IB Maths HL – Trigonometry

Questions and Answers

1. IB Mathematics HL– Trigonometry, Arc Length

How can we find the length of an arc with central angle $\frac{\pi}{3}$ in a circle with radius 4 cm?

Solution

Let $\theta$ (in radians) be the central angle in a circle with radius r, then the length $(l)$ of the arc cut off by $\theta$ is given by the following formula:

$l=r\theta$

Therefore the arc length in our case is

$l=4 \frac{\pi}{3} = \frac{4 \pi}{3} cm$

2. IB Mathematics HL– Trigonometry, Area of a sector

How can we find the area of a sector with central angle $\frac{\pi}{3}$ in a circle with radius 4 cm?

Solution

IB Maths HL – Trigonometry, Area of a sector

Let $\theta$ (in radians) be the central angle in a circle with radius r, then the area $(A)$ of the sector formed by angle $\theta$ is given by the following formula:

$A=\frac{1}{2}r^2 \theta$

Therefore the arc length in our case is

$A=\frac{1}{2}(4)^2 \frac{\pi}{3} =\frac{8\pi}{3} cm^2$

3. IB Maths HL Trigonometric equations

How can we solve the following trigonometric equation?

$2sinx=\sqrt 2$ for $- \pi \leq x \leq \pi$

Solution

Your primary goal in solving a trigonometric equation is to isolate the trigonometric function in the equation.

$2sinx=\sqrt 2 \Leftrightarrow sinx=\frac{\sqrt 2}{2}$

The above equation has two solutions in the interval $[-\pi , \pi]$

$x=\frac{\pi}{4} \ or\ x=\pi-\frac{\pi}{4}=\frac{3\pi}{4}$

4. IB Mathematics HL– Trigonometry, Trigonometric equations

How can we solve the following trigonometric equation?

$2sin^2x=sinx+1$ for $- \pi \leq x \leq \pi$

Solution

IB Maths HL – Trigonometry, Trigonometric equations

By treating the equation as a quadratic in $sinx$ and then factoring.

$2sin^2x=sinx+1\Leftrightarrow 2sin^2x-sinx-1=0 \Leftrightarrow$

$\Leftrightarrow (2sinx+1)(sinx-1)=0 \Leftrightarrow sinx-1=0 \ or \ 2sinx+1 =0 \Leftrightarrow$

Now, you have to solve two trigonometric equations

$sinx-1=0$ and $2sinx+1 =0$ in the interval $[-\pi , \pi]$

For the first equation we have:

$sinx-1=0\Leftrightarrow sinx=1$

The above equation has only one solution in the interval $[-\pi , \pi]$

$x=\frac{\pi}{2}$

For the second equation we have:

$2sinx+1 =0 \Leftrightarrow sinx=-\frac{1}{2}$

The above equation has two solutions in the interval $[-\pi , \pi]$

$x=-\frac{\pi}{6} \ or\ x=-\pi+\frac{\pi}{6}=-\frac{5\pi}{6}$

5. IB Mathematics HL – Trigonometry, Sine Rule

How can we find the side $a$ of a triangle ABC given that $\hat{A}=30^{\circ} , c=4 \ and \hat{C}=60^{\circ}$ ?

Solution

IB Maths HL – Trigonometry, Sine Rule

The Sine rule states

$\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}$

So in your case the side $a$ can be calculated as follows:

$\frac{a}{sin30^{\circ}}=\frac{4}{sin60^{\circ}}$

$a=\frac{1}{2} \frac{4}{\frac{sqrt 3}{2}}=\frac{4}{ sqrt 3}$

6. IB Mathematics HL – Trigonometry, Cosine Rule

How can we find the side $a$ of a triangle ABC given that $\hat{A}=60^{\circ} , c=4 \ and \ b=3$ ?

Solution

IB Maths HL – Trigonometry, Cosine Rule

The Cosine rule states

$a^2=b^2+c^2-2(bc)cosA$

So in your case the side $a$ can be calculated as follows:

$a^2=3^2+4^2-2(3(4))cos60^{\circ}$

$a^2=9+16-24\frac{1}{2}=25-12=13$

$a=\sqrt{13}$

7. IB Mathematics HL – Trigonometry, Area of a triangle

How can we find the area of a triangle ABC with sides $a=3 \ cm, b=4 \ cm$

and included angle $\hat{C}=30^{\circ}$ ?

Solution

IB Maths HL – Trigonometry, Area of a triangle

The area of a triangle is a half of the product of two sides and the sine of the included angle.

$A=\frac{1}{2}absinC$

Therefore, in this case, we have the following

$A=\frac{1}{2}(3)(4)sin30^{\circ} =6 \frac{1}{2}=3 \ cm^2$

8. IB Mathematics HL – Trigonometry, Trigonometric identities

How can we prove the following trigonometric equality?

$secx-secxsin^2x=cosx$

Solution

IB Maths HL – Trigonometry, Trigonometric identities

$secx-secxsin^2x=cosx$

L.H.S $\frac{1}{cosx}-\frac{sin^2x}{cosx}=\frac{1-sin^2x}{cosx}=$

$=\frac{cos^2x}{cosx}= cosx$ R.H.S

9. IB Mathematics HL – Trigonometry, Cosine Rule

How can we solve the triangle $ABC$ using Cosine Rule given that $\hat{A}=\frac{\pi}{3}$ , $AC=6$ and $AB=8$ .

Solution

IB Math HL – Trigonometry, Cosine Rule

Applying the Cosine Rule we have that

$(BC)^2=(AB)^2+(AC)^2-2(AB)(AC)cos \hat{A} \Rightarrow$

$(BC)^2=8^2+6^2-2\cdot 8\cdot 6 cos \frac{\pi}{3}\Rightarrow$

$(BC)^2=100-96\cdot \frac{1}{2}\Rightarrow$

$(BC)^2=100-48=52 \Rightarrow BC=\sqrt{52}$

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

$\hat{B}=\pi -\hat{A} -\hat{C}$

10. IB Mathematics HL – Trigonometry, Trigonometric equations

How can we solve the following trigonometric equation?

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

Solution

IB Math HL – Trigonometry, Trigonometric equations

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

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 $4x= 2\pi -(\pi-x) \Rightarrow 3x=\pi \Rightarrow x=\frac{\pi}{3}$ 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

$x=\frac{\pi}{5} , \frac{\pi}{3} , \frac{3\pi}{5} , \pi$

11. IB Mathematics HL – Trigonometry, Trigonometric equations, tanx

How can we solve the following trigonometric equation?

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

Solution

IB Math HL – Trigonometry, Trigonometric equations

$tan^2x=1 \Rightarrow tanx=1 \ or tanx=-1$

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

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

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

How can we simplify the following trigonometric expression?

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

Solution

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

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

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

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

$=\frac{2sin2xcos2x \cdot 2sin^2x }{cos2x \cdot 2sin^2(2x) }=$

$=\frac{2sin^2x }{sin(2x) }=\frac{2sin^2x }{2sinxcosx }=$

$=\frac{sinx }{cosx }=tanx$