pi }}{7}} \right)\sec \left( { – \frac{{8\pi }}{9}} \right) }\gt{ 0.}$