Section 1.3 Trigonometry
¶Subsection 1.3.1 Angles and Sectors of Circles
Mathematicians tend to deal mostly with radians and we will see later that some formulas are more elegant when using radians (rather than degrees). The relationship between degrees and radians is:Example 1.32. Degrees to Radians.
To convert 45∘ to radians, multiply by π180∘ to get π4.
Example 1.33. Radians to Degrees.
To convert 5π6 radians to degrees, multiply by 180∘π to get 150∘.
Sector Area.
The area of the sector is equal to:
Example 1.34. Angle Subtended by Arc.
If a circle has radius 3 cm, then an angle of 2 rad is subtended by an arc of 6 cm (s=rθ=3⋅2=6).
Example 1.35. Area of Circle.
If we substitute θ=2π (a complete revolution) into the sector area formula we get the area of a circle:
Subsection 1.3.2 Trigonometric Functions
There are six basic trigonometric functions:Sine (abbreviated by sin)
Cosecant (abbreviated by csc)
Cosine (abbreviated by cos)
Secant (abbreviated by sec)
Tangent (abbreviated by tan)
Cotangent (abbreviated by cot)
Definition 1.36. Basic Six Trigonometric Functions.
where opp stands for opposite, adj for adjacent, and hyp for hypotenuse.
Mnemonic.
The mnemonic SOH CAH TOA is useful in remembering how trigonometric functions of acute angles relate to the sides of a right triangle.
Subsection 1.3.3 Computing Exact Trigonometric Ratios
The unit circle is often used to determine the exact value of a particular trigonometric function.Mnemonic.
The first triangle should be easy to remember. To remember the second triangle, place the largest number (2) across from the largest angle (90∘=π/2). Place the smallest number (1) across from the smallest angle (30∘=π/6). Place the middle number (√3≈1.73) across from the middle angle (60∘=π/3). Double check using the Pythagorean Theorem that the sides satisfy a2+b2=c2.
The CAST Rule.
The CAST Rule says that in quadrant I all three of sinθ, cosθ, tanθ are positive. In quadrant II, only sinθ is positive, while cosθ, tanθ are negative. In quadrant III, only tanθ is positive, while sinθ, cosθ are negative. In quadrant IV, only cosθ is positive, while sinθ, tanθ are negative. To remember this, simply label the quadrants by the letters C-A-S-T starting in the bottom right and labelling counter-clockwise.
Example 1.37. Determining Trigonometric Ratios Without Unit Circle.
Determine sin5Ï€/6, cos5Ï€/6, tan5Ï€/6, sec5Ï€/6, csc5Ï€/6 and cot5Ï€/6 exactly by using the special triangles and CAST Rule.
We start by drawing the \(x\)-\(y\)-plane and indicating our angle of \(5\pi/6\) in standard position (positive angles rotate counterclockwise while negative angles rotate clockwise). Next, we drop a perpendicular to the \(x\)-axis (never drop it to the \(y\)-axis!).
Using \(\sin\theta=opp/hyp\) we find a value of \(1/2\text{.}\) Since \(\sin\theta\) is positive in quadrant II, we have
Using \(\cos\theta=adj/hyp\) we find a value of \(\sqrt 3/2\text{.}\) But \(\cos\theta\) is negative in quadrant II, therefore,
Using \(\tan\theta=opp/adj\) we find a value of \(1/\sqrt 3\text{.}\) But \(\tan\theta\) is negative in quadrant II, therefore,
To determine \(\sec\theta\text{,}\) \(\csc\theta\) and \(\cot\theta\) we use the definitions:
Example 1.38. CAST Rule.
If cosθ=3/7 and 3π/2<θ<2π, then find cotθ.
