IB Math SL – Trigonometry

Questions and Answers

1. IB Maths Arc Length

IB Mathematics SL– Trigonometry, Mensuration, Arc Length

How can we find the length of an arc with central angle $\frac{\pi}{6}$ in a circle with radius 7 cm?

Solution

IB Maths SL – Trigonometry, Arc Length

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 $l=7\frac{\pi}{6} = \frac{7 \pi}{6}cm$

2. Area of a sector IB Maths SL

IB Mathematics SL– Trigonometry, Mensuration, Area of a sector

How can we find the area of a sector with central angle $\frac{\pi}{6}$ in a circle with radius 7 cm?

Solution

IB Maths SL– Trigonometry, Mensuration, 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

$A=\frac{1}{2} (7)^2 \frac{\pi}{6} =\frac{49\pi}{12} cm^2$

3. Trigonometric equations IB Maths

IB Mathematics SL– Trigonometry, Trigonometric equations

How can we solve the following trigonometric equation?

$2cosx=\sqrt2$ for $- \pi \leq x \leq\pi$

Solution

IB Maths SL– Trigonometry, Trigonometric equations

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

$2cosx=\sqrt2 \Leftrightarrow cosx=\frac{\sqrt 2}{2}$

The above equation has two solutions in the interval

$[-\pi , \pi]$

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

4. IB Maths SL Trigonometric equations

IB Mathematics SL– Trigonometry, Trigonometric equations

How can we solve the following trigonometric equation?

$2cos^2x=cosx+1$ for $- \pi \leq x \leq \pi$

Solution

IB Maths SL– Trigonometry, Trigonometric equations

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

$2cos^2x=cosx+1\Leftrightarrow2cos^2x-cosx-1=0 \Leftrightarrow$

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

Now, you have to solve two trigonometric equations

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

For the first equation we have:

$cosx-1=0\Leftrightarrow cosx=1$

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

$x=0$

For the second equation we have:

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

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

$x=-\pi+\frac{\pi}{3}=-\frac{2\pi}{3} \ or\ x=\pi-\frac{\pi}{3}=\frac{2\pi}{3}$

5. Sine Rule IB Maths SL

IB Mathematics SL – Trigonometry, Sine Rule

How can we find the side $a$ of a triangle ABC given that

$\hat{A}=30^{\circ}, b=5 , and\ \hat{B}=60^{\circ}$ ?

Solution

IB Maths SL – Trigonometry, Sine Rule

The Sine rule states

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

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

$\frac{a}{sin30^{\circ}}=\frac{5}{sin60^{\circ}}$

$a=\frac{1}{2} \frac{5}{ \frac{\sqrt {3}}{2}}= \frac{5}{\sqrt {3}}$

6. IB Maths SL Cosine Rule

IB Mathematics SL – Trigonometry, Cosine Rule

How can we find the side $a$ of a triangle ABC given that $\hat{A}=60^{\circ}, c=2 \ and\ b=4\$ ?

Solution

IB Maths SL– 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=2^2+4^2-2(2(4))cos60^{\circ}\$

$a^2=4+16-16frac{1}{2}=20-8=12\$

$a=\sqrt{12}\$

7. Area of a triangle IB Maths

IB Mathematics SL – Trigonometry, Area of a triangle

How can we find the area of a triangle ABC with sides $a=4 \ cm, b=5 \ cm$

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

Solution

IB Maths SL – 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

$A=\frac{1}{2}(4)(5)sin30^{\circ}=10 \frac{1}{2}=5 \ cm^2$

8. IB Maths SLTrigonometric identities

IB Mathematics SL – Trigonometry, Trigonometric identities

How can we prove the following trigonometric equality?

