Pure 5.3 - Sine, cosine and tangent functions

Pure 5.3 - Sine, cosine and tangent functions

Build exact trigonometric values from two triangles, then use signs, symmetry and periodicity to reach any angle. Read the key features of the three graphs and see how a shift or a change of scale affects them.

Unit-circle meaning and exact values

Why does y=cos⁡(x+30∘)y=\cos(x+30^\circ) simply move sideways, while y=tan⁡2xy=\tan2x fits twice as many repeats into the same interval? We will read both graphs at the end. Begin with a smaller puzzle: why are sin⁡(π/3)\sin(\pi/3) and cos⁡(π/6)\cos(\pi/6) exactly equal?

You already know

The unit-circle point at angle θ\theta is (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta). Tangent is sin⁡θ/cos⁡θ\sin\theta/\cos\theta where cos⁡θ≠0\cos\theta\ne0. A half-turn is π\pi radians or 180∘180^\circ.

Cut an equilateral triangle of side 22 in half. Its right-triangle sides are 11, 3\sqrt3, 22, since the height is 22−12=3\sqrt{2^2-1^2}=\sqrt3. A right-angled isosceles triangle with legs 11 has hypotenuse 2\sqrt2. These two triangles generate all the special acute values below.

[DIAGRAM: asset_slug: p1_5_3_understand_and_use_the_sine_cosine_and_tangent_functions__diagram_01; asset_name: Exact-value triangles; caption: Two right triangles generate the special acute-angle values; description: A 30-60-90 triangle has sides 1, square root of 3 and 2. A 45-45-90 triangle has sides 1, 1 and square root of 2. Angles and sides are labelled separately.]
Two right triangles generate the special acute-angle values

Exact trigonometric value

An exact value uses integers, fractions or surds without rounding. For example, sin⁡(π/4)=2/2\sin(\pi/4)=\sqrt2/2 is exact; 0.7070.707 is an approximation. Exact does not mean “a long calculator decimal”.

Use the triangle ratios for acute angles and the unit-circle coordinates for the endpoints. These values are not in the formula booklet: learn them.

θ\theta00π/6\pi/6π/4\pi/4π/3\pi/3π/2\pi/2π\pi
sin⁡θ\sin\theta001/21/22/2\sqrt2/23/2\sqrt3/21100
cos⁡θ\cos\theta113/2\sqrt3/22/2\sqrt2/21/21/200−1-1
tan⁡θ\tan\theta003/3\sqrt3/3113\sqrt3undefined00

Worked example 1: reading the special triangles

Question. Find sin⁡(π/3)\sin(\pi/3), cos⁡(π/6)\cos(\pi/6) and tan⁡(π/4)\tan(\pi/4) exactly.

Plan. Use opposite over hypotenuse, adjacent over hypotenuse and opposite over adjacent.

sin⁡π3=32opposite over hypotenusecos⁡π6=32adjacent over hypotenusetan⁡π4=11=1equal legs\begin{aligned} \sin\frac\pi3&=\frac{\sqrt3}{2} &&\text{opposite over hypotenuse}\\ \cos\frac\pi6&=\frac{\sqrt3}{2} &&\text{adjacent over hypotenuse}\\ \tan\frac\pi4&=\frac11=1 &&\text{equal legs} \end{aligned}

Why this step? The two acute angles in a right triangle are complementary, so one angle's opposite side is the other's adjacent side.

Answer. 3/2\sqrt3/2, 3/2\sqrt3/2, 11, respectively.

Check. Both sine and cosine lie between 00 and 11, and tangent is 11 for equal legs.

Change only the acute angles to their complements. The opposite and adjacent roles swap.

Worked example 2: complementary acute angles

Question. Find sin⁡(π/6)\sin(\pi/6), cos⁡(π/3)\cos(\pi/3) and tan⁡(π/4)\tan(\pi/4) exactly.

Plan. Read the same two triangles from the other acute angle.

sin⁡π6=12short leg over hypotenusecos⁡π3=12short leg over hypotenusetan⁡π4=1the complement is still π/4\begin{aligned} \sin\frac\pi6&=\frac12 &&\text{short leg over hypotenuse}\\ \cos\frac\pi3&=\frac12 &&\text{short leg over hypotenuse}\\ \tan\frac\pi4&=1 &&\text{the complement is still }\pi/4 \end{aligned}

Why this step? π/6+π/3=π/2\pi/6+\pi/3=\pi/2, while π/4\pi/4 is its own complement.

Answer. 1/21/2, 1/21/2, 11, respectively.

Check. Dividing the π/3\pi/3 sine by its cosine gives (3/2)/(1/2)=3(\sqrt3/2)/(1/2)=\sqrt3, the table's tangent.

Here is the surprising connection: swapping the two coordinates of a unit-circle point replaces θ\theta by π/2−θ\pi/2-\theta. That is why complementary sine and cosine values match; it is a reflection, rather than a numerical coincidence.

Watch out

Wrong: tan⁡(π/2)=0\tan(\pi/2)=0. Correct: it is undefined, since cos⁡(π/2)=0\cos(\pi/2)=0. A vertical asymptote is not a very large plotted value.

Use the same reference triangles with a different pair of acute angles.

Multiples and signs

At 5π/65\pi/6, the reference angle is π/6\pi/6, but the point lies in quadrant II. Sine stays positive and cosine becomes negative. A reference angle gives the magnitude; a quadrant gives the sign.

