Cards In This Deck

• $\text{What are the exact values of }\cos 60^{\circ }\text{ and }\sin 60^{\circ }\text{?}$
  $$\begin{gathered} \cos 60^{\circ }=\frac{1}{2} \\ \sin 60^{\circ }=\frac{\sqrt{3}}{2} \end{gathered}$$
• $\text{The identity }\tan \theta \equiv \frac{\sin \theta }{\cos \theta }\text{ holds for all values of }\theta \text{ EXCEPT...}$
  $$\begin{gathered} \text{...when }\cos \theta =0\text{.} \\ \text{This occurs at }\theta =90^{\circ },270^{\circ },\dots \text{ (or }\frac{ \\ \pi }{2},\frac{3 \\ \pi }{2},\dots \text{), where }\tan \theta \text{ is undefined (asymptotes).} \end{gathered}$$
• $$\begin{gathered} \text{True or False? } \\ {\sin }^{2}3\theta +{\cos }^2\theta =1 \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Identity: }{\sin }^{2}A+{\cos }^{2}A\equiv 1\text{.} \\ \text{The angles must be the same. It is true that }{\sin }^{2}3\theta +{\cos }^{2}3\theta =1\text{, but you cannot mix }3\theta \text{ and }\theta \text{.} \end{gathered}$$
• $\text{When proving a trigonometric identity such as }\frac{{\cos }^2\theta }{1-\sin \theta }\equiv 1+\sin \theta \text{, which side should you generally start manipulating?}$
  $$\begin{gathered} \text{Start with the more complicated side (usually the Left Hand Side).} \\ \text{Here, start with LHS, use }{\cos }^2\theta \equiv 1-{\sin }^2\theta \text{, factorise via difference of two squares }(1-\sin \theta )(1+\sin \theta )\text{, and cancel.} \end{gathered}$$
• $\text{When solving an equation involving both }\sin \theta \text{ and }\cos \theta \text{ with no squared terms (e.g., }2\sin \theta =3\cos \theta \text{), which strategy should you use?}$
  $$\begin{gathered} \text{Divide both sides by }\cos \theta \text{ to convert the equation into }\tan \theta \text{.} \\ 2\frac{\sin \theta }{\cos \theta }=3\Rightarrow 2\tan \theta =3\Rightarrow \tan \theta =1.5 \end{gathered}$$
• $\text{When solving a trigonometric equation like }3{\sin }^{2}x+4\cos x-4=0\text{, how do you decide which identity to substitute?}$
  $$\begin{gathered} \text{Look at the linear term (the one without the power).} \\ \text{Since }\cos x\text{ is linear, you must substitute for the squared term using }{\sin }^{2}x\equiv 1-{\cos }^{2}x\text{ to create a quadratic equation entirely in terms of }\cos x\text{.} \end{gathered}$$
• $$\begin{gathered} \text{True or False? } \\ {\sin }^{-1}x\text{ is the same as }(\sin x{)}^{-1}\text{ or }\frac{1}{\sin x}\text{.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{- }{\sin }^{-1}x\text{ refers to the inverse function (}\arcsin x\text{), used to find angles.} \\ \text{- }(\sin x{)}^{-1}\text{ refers to the reciprocal function (}\frac{1}{\sin x}\text{ or }cosecx\text{).} \end{gathered}$$
• $$\begin{gathered} \text{True or False? } \\ \text{Since }\cos (-x)=\cos (x)\text{, it follows that }\sin (-x)=\sin (x)\text{.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Identity: }\sin (-x)=-\sin (x)\text{.} \\ \text{Cosine is an even function (symmetrical about y-axis), but Sine is an odd function (rotational symmetry).} \end{gathered}$$
• $$\begin{gathered} \text{True or False? } \\ \text{To solve the equation }5\sin \theta =2\cos \theta \sin \theta \text{ for }0\le \theta \le 360^{\circ }\text{, the first step is to divide both sides by }\sin \theta \text{.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Approach: Rearrange and factorise: }\sin \theta (5-2\cos \theta )=0\text{.} \\ \text{Dividing by }\sin \theta \text{ removes the solutions where }\sin \theta =0\text{ (i.e., }\theta =0^{\circ },180^{\circ },360^{\circ }\text{).} \end{gathered}$$
• $\text{What is the first step to solve an equation in the form }\sin (3\theta +45^{\circ })=0.5\text{ for the interval }0\le \theta \le 360^{\circ }\text{?}$
  $$\begin{gathered} \text{Adjust the interval for the whole angle }(3\theta +45^{\circ })\text{.} \\ \text{New interval: }45^{\circ }\le 3\theta +45^{\circ }\le 1125^{\circ }\text{. Only then find the principal value and subsequent solutions within this new range.} \end{gathered}$$
• $$\begin{gathered} \text{True or False? } \\ \text{Using a calculator, }{\sin }^{-1}(-0.5)=-30^{\circ }\text{. Since the interval is }0\le x\le 360^{\circ }\text{, this result can be ignored completely.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{Although }-30^{\circ }\text{ is outside the interval, it is the principal value needed to find the valid solutions using the CAST diagram or graph symmetries.} \\ \text{Valid solutions: }180^{\circ }-(-30^{\circ })=210^{\circ }\text{ and }360^{\circ }+(-30^{\circ })=330^{\circ }\text{.} \end{gathered}$$
• $\text{You are solving }4{\sin }^{2}x=3\text{ for }0\le x\le 360^{\circ }\text{. You find }\sin x=\frac{\sqrt{3}}{2}\text{. Are you finished?}$
  $$\begin{gathered} \text{No. You must also consider the negative square root.} \\ 4{\sin }^{2}x=3\Rightarrow {\sin }^{2}x=0.75\Rightarrow \sin x=\pm \frac{\sqrt{3}}{2} \\ \text{You need solutions for both positive and negative cases (4 solutions total).} \end{gathered}$$