# Precalculus

## Trigonometry # Signs of Trigonometric Functions

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Calculate sin α and cot α if cos α = 5/13 and the angle α lies in the 4th quadrant.

### Example 2

Calculate cos θ and tan θ if sin θ = −24/25 and the angle θ lies in the 3rd 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 $$\alpha = \frac{{2\pi }}{3}.$$

### Example 6

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

### 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 $$\cos \alpha = \frac{5}{{13}}$$ 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 $$\sin \theta = -\frac{24}{{25}}$$ 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 $$\alpha = \frac{{2\pi }}{3}.$$

Solution.

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

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

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

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

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

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

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

### Example 6.

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

Solution.

The terminal side of the angle $$\beta = \frac{{5\pi }}{4}$$ 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 $$\beta^\prime:$$

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

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

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

### 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 $$\frac{{13\pi }}{6}$$ 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 $$\sin \frac{{13\pi }}{6}$$ has a positive sign.

The angle $$\frac{{11\pi }}{7}$$ 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, $$\cos \frac{{11\pi }}{7}$$ has a positive sign.

The angle $$\frac{{9\pi }}{8}$$ is obviously in the $$3\text{rd}$$ quadrant, so $$\tan \frac{{9\pi }}{8}$$ 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 $${ - \frac{{2\pi }}{5}}$$ 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 $${ - \frac{{5\pi }}{7}}$$ 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 $${ - \frac{{8\pi }}{9}}$$ 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:

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