# Trigonometry Problems

Sub Topics
Trigonometry problems vary from simple problems related to right triangle to word problems depicting real life situations. While solving Calculus problems many steps involve trigonometric functions or equations. Hence a high level skill in solving trigonometry problem is essential to understand and master Calculus.

Let us now solve few problems using the trigonometric concepts, identities and equations.

## Simple Trigonometry Problems

Let us now see how a right triangle measures can be used to find the trigonometric functions of an angle.

Find the six trigonometric function values for $\angle$P

 In the diagram, lengths of sides PQ and QR and given.  We can find the length of the hypotenuse PR using   Pythagorean theorem  PR2 = PQ2 + QR2        = 82 + 62 = 64 + 36 = 100  PR = √100 = 10 cm  Now let us find the six trigonometric ratios of ∠P  Sin P = opp leg / hypotenuse = QR / PR = 6/10 = 0.6  Cos P = adj leg / hypotenuse = PQ / PR = 8/10 = 0.8  Tan P = opp leg / adj leg = PQ / QR = 6/8 = 0.75  Csc P = hypotenuse / opp leg = PR / QR = 10/6 ≈ 1.67  Sec P = hypotenuse / adj leg = PR / PQ = 10/8 = 1.25  Cot P = adj leg / opp leg = QR / PR = 8/6 ≈ 1.33 The reciprocal ratios csc, sec and cot can be found by taking the reciprocalsof sin, cos and tan ratios.

Now let us see how to use unit circle in determining the trigonometric function value of the given angle.

Use the unit circle to determine the six trigonometric functions of (1) $\frac{5\pi }{6}$  (2) $\frac{7\pi }{4}$

Suppose (x, y) is the point corresponding to the angle θ on unit circle, then the six trigonometric functions are given by

Sin $\theta$ = y,  Cos $\theta$ = x, Tan $\theta$ = $\frac{y}{x}$, Csc $\theta$ = $\frac{1}{y}$, Sec $\theta$ = $\frac{1}{x}$ and Cot $\theta$ = $\frac{x}{y}$.

(1) Corresponding to the angle $\frac{5\pi }{6}$, the ordered pair is $(-\frac{\sqrt{3}}{2},\frac{1}{2})$

Thus x = $-\frac{\sqrt{3}}{2}$   and y = 1/2

Sin $\frac{5\pi }{6}$ = 1/2               Cos $\frac{5\pi }{6}$ =  $-\frac{\sqrt{3}}{2}$

Tan $\frac{5\pi }{6}$ = $\frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=-\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{3}$

Csc $\frac{5\pi }{6}$ = 2                     Sec $\frac{5\pi }{6}$ = $-\frac{2\sqrt{3}}{3}$         Cot $\frac{5\pi }{6}$ = -√3

(2) Corresponding to the angle $\frac{7\pi }{4}$, the ordered pair is $(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})$

Thus x = $\frac{\sqrt{2}}{2}$     and    y = $-\frac{\sqrt{2}}{2}$

Sin $\frac{7\pi }{4}$ = $-\frac{\sqrt{2}}{2}$        Cos $\frac{7\pi }{4}$ = $\frac{\sqrt{2}}{2}$

Tan $\frac{7\pi }{4}$ = $\frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}$ = -1

Csc $\frac{\sqrt{2}}{2}$ = -√2                             Sec $\frac{\sqrt{2}}{2}$ = √2                   Cot $\frac{\sqrt{2}}{2}$ = -1.

## Sample Trigonometry Problems

We can evaluate trigonometric functions using unit circle only for few variable values. To evaluate trigonometric function of any angle Reference angles are used. A reference angle $\theta$' is an angle in the Quadrant I which is associated with the given angle $\theta$. The steps involved are as follows:
To find the function value of any angle $\theta$

1. Determine the function value for the reference angle $\theta$'.
2. Depending upon the quadrant in which the terminal side of $\theta$ lies, prefix the function value find in step 1 with the appropriate sign.