$(sinx+cosx)^2+(sinx-cosx)^2=2$

Solution

IB Maths SL–Trigonometry, Trigonometric identities

$(sinx+cosx)^2+(sinx-cosx)^2=2$

L.H.S $(sinx+cosx)^2+(sinx-cosx)^2=$

$=sin^2x+2sinxcosx+cos^2x + sin^2x-2sinxcosx+ cos^2x =$

$=2sin^2x+2cos^2x = 2(sin^2x+cos^2x)=2(1)=2$ R.H.S.

9. IB Mathematics SL – Trigonometry, Cosine Rule

How can we solve the triangle $ABC$ using Cosine Rule given that $\hat{A}=\frac{\pi}{3}$ , $AC=7$ and $AB=9$ .

Solution

IB Math SL – Trigonometry, Cosine Rule

Applying the Cosine Rule we have that

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

$(BC)^2=9^2+7^2-2\cdot 9\cdot 7 cos \frac{\pi}{3}\Rightarrow$

$(BC)^2=130-126\cdot \frac{1}{2}\Rightarrow$

$(BC)^2=67 \Rightarrow BC=\sqrt{67}$

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

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

$9^2=7^2+67-2(\sqrt{67})(7)cos \hat{C} \Rightarrow$

$81=49+67-14(\sqrt{67})cos \hat{C} \Rightarrow$

$-35=-14(\sqrt{67})cos \hat{C} \Rightarrow$

$\frac{35}{14}=\sqrt{67} cos \hat{C} \Rightarrow$

$cos \hat{C}=\frac{35}{14 \sqrt{67}}\Rightarrow$

$\hat{C}=arcos (\frac{35}{14 \sqrt{67}})$

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

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

10. IB Mathematics SL – Trigonometry, Trigonometric equations

How can we solve the following trigonometric equation?

$sin4x=-sinx$ in the interval $[0,\pi]$ .

Solution

IB Math SL – Trigonometry, Trigonometric equations

$sin4x=-sinx \Rightarrow sin4x=sin(-x) \Rightarrow$

From the unit circle we have the following:

$4x= -x \Rightarrow 5x=0 \Rightarrow x=0$ accepted

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

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

or $4x= 2\pi+\pi-(-x) \Rightarrow 3x=3\pi \Rightarrow x=\pi$ accepted

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

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

Therefore, the solutions are

$x=0 , \frac{\pi}{3} , \frac{2\pi}{5} , \frac{4\pi}{5} , \pi$

11. Trigonometry, Trigonometric equations, IB Math SL

IB Mathematics SL – Trigonometry, Trigonometric equations, tanx

How can we solve the following trigonometric equation?

$tan^3x=1$ in the interval $[0,2\pi]$ .

Solution

IB Math SL – Trigonometry, Trigonometric equations

$tan^3x=1 \Rightarrow tanx=1 \Rightarrow$

$\Rightarrow x=\frac{\pi}{4} \ or x=\pi+\frac{\pi}{4}=\frac{5\pi}{4}$

12. Trigonometry, Double angle identities, IB Math SL

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

How can we simplify the following trigonometric expression?

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

Solution

IB Math SL – Trigonometry, Trigonometric identities, double angle formulas

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

In order to simplify the aforementioned expression we are using the following double angle formulas:

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

So, the expression can be written as

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

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

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

$=\frac{cosx }{sinx }=\frac{1}{tanx}$

IB Math SL – Trigonometry

IB Math SL – Trigonometry

Questions and Answers

1. IB Maths Arc Length

2. Area of a sector IB Maths SL

Solution

3. Trigonometric equations IB Maths

4. IB Maths SL Trigonometric equations

5. Sine Rule IB Maths SL

6. IB Maths SL Cosine Rule

7. Area of a triangle IB Maths

8. IB Maths SLTrigonometric identities

9. IB Mathematics SL – Trigonometry, Cosine Rule

10. IB Mathematics SL – Trigonometry, Trigonometric equations

11. Trigonometry, Trigonometric equations, IB Math SL

12. Trigonometry, Double angle identities, IB Math SL

Determine an Unknown Length Using the Law of Sines

Application of the Law of Cosines