All revision notes topics

3.11 Symmetry properties of trigonometric graphsIB Maths: Analysis and Approaches HL: Revision notes

Section 1

Reflecting in the vertical axis: π−θ\pi-\theta

Think of the point P(cos⁡θ,sin⁡θ)P(\cos\theta,\sin\theta) on the unit circle. The angle π−θ\pi-\theta gives the point you get by reflecting PP in the yy-axis: (−cos⁡θ,sin⁡θ)(-\cos\theta,\sin\theta). Reading off the coordinates: sin⁡(π−θ)=sin⁡θ,cos⁡(π−θ)=−cos⁡θ,tan⁡(π−θ)=−tan⁡θ.\sin(\pi-\theta)=\sin\theta,\qquad \cos(\pi-\theta)=-\cos\theta,\qquad \tan(\pi-\theta)=-\tan\theta. So sin⁡150∘=sin⁡30∘=12\sin150^\circ=\sin30^\circ=\frac12 but cos⁡150∘=−cos⁡30∘=−32\cos150^\circ=-\cos30^\circ=-\frac{\sqrt3}{2}. On the graph of y=sin⁡xy=\sin x this is the line of symmetry x=π2x=\frac{\pi}{2}.

Key termsunit circleline of symmetry
Common mistake

cos⁡(π−θ)\cos(\pi-\theta) is not cos⁡θ\cos\theta. Only sine keeps its sign under π−θ\pi-\theta.

Section 2

Half-turns and negative angles: π+θ\pi+\theta, −θ-\theta, 2π−θ2\pi-\theta

Adding π\pi is a half-turn about the origin: PP goes to (−cos⁡θ,−sin⁡θ)(-\cos\theta,-\sin\theta), so sin⁡(π+θ)=−sin⁡θ,cos⁡(π+θ)=−cos⁡θ,tan⁡(π+θ)=tan⁡θ.\sin(\pi+\theta)=-\sin\theta,\quad \cos(\pi+\theta)=-\cos\theta,\quad \tan(\pi+\theta)=\tan\theta. Replacing θ\theta by −θ-\theta reflects in the xx-axis: (cos⁡θ,−sin⁡θ)(\cos\theta,-\sin\theta), so sin⁡(−θ)=−sin⁡θ\sin(-\theta)=-\sin\theta, cos⁡(−θ)=cos⁡θ\cos(-\theta)=\cos\theta, tan⁡(−θ)=−tan⁡θ\tan(-\theta)=-\tan\theta. Because the functions repeat every 2π2\pi, the angle 2π−θ2\pi-\theta behaves exactly like −θ-\theta.

Key termshalf-turnperiod
Exam tip

Tangent has period π\pi, so tan⁡(π+θ)=tan⁡θ\tan(\pi+\theta)=\tan\theta and tan⁡200∘=tan⁡20∘\tan200^\circ=\tan20^\circ.

Section 3

Complementary angles: π2±θ\frac{\pi}{2}\pm\theta

Reflecting PP in the line y=xy=x swaps its coordinates, which gives the angle π2−θ\frac{\pi}{2}-\theta: sin⁡(π2−θ)=cos⁡θ,cos⁡(π2−θ)=sin⁡θ.\sin\left(\frac{\pi}{2}-\theta\right)=\cos\theta,\qquad \cos\left(\frac{\pi}{2}-\theta\right)=\sin\theta. For π2+θ\frac{\pi}{2}+\theta: sin⁡(π2+θ)=cos⁡θ\sin\left(\frac{\pi}{2}+\theta\right)=\cos\theta and cos⁡(π2+θ)=−sin⁡θ\cos\left(\frac{\pi}{2}+\theta\right)=-\sin\theta. Graphically, y=cos⁡xy=\cos x is y=sin⁡xy=\sin x translated π2\frac{\pi}{2} to the left.

