Unit circle and trig functions

Having just introduced the trigonometric functions a few pages ago, it is time to throw that all away and redefine them.

Contents

New definitions

On the xyxy-plane, the circle centered at (0,0)(0, 0), the origin, with radius 11 is

x2+y2=1x^2 + y^2 = 1

Let’s draw a right triangle with base on the positive xx-axis, with hypotenuse from the origin to the unit circle in the first quadrant (x>0,y>0x > 0, y > 0). Let the central angle be θ\theta. Since the hypotenuse is 11, we have sin⁡θ=y\sin\theta = y, and cos⁡θ=x\cos\theta = x.

We now redefine the trigonometric ratios to be

sin⁡θ=y\sin\theta = y
cos⁡θ=x\cos\theta = x

where x2+y2=1x^2 + y^2 = 1, and the angle θ\theta is the central angle between (1,0)(1, 0) to (x,y)(x, y) Note that

oppositehypotenuse÷adjacenthypotenuse=oppositeadjacent\frac{\text{opposite}}{\text{hypotenuse}} \div \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\text{opposite}}{\text{adjacent}}

Then

tan⁡θ=sin⁡θcos⁡θ=yx\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{y}{x}

From the definition of gradient, tan⁡θ\tan\theta is the gradient of the line containing the hypotenuse.

Pythagorean identity

Combining the Pythagorean theorem and the unit circle definitions of trig functions, we get

sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1

sin⁡2θ=(sin⁡θ)2\sin^2\theta = (\sin\theta)^2, by definition. But sin⁡−1θ≠(sin⁡θ)−1\sin^{-1}\theta \neq (\sin\theta)^{-1}. sin⁡−1\sin^{-1} is the inverse function. (sin⁡θ)−1(\sin\theta)^{-1} is the reciprocal of sin⁡θ\sin\theta.

Exact angles

A right triangle with legs length 11 and hypotenuse 2\sqrt{2} is called a 1−1−21-1-\sqrt{2} triangle, with two 45°45\degree.

Splitting an equilateral triangle in half results in a 1−3−21-\sqrt{3}-2 triangle with a 60°60\degree and a 30°30\degree.

Using these two triangles, and the xx- and yy-intercepts of the unit circle we obtain

θ\thetasin⁡θ\sin\thetacos⁡θ\cos\thetatan⁡θ\tan\theta
0°,00\degree, 00042=1\displaystyle \frac{\sqrt4}{2} = 100
30°,π6\displaystyle 30\degree, \frac{\pi}{6}\quad\quad12\displaystyle \frac{1}{2}32\displaystyle \frac{\sqrt3}{2}33\displaystyle \frac{\sqrt3}{3}
45°,π4\displaystyle 45\degree, \frac{\pi}{4}22\displaystyle \frac{\sqrt{2}}{2}22\displaystyle \frac{\sqrt2}{2}11
60°,π3\displaystyle 60\degree, \frac{\pi}{3}32\displaystyle \frac{\sqrt3}{2}12\displaystyle \frac{1}{2}3\sqrt3
90°,π2\displaystyle 90\degree, \frac{\pi}{2}42=1\displaystyle \frac{\sqrt4}{2} = 100undefined

The easiest way to remember is

  1. remember unit circle definition and the intercepts
  2. remember the 00, 11, 2\sqrt2, 3\sqrt3, 4\sqrt{4}, all over 22 pattern in sin⁡\sin
  3. reverse the pattern for cos⁡\cos
  4. remember gradient is increasing so tan⁡\tan also increases

Signs

The signs of the trig functions are directly related to the signs of the yy, xx, and the gradients. In summary, sin⁡\sin is positive in quadrants I and II, cos⁡\cos is positive in I, IV, tan⁡\tan is positive in I, III.

In some places this is known as the CAST rule.

Angles outside quadrant I

We shall mainly focus on multiples of the above exact angles, which is what you will in encounter in almost all the paper 1 trig questions.

For such angles, follow these steps for an angle in radians

  1. Simplify the angle, eg 16π6=8π3\frac{16\pi}{6} = \frac{8\pi}{3}
  2. Identify the quadrant and the associated sign for the function, eg cos⁡8π3\cos\frac{8\pi}{3} in quadrant II, where cos⁡\cos is negative, so cos⁡8π3<0\cos\frac{8\pi}{3} < 0
  3. Ignore numerator, find the angle associated with the denominator and the function, and takes its value, eg cos⁡8π3=−cos⁡π3=−12\cos\frac{8\pi}{3} = -\cos\frac{\pi}{3} = -\frac 12

Tip: the only non-simplified fractions of π\pi you will encounter are from your own calculations.

Example: Find tan⁡7π4\tan\frac{7\pi}{4}.

