IB Math SL – Trigonometry

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\Leftrightarrow 2cos^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}

 

Determine an Unknown Length Using the Law of Sines


Application of the Law of Cosines