IB Math HL – Trigonometry

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