Key termscomplementary angles
Exam tip

If unsure, check with a value: put θ=0\theta=0. cos⁡(π2+0)=0\cos\left(\frac{\pi}{2}+0\right)=0 and −sin⁡0=0-\sin0=0, then try θ=π2\theta=\frac{\pi}{2}: cos⁡π=−1=−sin⁡π2\cos\pi=-1=-\sin\frac{\pi}{2}.

Section 4

Symmetry of the graphs

  • y=sin⁡xy=\sin x is an odd function (rotational symmetry of order 2 about the origin). Its lines of symmetry are x=π2+kπx=\frac{\pi}{2}+k\pi.
  • y=cos⁡xy=\cos x is an even function (symmetric in the yy-axis). Its lines of symmetry are x=kπx=k\pi.
  • y=tan⁡xy=\tan x is odd, has period π\pi and vertical asymptotes at x=π2+kπx=\frac{\pi}{2}+k\pi; it has no lines of symmetry.

To prove a graph has the line of symmetry x=cx=c, show f(2c−x)=f(x)f(2c-x)=f(x). For example, for f(x)=sin⁡x+sin⁡3xf(x)=\sin x+\sin3x, f(π−x)=f(x)f(\pi-x)=f(x), so x=π2x=\frac{\pi}{2} is a line of symmetry.

Key termsodd functioneven function

Section 5

Using symmetry to find values and solve equations

Once you know one solution α\alpha in [0,2π)[0,2\pi), symmetry gives the others:

  • sin⁡x=k\sin x=k: x=αx=\alpha or x=π−αx=\pi-\alpha.
  • cos⁡x=k\cos x=k: x=αx=\alpha or x=2π−αx=2\pi-\alpha.
  • tan⁡x=k\tan x=k: x=αx=\alpha or x=π+αx=\pi+\alpha.

Then add multiples of the period to reach the required interval.

In a model such as h(t)=20−18cos⁡(πt15)h(t)=20-18\cos\left(\frac{\pi t}{15}\right), h(30−t)=h(t)h(30-t)=h(t) says the ride is symmetric about the top at t=15t=15: once you find that the height is 29 m at t=10t=10, you get t=20t=20 for free.

Key termsprincipal value
Example

cos⁡x=−12\cos x=-\frac12 on [0,2π][0,2\pi]: α=π−π3=2π3\alpha=\pi-\frac{\pi}{3}=\frac{2\pi}{3}, and 2π−2π3=4π32\pi-\frac{2\pi}{3}=\frac{4\pi}{3}.

Common mistake

Stopping at the calculator value. arcsin⁡\arcsin gives only one angle; the second comes from π−α\pi-\alpha.

Must know

  • sin⁡(π−θ)=sin⁡θ\sin(\pi-\theta)=\sin\theta, cos⁡(π−θ)=−cos⁡θ\cos(\pi-\theta)=-\cos\theta, tan⁡(π−θ)=−tan⁡θ\tan(\pi-\theta)=-\tan\theta.
  • sin⁡(π+θ)=−sin⁡θ\sin(\pi+\theta)=-\sin\theta, cos⁡(π+θ)=−cos⁡θ\cos(\pi+\theta)=-\cos\theta, tan⁡(π+θ)=tan⁡θ\tan(\pi+\theta)=\tan\theta.
  • sin⁡(−θ)=−sin⁡θ\sin(-\theta)=-\sin\theta, cos⁡(−θ)=cos⁡θ\cos(-\theta)=\cos\theta: sine is odd, cosine is even.
  • sin⁡(π2−θ)=cos⁡θ\sin\left(\frac{\pi}{2}-\theta\right)=\cos\theta, cos⁡(π2−θ)=sin⁡θ\cos\left(\frac{\pi}{2}-\theta\right)=\sin\theta.
  • Show f(2c−x)=f(x)f(2c-x)=f(x) to prove a line of symmetry x=cx=c.

That's the notes covered.

Carry on to the next subtopic.