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IB Mathematics AI SL Use of sine, cosine and tangent ratios MAI Study Notes - New Syllabus

IB Mathematics AI SL Use of sine, cosine and tangent ratios MAI Study Notes

LEARNING OBJECTIVE

  • Use of sine, cosine and tangent ratios to find the sides and angles of right-angled triangles

Key Concepts: 

  • Pythagoras & Right-Angled Trigonometry

MAI HL and SL Notes – All topics

USE OF SINE, COSINE AND TANGENT RATIOS

For any right–angled triangle:

 

We define the sine, the cosine, and the tangent of angle \( \theta \) by:

$ \sin\theta = \frac{b}{a} = \frac{\text{opposite}}{\text{hypotenuse}} $
$ \cos\theta = \frac{c}{a} = \frac{\text{adjacent}}{\text{hypotenuse}} $
$ \tan\theta = \frac{b}{c} = \frac{\text{opposite}}{\text{adjacent}} $

It also holds: Pythagoras’ theorem: \( a^2 = b^2 + c^2 \).

$\begin{aligned} & \sin B=\frac{4}{5} \\ & \cos B=\frac{3}{5} \\ & \tan B=\frac{4}{3}\end{aligned}$
$ 5^2 = 3^2 + 4^2 \quad (\text{both sides give } 25). $

Every angle has a fixed sine, cosine, and tangent.

$\text{Step-by-step GDC instructions}$

Ensure the calculator is in Degree mode:

TI-84: Press `MODE` → scroll to `Angle` → select `Degree`.
TI-Nspire: Press `Home` → `Settings` → `Document Settings` → set Angle to `Degree`.
Casio: Press `SHIFT` → `MODE (Setup)` → choose `3: Degree`.

Calculate $\sin(30^\circ)$

Enter: sin(30)
Press ENTER or =
Result:

$\sin(30^\circ) = 0.5 = \frac{1}{2}$

Calculate $\cos(30^\circ)$

Enter: cos(30)
Press ENTER or =
Result:

$\cos(30^\circ) \approx 0.8660254038$

This equals:

$\cos(30^\circ) = \frac{\sqrt{3}}{2} \approx \frac{1.73205}{2} \approx 0.8660$

Calculate $\tan(30^\circ)$

Enter: tan(30)
Press ENTER or =
Result:

$\tan(30^\circ) \approx 0.5773502692$

This equals:

$\tan(30^\circ) = \frac{1}{\sqrt{3}} \approx \frac{1}{1.73205} \approx 0.5774$
And:

$\frac{\sqrt{3}}{3} \approx \frac{1.73205}{3} \approx 0.5774$

If we know the sine, cosine, or tangent of an acute angle \( \theta \) (i.e., \( \theta < 90^\circ \)), we can find the angle \( \theta \) by using the inverse functions of our GDC:

$ \sin^{-1}, \quad \cos^{-1}, \quad \text{and} \quad \tan^{-1}. $

For example:

If we know that \( \sin\theta = \frac{1}{2} \), then \( \sin^{-1}\frac{1}{2} = 30^\circ \).

Example

In the triangle

Find all the angle (A,B,C) and theta.

,

▶️Answer/Explanation

Solution:

$ \sin B = \frac{4}{5}, \quad \cos B = \frac{3}{5}, \quad \tan B = \frac{4}{3}. $
Therefore:
$ B = \sin^{-1}\frac{4}{5} = 53.1^\circ. $
Notice that \( \cos^{-1}\frac{3}{5} \) and \( \tan^{-1}\frac{4}{3} \) give the same result.
Then:
$ C = 90^\circ – 53.1^\circ = 36.9^\circ. $

For two angles \( A \) and \( B \):

 \( A \) and \( B \) are called complementary if \( A + B = 90^\circ \).
 \( A \) and \( B \) are called supplementary if \( A + B = 180^\circ \).

SIN, COS, TAN for basic angles: \( 0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ \).

Although we can find these values using a GDC, it is worth mentioning a practical way for their calculation:

For \( \cos\theta \), we obtain the same values in the opposite order.
For \( \tan\theta \), we simply divide \( \sin\theta \) by \( \cos\theta \).

REMARKS:

\( \sin\theta, \cos\theta, \tan\theta, \cot\theta \) are also defined for obtuse angles (\( \theta > 90^\circ \)).
At the moment, it is enough to know that supplementary angles have:
Equal sines but opposite cosines.

Example:

$ \sin 30^\circ = \frac{1}{2}, \quad \sin 150^\circ = \frac{1}{2} $
$ \cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \cos 150^\circ = -\frac{\sqrt{3}}{2} $
 The values of \( \sin\theta \) and \( \cos\theta \) range between \(-1\) and \(1\).

THE SINE RULE AND THE COSINE RULE

For any triangle, two rules always hold:

SINE RULE:

$ \boxed{\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} }$

COSINE RULE:

$\boxed{ a^2 = b^2 + c^2 – 2bc \cos A }$

$ \boxed{b^2 = c^2 + a^2 – 2ca \cos B }$

$\boxed{c^2 = a^2 + b^2 – 2ab \cos C}$

Another Form:

$\boxed{\cos A = \frac{b^2 + c^2 – a^2}{2bc}}$

$\boxed{\cos B = \frac{c^2 + a^2 – b^2}{2ca}}$

$\boxed{\cos C = \frac{a^2 + b^2 – c^2}{2ab}}$

Example

Consider the following triangle:

Use sine and cosine rule.

