# 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

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