We first draw a right angle triangle. Since \(\cos\theta=adj/hyp=3/7\text{,}\) we let the adjacent side have length \(3\) and the hypotenuse have length \(7\text{.}\)
Since we are given \(3\pi/2\lt \theta\lt 2\pi\text{,}\) we are in the fourth quadrant. By the CAST Rule, \(\tan\theta\) is negative in this quadrant. As \(\tan\theta=opp/adj\text{,}\) it has a value of \(\sqrt{40}/3\text{,}\) but by the CAST Rule it is negative, that is,
Therefore,
Subsection 1.3.4 Graphs of Trigonometric Functions
¶The graph of the functions sinx and cosx can be visually represented as:Subsection 1.3.5 Trigonometric Identities
There are numerous trigonometric identities, including those relating to shift/periodicity, Pythagoras type identities, double-angle formulas, half-angle formulas and addition formulas. We list these below.Shifts and Periodicity.
sin(θ+2π)=sinθ cos(θ+2π)=cosθ tan(θ+2π)=tanθ sin(θ+π)=−sinθ cos(θ+π)=−cosθ tan(θ+π)=tanθ sin(−θ)=−sinθ cos(−θ)=cosθ tan(−θ)=−tanθ sin(π2−θ)=cosθ cos(π2−θ)=sinθ tan(π2−θ)=cotθ
Pythagoras Type Formulas.
Double-angle Formulas.
Half-angle Formulas.
Addition Formulas.
Example 1.39. Double Angle.
Find all values of x with 0≤x≤π such that sin2x=sinx.
Using the double-angle formula \(\sin 2x = 2\sin x\cos x\) we have:
Thus, either \(\sin x=0\) or \(\cos x = 1/2\text{.}\) For the first case when \(\sin x = 0\text{,}\) we get \(x=0\) or \(x=\pi\text{.}\) For the second case when \(\cos x=1/2\text{,}\) we get \(x=\pi/3\) (use the special triangles and CAST rule to get this). Thus, we have three solutions: \(x=0\text{,}\) \(x=\pi/3\text{,}\) \(x=\pi\text{.}\)
Exercises for Section 1.3.
Exercise 1.3.1.
Find all values of \(\theta\) such that \(\sin(\theta) = -1\text{;}\) give your answer in radians.
AnswerExercise 1.3.2.
Find all values of \(\theta\) such that \(\cos(2\theta) = 1/2\text{;}\) give your answer in radians.
AnswerExercise 1.3.3.
Compute the following:
\(\sin(3\pi)\)
\(\sec(5\pi/6)\)
\(\cos(-\pi/3)\)
\(\csc(4\pi/3)\)
\(\tan(7\pi/4)\)
\(\cot(13\pi/4)\)
Exercise 1.3.4.
If \(\sin\theta=\frac{3}{5}\) and \(\frac{\pi}{2}\lt \theta\lt \pi\text{,}\) then find \(\sec\theta\text{.}\)
Answer\(-\frac{5}{4}\)
Exercise 1.3.5.
Suppose that \(\tan\theta=x\) and \(\pi\lt \theta\lt \frac{3\pi}{2}\text{,}\) find \(\sin\theta\) and \(\cos\theta\) in terms of \(x\text{.}\)
AnswerExercise 1.3.6.
Find an angle \(\theta\) such that \(-\frac{\pi}{2}\lt \theta\lt \frac{\pi}{2}\) and \(\sin\theta=\sin\frac{23\pi}{7}\text{.}\)
AnswerExercise 1.3.7.
Use an angle sum identity to compute \(\cos(\pi/12)\text{.}\)
AnswerExercise 1.3.8.
Use an angle sum identity to compute \(\tan(5\pi/12)\text{.}\)
AnswerExercise 1.3.9.
Verify the following identities
\(\ds \cos^2(t)/(1-\sin(t)) = 1+\sin(t)\)
\(2\csc(2\theta)=\sec(\theta)\csc(\theta)\)
\(\sin(3\theta) - \sin(\theta) = 2\cos(2\theta)\sin(\theta)\)
Exercise 1.3.10.
Sketch the following functions:
\(y=2\sin(x)\)
\(y=\sin(3x)\)
\(y=\sin(-x)\)
Exercise 1.3.11.
Find all of the solutions of \(\ds 2\sin(t) -1 -\sin^2(t) =0\) in the interval \([0,2\pi]\text{.}\)
AnswerWe want to find all solutions to the equation
We first make the substitution, \(w = \sin t\text{.}\) Then our equation becomes,
We have thus reduced the problem to solving \(\sin t = 1\) on \(\left[0,2\pi\right]\text{.}\) Therefore, \(t = \frac{\pi}{2}\text{.}\)