TRIGONOMETRY · HL
Trig Graphs
Sine, cosine and tangent graphs — range and period.
Section 1 of 8
Draw $y = \sin x$
Draw $y = \sin x$, $0 \leq x \leq 360^\circ$.
$x$ = angle $y$ = height
| $x$ | $0$ | $90$ | $180$ | $270$ | $360$ |
|---|---|---|---|---|---|
| $\sin x$ | $0$ | $1$ | $0$ | $-1$ | $0$ |
$\sin A = \tfrac{1}{2}$ → $A = 30^\circ$ or $150^\circ$
Draw on above $y = 2\sin x$, $0 \leq x \leq 360^\circ$.
| $x$ | $0$ | $90$ | $180$ | $270$ | $360$ |
|---|---|---|---|---|---|
| $\sin x$ | $0$ | $1$ | $0$ | $-1$ | $0$ |
| $2\sin x$ | $0$ | $2$ | $0$ | $-2$ | $0$ |
Range
1.Range $=$ [ how low , how high ]
Section 2 of 8
Period
Period
1.Period $=$ how quick it repeats $= 360^\circ$
Section 3 of 8
Plot $y = 1 + 2\sin x$
Plot $y = 1 + 2\sin x$ for $0 \leq x \leq 360^\circ$.
| $x$ | $0$ | $90$ | $180$ | $270$ | $360$ |
|---|---|---|---|---|---|
| $\sin x$ | $0$ | $1$ | $0$ | $-1$ | $0$ |
| $2\sin x$ | $0$ | $2$ | $0$ | $-2$ | $0$ |
| $1+2\sin x$ | $1$ | $3$ | $1$ | $-1$ | $1$ |
Section 4 of 8
Draw $y = \sin 2x$
Draw $y = \sin 2x$ for $0 \leq x \leq 180$.
| $x$ | $0$ | $45$ | $90$ | $135$ | $180$ |
|---|---|---|---|---|---|
| $2x$ | $0$ | $90$ | $180$ | $270$ | $360$ |
| $\sin 2x$ | $0$ | $1$ | $0$ | $-1$ | $0$ |
Your getting $\sin$ of $2x$. So $2x$ values must be $0, 90, 180, 270, 360$. Put in first.
Section 5 of 8
$f(x) = a + b\sin cx$
Formula
1.$f(x) = a + b\sin cx$
2.Range $= [\, a-b \, , \; a+b \,]$
3.Period $= \dfrac{360}{c}$
Draw $y = 2\sin 3x$ on the domain $0 \leq x \leq 120$.
Must say D
Mode — Set up first.
Hit $3$
$f(x) =$ → Hit $\;2\;\sin\;3\;x\;)$
Hit $=$
$g(x) =$ → Ignore, hit $=$
Table Range: Start: hit $0$ · End: hit $120$ · Step: Period $\tfrac{120}{4}$
| $x$ | $0$ | $30$ | $60$ | $90$ | $120$ |
|---|---|---|---|---|---|
| $f(x)$ | $0$ | $2$ | $0$ | $-2$ | $0$ |
$y = 2\sin 3x$ Range $[-2, 2]$ Period $= \dfrac{360}{3} = 120^\circ$
| $x$ | $0$ | $30$ | $60$ | $90$ | $120$ |
|---|---|---|---|---|---|
| $3x$ | $0$ | $90$ | $180$ | $270$ | $360$ |
| $\sin 3x$ | $0$ | $1$ | $0$ | $-1$ | $0$ |
| $2\sin 3x$ | $0$ | $2$ | $0$ | $-2$ | $0$ |
Section 6 of 8
Draw $y = \cos x$
Draw $y = \cos x$, $0 \leq x \leq 360^\circ$.
| $x$ | $0$ | $90$ | $180$ | $270$ | $360^\circ$ |
|---|---|---|---|---|---|
| $\cos x$ | $1$ | $0$ | $-1$ | $0$ | $1$ |
Find $x$ where $\cos x = -\tfrac{1}{2}$, $0 \leq x \leq 360^\circ$.
CAST: $\cos$ negative → quadrants S and T.
$\cos x = \tfrac{1}{2}$ → $x = 60^\circ$
$x = 120^\circ$ or $240^\circ$
Section 7 of 8
Draw $f(x) = \cos 2x$
$\sin$ goes up, $\cos$ goes down from the $y$-axis.