Quadrant IV, tan⁡\tan is negative, tan⁡π4=1\tan\frac{\pi}{4} = 1, so tan⁡7π4=−1■\tan\frac{7\pi}{4} = -1 \qed

Example: Find sin⁡5π6\sin\frac{5\pi}{6}.

Quadrant II, sin⁡\sin is positive, sin⁡π6=12\sin\frac{\pi}{6} = \frac 12, so sin⁡5π6=12■\sin\frac{5\pi}{6} = \frac 12 \qed

Example: Find cos⁡210°\cos 210\degree.

210°210\degree is 30°30\degree more than π rad=180°\pi\text{ rad} = 180\degree

30°30\degree is π6\frac{\pi}{6} so use that

210°210\degree is Quadrant III, cos⁡\cos is negative, so cos⁡210°=−cos⁡π6=−32■\cos 210\degree = -\cos\frac{\pi}{6} = -\frac{\sqrt 3}{2} \qed

Tip: When converting degree to radians, you can simply do plus/minus from the nearest integer powers of π rad=180°\pi \text{ rad} = 180\degree, as that will preserve the denominator. But remember to use the quadrant of the original angle.

Questions involving special angles that you just found in a previous part likely require an identity for this part.

Converting trig functions with unknown angle

Step 1. Convert the fraction to opposite\text{opposite}, adjacent\text{adjacent}, and/or hypotenuse\text{hypotenuse} (ok maybe we are not throwing able the old definitions)

Tip: if you are not given a fraction or division, the denominator is 11.

Step 2. Find the missing one of the three, using Pythagorean Theorem

Step 3. Express the desired function using the expression derived in Steps 1 and 2

Step 4. Use the quadrant given in the question and apply the appropriate sign

Example: Given cos⁡x=−0.6,x∈[0,π]\cos x = -0.6, x\in \left[0, \pi\right], find sin⁡x\sin x and tan⁡x\tan x.

Depending on if you are more comfortable with fractions or decimals, convert to what you are comfortable with. −0.6=−610=−35-0.6 = -\frac{6}{10} = -\frac{3}{5}

Notice some 3−4−53-4-5 Pythagorean triple action.

adjacent: 3,hypotenuse: 5,opposite: 4\text{adjacent: }3, \text{hypotenuse: }5, \text{opposite: }4

The −0.6-0.6 for cos⁡\cos means quadrant II, where sin⁡\sin is positive and tan⁡\tan is negative.

sin⁡x=45,tan⁡x=−43■\sin x = \frac{4}{5}, \tan x = -\frac 43 \qed

Example: Let y=tan⁡x,x∈]π2,π[y = \tan x, x \in\left]\frac{\pi}{2}, \pi\right[, find sin⁡x\sin x and cos⁡x\cos x in terms of yy.

opposite: y,adjacent: 1,hypotenuse: y2+1\text{opposite: }y, \text{adjacent: }1, \text{hypotenuse: }\sqrt{y^2 + 1}

In quadrant II, where tan⁡\tan is negative, so y<0y < 0, where sin⁡\sin is positive but cos⁡\cos is negative.

sin⁡x=−yy2+1,cos⁡x=−1y2+1■\sin x = -\frac{y}{\sqrt{y^2+1}}, \cos x = -\frac{1}{\sqrt{y^2+1}} \qed

Note that we need a negative sign in sin⁡\sin to cancel out the negative yy.

We largely ignored signs, until at the end. This is ok because steps like adjacent: 3,hypotenuse: 5,opposite: 4\text{adjacent: }3, \text{hypotenuse: }5, \text{opposite: }4 are method marks (or no marks) not accuracy marks.

Inverse trig functions

As previously explored, a function is invertible if and only if it is one-to-one. We need to select part of the domain to define an inverse function. We want the domain

  • to produce an one-to-one function
  • to cover the entire range of the trig function
  • to involve quadrant I and another quadrant
  • to be continuous

The domain of an invertible function is the range of the inverse function, and vice versa. The inverse trigonometric functions satisfying these conditions are

inverse trig functiondomainrange
sin⁡−1,arcsin⁡\sin^{-1}, \arcsin[−1,1]\left[-1, 1\right][−π2,π2]\displaystyle \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
cos⁡−1,arccos⁡\cos^{-1}, \arccos[−1,1]\left[-1, 1\right][0,π]\left[0, \pi\right]
tan⁡−1,arctan⁡\tan^{-1}, \arctan]−∞,∞[\left]-\infty, \infty\right[]−π2,π2[\displaystyle \left]-\frac{\pi}{2}, \frac{\pi}{2}\right[

The inverse functions of other portions of trigonometric functions are discussed here (beyond HL?)