▶️Answer/Explanation

Solution:

We confirm by GDC that the SINE RULE holds:
$ \frac{4}{\sin 104.5^\circ} \approx 4.13 $
$ \frac{3}{\sin 46.6^\circ} \approx 4.13 $
$ \frac{2}{\sin 28.9^\circ} \approx 4.13 $

We can also confirm the three versions of the COSINE RULE:
$ 4^2 = 3^2 + 2^2 – 2(3)(2) \cos 104.5^\circ $
(LHS = 16, RHS = 16)
$ 3^2 = 2^2 + 4^2 – 2(2)(4) \cos 46.6^\circ $
(LHS = 9, RHS = 9)
$ 2^2 = 4^2 + 3^2 – 2(4)(3) \cos 28.9^\circ $
(LHS = 4, RHS = 4)

Consider the following right-angled triangle:

Then:

$ \frac{a}{\sin 90^\circ} = \frac{b}{\sin B} = \frac{c}{\sin C} \Rightarrow a = \frac{b}{\sin B} = \frac{c}{\sin C} $

and so:

$ \sin B = \frac{b}{a} \quad \text{and} \quad \sin C = \frac{c}{a} $ as expected by the definition of \( \sin \theta \).

Also, \( a^2 = b^2 + c^2 – 2bc \cdot \cos 90^\circ \) implies:

$ a^2 = b^2 + c^2 $

which is the Pythagoras’ theorem, since \( \cos 90^\circ = 0 \).

Moreover:

$ b^2 = c^2 + a^2 – 2ca \cdot \cos B \Rightarrow b^2 = c^2 + (b^2 + c^2) – 2ca \cdot \cos B $
$ \Rightarrow -2c^2 = -2ca \cdot \cos B $
$ \cos B = \frac{c}{a} $as expected by the definition of \( \cos \theta \).

Similarly, we get \( \cos C = \frac{b}{a} \).

Consequently:

 The SINE RULE generalizes the definition of \( \sin \theta \).
 The COSINE RULE generalizes the definition of \( \cos \theta \) and Pythagoras’ theorem.

THE AREA OF A TRIANGLE

Consider the following (not right-angled) triangle:

The area of the triangle is given by:

$ \text{Area} = \frac{1}{2} bc \sin A $

Notice that two sides and an included angle are involved in the formula! We can derive two similar versions for this formula:

$ \text{Area} = \frac{1}{2} ab \sin C $
$ \text{Area} = \frac{1}{2} ac \sin B $

Example

Find the Area of triangle.

▶️Answer/Explanation

Solution:

$ \text{Area} = \frac{1}{2} \cdot 2 \cdot 3 \cdot \sin 104.5^\circ \approx 2.90 $
The other two versions give the same result:
$ \text{Area} = \frac{1}{2} \cdot 2 \cdot 4 \cdot \sin 46.6^\circ \approx 2.90 $
$ \text{Area} = \frac{1}{2} \cdot 3 \cdot 4 \cdot \sin 28.9^\circ \approx 2.90 $

THE SOLUTION OF A TRIANGLE

Any triangle has 6 basic elements: 3 angles and 3 sides.
If we are given any 3 among those 6 elements (except 3 angles!), we are able to find the remaining 3 elements by using the sine rule or the cosine rule appropriately.

Example(given three sides)

Find angle $C$

▶️Answer/Explanation

Solution:

We use COSINE RULE:
$ 4^2 = 2^2 + 3^2 – 12 \cos A $
$ \Rightarrow 3 = -12 \cos A $
$ \Rightarrow \cos A = -0.25 $
$ \Rightarrow A = 104.5^\circ $

$\begin{aligned} 3^2=2^2 & +4^2-16 \cos B \\ & \Rightarrow-11=-16 \cos B \\ & \Rightarrow \cos B=0.6875 \\ & \Rightarrow B=46.6^{\circ}\end{aligned}$

Finally:
$ C = 180^\circ – A – B = 180^\circ – 104.5^\circ – 46.6^\circ, $
Thus:
$ C = 28.9^\circ. $

Notice: We may sometimes have no solutions at all. For example, if \( a=1.0 \), \( b=3 \), \( c=2 \), it is not possible to construct such a triangle! Indeed, the cosine rule gives us \( \cos A = -7.25 \), which is not possible!

Example (given two sides and an included angle)

Find angle $B$ and $C$

 

▶️Answer/Explanation

Solution:

We use COSINE RULE:
$ BC^2 = 2^2 + 3^2 – 12 \cos 104.5^\circ = 16 $
Thus \( BC = 4 \).

Then we know all three sides, and hence \( B \) and \( C \) can be found as above: \( B = 46.6^\circ \) and \( C = 28.9^\circ \).

Example (given one side and two angles)

Find the length $BC$ and $AB$.

▶️Answer/Explanation

Solution:

In fact, we know the third angle as well:
$ C = 180^\circ – A – B = 180^\circ – 104.5^\circ – 46.6^\circ, \text{ thus } C = 28.9^\circ. $

Now we can use the sine rule twice:
$ \frac{3}{\sin 46.6^\circ} = \frac{BC}{\sin 104.5^\circ} \Rightarrow BC = 4 $
$ \frac{3}{\sin 46.6^\circ} = \frac{AB}{\sin 28.9^\circ} \Rightarrow AB = 2 $

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