Reference angle

The reference angle is the acute angle between the terminal ray and the horizontal axis. For example, 4π/34\pi/3 has reference angle π/3\pi/3; 11π/611\pi/6 has reference angle π/6\pi/6. Axis angles use their coordinates directly rather than an acute reference triangle.

In quadrant I all three functions are positive; in II only sine is positive; in III only tangent is positive; in IV only cosine is positive. These signs follow from horizontal and vertical coordinates, so you can reconstruct them if a mnemonic slips your mind.

Worked example 3: exact values outside quadrant I

Question. Find sin⁡(4π/3)\sin(4\pi/3), cos⁡(11π/6)\cos(11\pi/6) and tan⁡(5π/3)\tan(5\pi/3) exactly.

Plan. Identify a reference angle and a sign for each value.

sin⁡4π3=−sin⁡π3=−32III: sine negativecos⁡11π6=cos⁡π6=32IV: cosine positivetan⁡5π3=−tan⁡π3=−3IV: tangent negative\begin{aligned} \sin\frac{4\pi}{3}&=-\sin\frac\pi3=-\frac{\sqrt3}{2} &&\text{III: sine negative}\\ \cos\frac{11\pi}{6}&=\cos\frac\pi6=\frac{\sqrt3}{2} &&\text{IV: cosine positive}\\ \tan\frac{5\pi}{3}&=-\tan\frac\pi3=-\sqrt3 &&\text{IV: tangent negative} \end{aligned}

Why this step? Copying a first-quadrant value without its sign loses the actual coordinate or ratio.

Answer. −3/2-\sqrt3/2, 3/2\sqrt3/2, −3-\sqrt3, respectively.

Check. The first and third angles have negative vertical coordinates, agreeing with their sine or tangent signs.

Change only the quadrants while retaining the same reference angles in the same order.

Worked example 4: the same magnitudes with different signs

Question. Find sin⁡(2π/3)\sin(2\pi/3), cos⁡(7π/6)\cos(7\pi/6) and tan⁡(4π/3)\tan(4\pi/3) exactly.

Plan. Reuse the special magnitudes and read the new quadrant signs.

sin⁡2π3=sin⁡π3=32II: sine positivecos⁡7π6=−cos⁡π6=−32III: cosine negativetan⁡4π3=tan⁡π3=3III: tangent positive\begin{aligned} \sin\frac{2\pi}{3}&=\sin\frac\pi3=\frac{\sqrt3}{2} &&\text{II: sine positive}\\ \cos\frac{7\pi}{6}&=-\cos\frac\pi6=-\frac{\sqrt3}{2} &&\text{III: cosine negative}\\ \tan\frac{4\pi}{3}&=\tan\frac\pi3=\sqrt3 &&\text{III: tangent positive} \end{aligned}

Why this step? Tangent in quadrant III is positive because both sine and cosine are negative.

Answer. 3/2\sqrt3/2, −3/2-\sqrt3/2, 3\sqrt3, respectively.

Check. tan⁡(4π/3)=(−3/2)/(−1/2)=3\tan(4\pi/3)=(-\sqrt3/2)/(-1/2)=\sqrt3.

A negative sign belongs to the value, not automatically to every angle greater than π\pi.

Watch out

Wrong: cos⁡(11π/6)=−3/2\cos(11\pi/6)=-\sqrt3/2 because the angle is “large”. Correct: quadrant IV has positive horizontal coordinate, so the cosine is 3/2\sqrt3/2.

Move the same reference angles to new quadrants once more.

Base graphs and symmetry

The cosine graph starts at (0,1)(0, 1), while sine starts at (0,0)(0, 0) and rises. Tangent also passes through (0,0)(0, 0), but has breaks wherever its cosine denominator is zero. Those starting points help you distinguish the graphs immediately.

[DIAGRAM: asset_slug: p1_5_3_understand_and_use_the_sine_cosine_and_tangent_functions__diagram_02; asset_name: Base trigonometric graphs; caption: Sine and cosine repeat after a full turn; tangent after half a turn; description: Three aligned graphs show sine, cosine and tangent between minus 2 pi and 2 pi. The tangent graph has separate branches and dashed vertical asymptotes.]
Sine and cosine repeat after a full turn

Period and symmetry

A period p>0p>0 satisfies f(x+p)=f(x)f(x+p)=f(x) throughout the domain; the fundamental period is the smallest such positive repeat. An even function satisfies f(−x)=f(x)f(-x)=f(x), giving reflection symmetry in the vertical axis. An odd function satisfies f(−x)=−f(x)f(-x)=-f(x), giving half-turn symmetry about the origin.

Reflection in the horizontal axis changes the sine coordinate's sign but preserves cosine. Hence cosine is even, while sine and tangent are odd. Rotating a full turn restores both coordinates; a half-turn restores their ratio.

FunctionDomainRangeFundamental periodZerosSymmetry
sin⁡x\sin xR\mathbb R[−1,1][-1, 1]2π2\pi or 360∘360^\circx=kπx=k\piodd
cos⁡x\cos xR\mathbb R[−1,1][-1, 1]2π2\pi or 360∘360^\circx=π/2+kπx=\pi/2+k\pieven
tan⁡x\tan xx≠π/2+kπx\ne\pi/2+k\piR\mathbb Rπ\pi or 180∘180^\circx=kπx=k\piodd

Here k∈Zk\in\mathbb Z. Tangent's excluded inputs are vertical asymptotes. Sine and cosine attain maximum 11 and minimum −1-1; tangent has neither. These graph features and symmetry results are not separately supplied in the formula booklet: learn them.

Key idea

Use the graph's zeros, range, breaks and smallest repeat together; a single point does not identify a function.

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