Example: Evaluate (a) Cos $\frac{4\pi }{3}$     (b) tan (-315º)

(a) Cos $\frac{4\pi }{3}$

Angle $\frac{4\pi }{3}$ can be written as $\frac{4\pi }{3}=\pi +\frac{\pi }{3}$

This means the terminal side of the angle falls in Quadrant III.

The Reference angle for an angle in Quadrant III is the measure by which it exceeds the measure π radians or 180º.

Hence the Reference angle of $\frac{4\pi }{3}$  is $\frac{\pi }{3}$

Cosine is negative in Quadrant III.

Hence Cos $\frac{4\pi }{3}$ = - Cos $\frac{\pi }{3}$ = -$\frac{1}{2}$.

(b) tan (-315º)  = - tan 315º    tan (-$\theta$) = -tan $\theta$.

315º = 270º + 45º.  The terminal side of the angle falls in Quadrant IV.

The reference angle for an angle in Quadrant IV is the measure by which it falls short of 2π radians or 360º.

Hence the reference angle for 315º = 360 - 315 = 45º.

Tangent function is negative in Quadrant IV.

Hence tan (-315º) = - tan 315º = -(-tan 45º) = -(-1) = 1.

Now here is a sample problem for verifying the identity:

Prove: $\frac{CosA}{1+Sina}+\frac{1+SinA}{CosA}=2SecA$

For verifying identities we generally simplify the left side of the equation given to arrive at the expression given on the right side.

Left side = $\frac{CosA}{1+Sina}+\frac{1+SinA}{CosA}$

= $\frac{Cos^{2}A +(1+SinA)^{2}}{CosA(1+SinA)}$                        The fractions are added using LCD

= $\frac{Cos^{2}A+1+2SinA+Sin^{2}A}{CosA(1+SinA)}$                  Binomial square expanded

= $\frac{Cos^{2}A+Sin^{2}A+1+2SinA}{CosA(1+SinA)}$

= $\frac{1+1+2SinA}{CosA(1+SinA)}$                                             Pythagorean Identity Sin2 $\theta$ + Cos2 $\theta$ = 1

= $\frac{2(1+SinA)}{CosA(1+SinA)}$                                              Numerator factored

= $\frac{2}{CosA}$                                                                        The common factor 1 + Sin A is canceled off.

= 2 Sec A                                                                                       Reciprocal Identity

= Right side.

Hence the identity is proved.

## Solve Trigonometry Problems

Below you could see trigonometry problems

1. Let us evaluate the trigonometric functions when the value of one function is given and a hint for the quadrant in which the terminal side of the angle falls.

Evaluate the other trigonometric functions when Cos $\theta$ = $\frac{8}{17}$ and given the constraint tan $\theta$ < 0.

The cosine value is positive and the constraint is that the tan is negative. Hence the angle $\theta$ falls in Quadrant IV.

The ordered pair representing the angle in Quadrant IV is of the form (x, -y) and Cos $\theta$ = $\frac{x}{r}$ where r is the radius of the circle.

As x is positive,  we can consider x = 8 and r = 17.  Using Pythagorean theorem the y value can be found as 15. We can express these in a right angle as follows:

Sin $\theta$ = $\frac{y}{r}$ = -$\frac{15}{17}$ and tan $\theta$ = -$\frac{15}{8}$  Csc $\theta$ = -$\frac{17}{15}$  Sec $\theta$ = $\frac{17}{8}$ and Cot $\theta$ = -$\frac{8}{5}$.

The value of Sin $\theta$ can also be found using Pythagorean identity and the quadrant information and the rest of the functions can be evaluated using quotient identity and reciprocal identities.

2. Trigonometric equations contain trigonometric functions of a variable. Trigonometric functions are solved using trigonometric identities, inverse trigonometric functions
and algebraic methods.
Let us solve one trigonometric equation using algebraic technique and inverse trigonometric functions:

Solve:  3 tan3 x = tan x

3 tan3 x - tan x = 0                              Subtract tan x on either side of the equation

tan x (3 tan2 x  - 1) = 0                        The common factor is factored out.