Draw $f(x) = \cos 2x$ on the domain $0 \leq x \leq 360$.
| $x$ | $0$ | $45$ | $90$ | $135$ | $180$ |
|---|---|---|---|---|---|
| $2x$ | $0$ | $90$ | $180$ | $270$ | $360$ |
| $\cos 2x$ | $1$ | $0$ | $-1$ | $0$ | $1$ |
Period $\dfrac{360}{2} = 180$
Solve $\cos 2x = \dfrac{\sqrt{3}}{2}$, $0 \leq x \leq 360^\circ$.
CAST: $\cos$ positive → quadrants A and C.
$2x = 30$ → $x = 15^\circ$ $2x = 330$ → $x = 165^\circ$
$2x = 390$ → $x = 195^\circ$ $2x = 690$ → $x = 345^\circ$
Section 8 of 8
Tan graph
Find $\tan 90 = \dfrac{\sin 90}{\cos 90} = \dfrac{1}{0} = \infty$
| $x$ | $0$ | $45$ | $90$ | $135$ | $180$ | $225$ | $270$ | $315$ | $360$ |
|---|---|---|---|---|---|---|---|---|---|
| $\tan x$ | $0$ | $1$ | $\infty$ | $-1$ | $0$ | $1$ | $\infty$ | $-1$ | $0$ |
Note
1.$-1 \leq \cos\theta \leq 1$
2.$-1 \leq \sin\theta \leq 1$
3.$-\infty \leq \tan\theta \leq \infty$
Prior knowledge
1.$f(x) = a + b\cos cx$
2.Range $= [\, a-b \, , \; a+b \,]$
3.Period $= \dfrac{360}{c} = \dfrac{2\pi}{c}$
The height of the water in a port was measured over a period of time. The average height was found to be $1{\cdot}6$ m. The height measured in metres, $h(t)$, was modelled using the function
$h(t) = 1{\cdot}6 + 1{\cdot}5\cos\!\left(\dfrac{\pi}{6}t\right)$
where $t$ represents the number of hours since the last recorded high tide and $\left(\dfrac{\pi}{6}t\right)$ is expressed in radians.
$a = 1.6$ $b = 1.5$ $c = \dfrac{\pi}{6}$
(a) Find the period and range of $h(t)$.
Period: $\dfrac{2\pi}{c} = \dfrac{2\pi}{\frac{\pi}{6}} = 12$
Range: $[\,1.6-1.5\, , \; 1.6+1.5\,] = [\,0.1\, , \; 3.1\,]$
(b) Find the maximum height of the water in the port.
$3.1$ m
(d)(i) High tide at midnight ($t = 0$). Complete the table.
| Time | Mid | 3a | 6a | 9a | 12 | 3p | 6p | 9p | Mid |
|---|---|---|---|---|---|---|---|---|---|
| $t$ (hours) | $0$ | $3$ | $6$ | $9$ | $12$ | $15$ | $18$ | $21$ | $24$ |
| $h(t)$ (m) | $3.1$ | $1.6$ | $0.1$ | $1.6$ | $3.1$ | $1.6$ | $0.1$ | $1.6$ | $3.1$ |
(d)(ii) Sketch the graph of $h(t)$ between midnight and the following midnight.
$t = 0$: $h(0) = 1.6 + 1.5\cos 0 = 3.1$
$h(3) = 1.6 + 1.5\cos\dfrac{\pi}{2} = 1.6$
(e) Find, from your sketch, the difference in water height between low tide and high tide.
$3.1 - 0.1 = 3$ m
(f) A fully loaded barge requires a minimum water level of $2$ m; unloaded it needs $1{\cdot}5$ m. Estimate the maximum time the barge can spend in port without resting on the sea-bed.
$3.10\text{pm} - 9\text{am} = 6$ hrs $10$ mins
SUM
The lot in one box
Trig graphs toolkit
1.$f(x) = a + b\sin cx$ → Range $=[\,a-b\,,\;a+b\,]$, Period $=\dfrac{360}{c}$
2.$f(x) = a + b\cos cx$ → Range $=[\,a-b\,,\;a+b\,]$, Period $=\dfrac{360}{c}=\dfrac{2\pi}{c}$
3.Range $=$ [ how low , how high ]
4.Period $=$ how quick it repeats $=360^\circ$
5.$-1 \leq \sin\theta \leq 1$, $-1 \leq \cos\theta \leq 1$, $-\infty \leq \tan\theta \leq \infty$
End of lesson
Trig Graphs — HL · Mathslive.ie