IB μαθηματικά HL – Trigonometry

IB Mathematics HL – Trigonometry, Cosine Rule

Solve the triangle \(ABC\)using Cosine Rule given that \(\hat{A}=\frac{\pi}{3}\) , \(AC=6\) and \(AB=8\).

Solution

Applying the Cosine Rule we have that

\((BC)^2=(AB)^2+(AC)^2-2(AB)(AC)cos \hat{A}\Rightarrow \) \((BC)^2=8^2+6^2-2\cdot8\cdot6cos \frac{\pi}{3}\Rightarrow \) \((BC)^2=100-96\cdot \frac{1}{2}\Rightarrow \) \((BC)^2=100-48=52\Rightarrow BC=\sqrt{52}\)

Now, in order to find another angle, for example \(\hat{C}\), we could apply again cosine rule and solve for \(cos\hat{C}\) as following

\((AB)^2=(AC)^2+(BC)^2-2(BC)(AC)cos \hat{C} \Rightarrow \) \(8^2=6^2+52-2(\sqrt{52})(6)cos \hat{C} \Rightarrow \) \(64=36+52-12(\sqrt{52})cos \hat{C} \Rightarrow \) \(-24=-12(\sqrt{52})cos \hat{C} \Rightarrow \) \(2=\sqrt{52} cos \hat{C}\Rightarrow\) \(cos \hat{C}=\frac{2}{\sqrt{52}}\Rightarrow \) \( \hat{C}=arcos (\frac{2}{\sqrt{52}}) \)

Finally, for the angle \(  \hat{B}\) we have that

\(  \hat{B}=\pi -\hat{A} -\hat{C} \)

 

Trigonometry,Trigonometric equations, IB Math HL

 

Solve the following trigonometric equation

\(cos4x=-cosx\) in the interval \([0,\pi]\).

 

Solution

\(cos4x=-cosx \Rightarrow cos4x=cos(\pi-x) \Rightarrow \)

From unit circle we have the following:

\(4x=\pi-x \Rightarrow 5x=\pi \Rightarrow x=\frac{\pi}{5}\) accepted

or  \(4x=- (\pi-x) \Rightarrow 3x=-\pi \Rightarrow x=-\frac{\pi}{3}\) rejected

or \(4x= 2\pi –(\pi-x) \Rightarrow3x=\pi\Rightarrow x=\frac{\pi}{3}\) accepted

or \(4x= 2\pi+\pi-x \Rightarrow 5x=3\pi \Rightarrow x=\frac{3\pi}{5}\) accepted

or  \(4x= 4\pi –(\pi-x)\Rightarrow 3x=3\pi \Rightarrow x=\pi\)accepted

or \(4x= 4\pi+\pi-x \Rightarrow 5x=5\pi \Rightarrow x=\pi \) accepted

or  \(4x= 6\pi –(\pi-x)\Rightarrow 3x=5\pi \Rightarrow x=\frac{5\pi}{3} \)rejected

Therefore, the solutions are

\(x=\frac{\pi}{5} , \frac{\pi}{3} , \frac{3\pi}{5} ,\pi\)

 

Trigonometry, Trigonometric equations, IB Math HL

 

Solve the following trigonometric equation

\(tan^2x=1\) in the interval \([0,2\pi]\).

 

Solution

\(tan^2x=1 \Rightarrow tanx=1\ or tanx=-1 \)

First we’ll solve the trigonometric equation \(tanx=1\)

\(tanx=1 \Rightarrow x=\frac{\pi}{4} \ or x=\pi+\frac{\pi}{4}=\frac{5\pi}{4}\)

and then the other trigonometric equation \(tanx=-1\).

\(x=\frac{3\pi}{4} \ or x=2\pi-\frac{\pi}{4}=\frac{7\pi}{4}\)

Therefore, the solutions are

\(x=\frac{\pi}{4} , \frac{5\pi}{4} , \frac{3\pi}{4} ,\frac{7\pi}{4}\)

Trigonometry, Trigonometric formulas, IB Math HL

IB Mathematics HL – Trigonometry, Trigonometric identities, double angle formula

 

Simplify the following trigonometric expression

\(\frac{sin4x(1-cos2x)}{cos2x(1-cos4x)}\)

 

Solution

 

\(\frac{sin4x(1-cos2x)}{cos2x(1-cos4x)}\)

We’ll use the following double angle formulas:

\(sin2x=2sinxcosx\) and \(sin^2x=\frac{1}{2}(1-cos2x)\)

So, the expression can be written as

\(\frac{sin4x(1-cos2x)}{cos2x(1-cos4x)}=\) \(=\frac{2sin2xcos2x\cdot 2sin^2x }{cos2x \cdot 2sin^2(2x) }=\) \(=\frac{2sin^2x }{sin(2x) }=\frac{2sin^2x }{2sinxcosx }= \) \(=\frac{sinx }{cosx }=tanx \)