Using zero factor property,

tan x = 0                                ....................(1)

and

3 tan2 x - 1 = 0

tan2 x = $\frac{1}{3}$

tan x = $\pm$ $\frac{\sqrt{3}}{3}$   ....................(2)

The solutions for the equations can be written as

x = arctan 0,        x = arctan $\frac{\sqrt{3}}{3}$      and x = arctan -$\frac{\sqrt{3}}{3}$

Since the range of inverse tan function is (-$\frac{\pi}{2}$, $\frac{\pi}{2}$), the three principle solutions are

x = 0, x = $\frac{\pi }{6}$   and x = -$\frac{\pi }{6}$

As the period of tan function is $\pi$, the general solution can be written as,

x = 0 + n$\pi$,  x = $\frac{\pi }{6}$ + n$\pi$    and x = - $\frac{\pi }{6}$ + n$\pi$.

## Trigonometry Problem with solutions

Here are two trigonometry word problems with solutions.

1.  On the descent of an airplane from an altitude of 2.5 Km the airfield as well as a Foot Ball ground in the same line with the airfield are viewed by the Pilot.  It is noted the angles of  depression of the airfield and the Foot ball ground are 29º and 42º.  How far is the airfield from the Foot ball ground?

The following sketch depicts the situation described in the problem.

The positions of the airfield and the ground are marked with A and B. The alternate interior angles to the angles of depression and the law of sines can be used to solve the problem.

m $\angle$ADC = 90 - 29 = 61º              Complement of angle A

Using law of sines

$\frac{2.5}{sin29^{\circ}}=\frac{x+y}{sin61^{\circ}}$

x + y = $\frac{2.5}{sin29^{\circ}}\times sin61^{\circ}$ ≈ 4.51 Km   ----------------(1)

In right triangle BCD,

m $\angle$BDC = 90 - 42 = 48º             Complement of angle B

Using law of sines

$\frac{2.5}{sin42^{\circ}}=\frac{y}{sin48^{\circ}}$

y = $\frac{2.5}{sin42^{\circ}}\times sin48^{\circ}$ ≈ 2.78

Hence the distance between the airstrip and the foot ball ground x = (x + y) - y ≈ 4.51 - 2.78 ≈ 1.73 Km

The above problem can also be solved using right triangle trigonometry and eliminating y.

2.  The entrance to a Library is 10 ft above the level ground and one has to climb the steps built to reach it. The Museum administration wants to build a ram to make the Library Wheel chair accessible. The Safety regulations requires the slope of the ramp is not more than 1 inch rise for every 1 ft run. Find the maximum angle the ramp can make with the horizontal in order not to violate the regulations. Also find the length of the ramp built with the maximum elevation.

Let θ be the angle the ramp make with the horizontal. The maximum slope allowed by regulations = 1 inch / 1 ft   = 1 inch / 12 inches = $\frac{1}{12}$.

Hence the maximum value of tan $\theta$ = $\frac{1}{12}$ = 0.08333

The maximum value of $\theta$ = arc tan (0.083333) ≈ 4.76º                  The angle measure found using calculator.

Hence the ramp can make maximum an angle = 4.76º to meet the Safety regulations.

Now we use the Sine function to find the length of the ramp with maximum elevation.

Sin $\theta$ = $\frac{AB}{BC}$                                           Definition of sine ratio

Sin 4.76 = $\frac{10}{BC}$

BC = $\frac{10}{Sin4.76^{\circ}}$ ≈ 120.5 ft

## Practice Trigonometry Problems

Finally few problems for you to practice

1.  Prove the identity: $\frac{1}{1+sinx}= sec^{2}x-tanxsecx$
(Hint: Try simplifying the Right side to get the expression on the left. Else multiply and divide the left side by 1-sin x  and simplify.)

2.  Solve the equation ln(csc $\theta$) + ln(tan $\theta$) = ln 2 given than 0 ≤ $\theta$ ≤ 90º.
(Answer $\theta$ = 60º      Hint: Use the properties of logarithm)

3.  Prove the co function identity cos($\frac{\pi}{2}$ - x) = sin x using the Cosine difference formula.

4.  A roadway sign on a mountain indicates that the grade is 11º for the next  mile down the road. If a car is descending find the height climbed down to the nearest feet
during the one mile drive.