Trigonometry in contextEdexcel A-Level Maths: Revision notes

Periodic situations, such as tides, daylight hours, a point on a wheel or a vibrating spring, are modelled by $y=a+b\sin(ct+d)\quad\text{or}\quad y=a+b\cos(ct+d).$

• $a$ is the mean (midline) value; the greatest value is $a+|b|$ and the least is $a-|b|$.
• $|b|$ is the amplitude.
• The period is $\frac{360^\circ}{c}$ or $\frac{2\pi}{c}$, and $d$ shifts the graph horizontally. Angles may be in degrees or radians: check the unit given in the question and set your calculator to match.

A point on a wheel of radius $5$ m with centre $6$ m above the ground has height $h=6-5\cos(30t)^\circ$: the minus sign starts the point at the lowest position, $h(0)=1$. The period is $\frac{360}{30}=12$ s. To find when $h=9$ solve $\cos(30t)^\circ=-0.6$ and take the smallest positive $t$. For the daylight model $D=12+4\sin\frac{2\pi(n-80)}{365}$ the greatest value is $16$ hours, on day $171$. $D>14$ means $\sin(\cdots)>\frac12$, which happens for one third of the cycle, about $122$ days.

Solve the trigonometric equation for the angle, list all values of the angle in the range the context allows (extend the range if the angle is $ct+d$, not $t$), then convert back to $t$. For $d=7+5\sin\left(\frac{\pi t}6+0.927\right)=9$ with $0\le t\le12$ the angle lies in $[0.927,\,7.210]$, so use $2.730$ and $6.695$ giving $t=3.4$ and $t=11.0$. Then interpret: the depth is at least $9$ m for $0\le t\le3.44$ and $11.0\le t\le12$, a total of $4.4$ hours. Always give the answer in the units of the context and to a sensible accuracy.

A force $F$ at angle $\theta$ above the horizontal has horizontal component $F\cos\theta$ and vertical component $F\sin\theta$. The resultant of perpendicular forces has size $\sqrt{X^2+Y^2}$ and direction $\tan^{-1}\frac YX$. For a projectile with speed $u$ at angle $\alpha$, the range on level ground is $R=\frac{u^2\sin2\alpha}{g}$, so the greatest range is at $\alpha=45^\circ$, and two angles ($\alpha$ and $90^\circ-\alpha$) give the same range. Example: $\tan\alpha=\frac34$ gives $\sin2\alpha=2\cdot\frac35\cdot\frac45=\frac{24}{25}$; with $u=20$, $R=\frac{400\times24/25}{9.8}=39.2$ m.

Models such as $d=7+3\sin x+4\cos x$ with $x=\frac{\pi t}6$ are easier to use as a single wave: $3\sin x+4\cos x=5\sin(x+0.927)$, so $d=7+5\sin(x+0.927)$. The greatest depth is $12$ m and the least is $2$ m. The first maximum is when $x+0.927=\frac\pi2$, giving $t=1.23$ hours. Remember the limits of a model: a sine model repeats forever, but real tides, daylight and weather vary from cycle to cycle.