trigonometry

Assumed Knowledge (from methods) §

• Angles in degrees in radians (and converting between units).
• Defining sin and cos using the unit circle.
• Graphs of y = sinx, y = cosx and y = tanx.
• The pythagorean identity.
• Symmetry properties of trig functions.
• Exact values of trig functions for integer multiples of $\frac{\pi }{2}$, $\frac{\pi }{4}$, and $\frac{\pi }{6}$ radians.

Sketch the graph of the function §

h: [0, 2pi] -> $\mathbb{R}$, h(x) = $3\cos \left(2x+\frac{\pi }{3}\right)+1$ dilations first, followed by translations, provided it is written in the factorised form.

General solutions to trig equations. §

recap §

• sin (x) = 1/2
  • we said within one revolution of a unit circle, there exists two solutions for this statement.
  • 2 out of infinity many solutions, with other solutions that can be obtained by some number of revolutions around the unit circle.
  • $\therefore$ in general, $x=\frac{\pi }{6}+k2\pi$, $\frac{5\pi }{6}+k2\pi$ for some integer k

more on general solutions §

• for the trigonometric equation such as $\cos \theta =0.3$. there exists infinitely many solutions.

• if you put the equation in a calculator, ${\cos }^{-1}(0.3)=72.54°$ <- only 1 definite value, out of infinite values is given, not even two values of the unit circle.

  • how does the calculator decide which value to give?
  • there exists a rule/notion of what an inverse of a trig function gives you.

• think of ${\cos }^{-1}$ as a function, such value -> ${\cos }^{-1}$ -> angle

  • we need to define a range for the possible outputs
    • by convention, ${\cos }^{-1}$ has range $\{\theta |0\le \theta \le \pi \}$
    • by convention, the graph of $co{s}^{-1}x$ is:
      • 
    • ${\sin }^{-1}$ has range $\left\{\theta |-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}\right\}$
      • 
    • ${\tan }^{-1}$ has range $\{\theta |-\frac{\pi }{2}<\theta <\frac{\pi }{2}\}$
      • 

• consider the trig equation $\cos \theta =\frac{1}{2}$

• by convention, ${\cos }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{3}$

• the solution in the reflected position can be negative

• this can be expressed as

  • $\theta =2k\pi \pm {\cos }^{-1}\left(\frac{1}{2}\right)$
  • $=2k\pi \pm \frac{\pi }{3}$
  • $=\frac{(6k\pm 1)\pi }{3},k\in \mathbb{Z}$

• consider the trig equation $\sin \theta =\frac{1}{2}$ - a calculator gives ${\sin }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{6}$ - the other solution is $\pi -{\sin }^{-1}(\frac{1}{2})$ - so general solution is - $\theta =k2\pi +{\sin }^{-1}(\frac{1}{2})$ or $\theta =(2k+1)\pi -{\sin }^{-1}(\frac{1}{2})$

  dont have to write in this form, but you may see this form in questions!

• relationship between b and period is that b = 2pi/period for y = cos(bx)

proving trigonometric identities §

• prove LHS and RHS
• e.g. prove that $\frac{1}{1+{\tan }^2(\theta )}={\cos }^2(\theta )$
• write $LHS=\frac{1}{1+{\tan }^2(\theta )}$
• $\frac{1}{1+\frac{{\sin }^2\theta }{{\cos }^2\theta }}$ (using $\tan (x)=\frac{\sin (x)}{\cos (x)}$)
• $=\frac{\frac{1}{{\cos }^2\theta +{\sin }^2\theta }}{\cos \theta }$
• $=\frac{1}{\frac{1}{{\cos }^2\theta }}$ (using pythagorean identity)
• $={\cos }^2\theta$
• $=RHS$
• Thus $\frac{1}{1+{\tan }^2\theta }={\cos }^2\theta$ for all $\theta$
• usually its better to start with the side that looks more complex as the LHS.
  • there are exceptions though.

reciprocal trigonometric functions §

• sec theta = 1/costheta

• cosec theta = 1/sintheta

• cot theta = 1/tan theta (= costheta/sintheta)

• an untranslated sec graph will have a minimum point on the y axis, so we can say there has been no hroizontal translation. so we can say the value of a is positive, as we expect the value to be a positive 3

  • if a = 3, then c = 0 (what is going on)

• sec x = 1/cos(x)

• however, 3sec(2x)+1 != 1/3(cos(2x))+1

  • you cant do the translation on cosine, then getting the reciprocal - it just wont work XDDDDDDDD.

9C: compound and double angle formula §

• sin(A+B) = sinA + sinB??????
• f(x) = x + 3
• f(a+b) = a + b + 3
• f(a) + f(b)
  • = a + 3 + b + 3
  • a + b + b
  • != f(a + b)
  • therefore it is not equal

angle sum and difference identities §

• how to prove (or derive) them:
  • start by deriving cos(A-b) = cosAcosB + sinAsinB
  • who knows now

double angle formulae §

• cos(A+A) = cosAcosA - sinAsinA
• cos(2A) = ${\cos }^{2}A-{\sin }^{2}A$
• $\cos (2A)=1-2{\sin }^{2}A$
• do 9C yayayayayayayayayay

how to use the compound angle formulas §

• determine exact values for sin 15
• sin 15 = sin (45 - 30!!!!!!!!!!!!!)
• find exact value of sin (A+ B) = sinAcosB + cosAsinB :DDDDD yay
• since A is acute, sin A = 12/13 (for another question)

9D sum of trigonometric functions §

• very important later on in the year 12 course apparently. (auxilliary angle)
• consider $y=2\sin x$ and $y=3\sin x$
• what does the graph of $y=2\sin x+3$ look like?
  • 
• wave interference / superposition oh my god!!!!!!!
• very easy what youd get if you add these two graphs, but it becomes complicated when two graphs with different phase shifts get added!
• consider a sine and cosine graph:
• maximum is $2\times \frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$
• $y=\sin (x)+\cos (x)$
• $y=\sqrt{2}\sin \left(x+\frac{\pi }{4}\right)$
• how to prove that \sin x+\cos x=\sqrt{ 2 } \sin \left( x + \frac{\pi}{4} \right)$$$$\sin x + \cos x = \sqrt{ 2 } \left( \frac{1}{\sqrt{ 2 } } \sin x+\frac{1}{\sqrt{ 2 }} \cos x\right)$$$$=\sqrt{ 2 }\left( \cos \frac{\pi}{4} \sin x + \sin \frac{\pi}{4} \cos(x) \right)$$$$=\sqrt{ 2 } \left( \sin x \cos \frac{\pi}{4} +\cos x\sin \frac{\pi}{4} \right)$$$$=\sqrt{ 2 }\sin\left( x+\frac{\pi}{4} \right)
• general version, if you dont know sqrt(2)a\sin x+b \cos x = \sqrt{ a^2+b^2 } \left( \frac{a}{\sqrt{ a^2 + b^2 }} \sin x + \frac{b}{\sqrt{ a^2 + b^2 }} \cos x \right)$$$$=\sqrt{ a^2+b^2 } (\cos \alpha \sin x + \sin \alpha \cos x)$$$$=\sqrt{ a^2 + b^2 } \sin(2x+\alpha) (?)