Which jobs use trigonometry and geometry

Content:
»Sin (x), cos (x) and tan (x) in a right triangle
»Relationships of Trigonometric Functions
»Law of sine
»Cosine law

You will encounter the trigonometric functions sine (\ (sin (x) \)), cos (\ (cos (x) \)) and tangent (\ (tan (x) \)) very often in many mathematical areas. These are the most important trigonometric functions. In this article, we'll look at the geometric statements that relate to right triangles.

sin (x), cos (x) and tan (x) in the right triangle

With the help of these functions we can describe the aspect ratio in a right triangle as a function of one of the angles. However, the right angle must not be used. We have already defined the sides of a triangle. The formulas are therefore defined as follows:

\ begin {align}
& sin (\ alpha) & = && \ dfrac {opposite cathetus} {hypotenuse} \
& cos (\ alpha) & = && \ dfrac {adjacent} {hypotenuse} \
& tan (\ alpha) & = && \ dfrac {opposite side} {adjacent side} \
\ end {align}

If one of the acute angles is given and a side of the triangle is given, the remaining sides can be determined by rearranging the above formulas. Due to the sum of the interior angles, the second acute angle can also be easily determined.
In the following pictures you can see how you can switch over and calculate the rest of a given page.

Let us illustrate this again with a concrete example: Given a right-angled triangle. An angle has the size \ (\ alpha = 40 ° \), the corresponding adjacent cathete has the length \ (A = 4cm \). So how do we calculate the two remaining sides?

solution

First we draw a triangle and label the sizes we are looking for and the given sizes.

Now we look for a formula in which the adjacent as well as the hypotenuse occur, that would be \ (cos (x) \). Then the hyptenuse can be used

\ (H = \ dfrac {A} {cos (\ alpha)} = \ dfrac {4cm} {cos (40 °)} = 5.22cm \)

to calculate. Since we only lack the opposite side, we can choose whether we want to calculate it with the help of the adjacent side or the hypotenuse. If we choose the next one, then we get through

\ (G = tan (\ alpha) \ times A = tan (40 °) \ times 4cm = 3.36cm \)

the length of the opposite cathetus.

Relationships of trigonometric functions

If we take a closer look at the formulas, we can see that sine, cosine and tangent are related to one another in certain ways. To do this, we first draw another right-angled triangle and label it. If one of the acute angles is designated as \ (\ alpha \), we can label the remaining angle as \ (90 ° - \ alpha \) based on the sum of the interior angles.

If we apply sine and cosine, we get
\ begin {align *} sin (90 ° - \ alpha) = \ dfrac {opposite cathetus} {hypotenuse} = \ dfrac {b} {c} \
cos (\ alpha) = \ dfrac {adjacent} {hypotenuse} = \ dfrac {b} {c} \ end {align *}


We can now equate these two equations and obtain them
\ begin {align *} sin (90 ° - \ alpha) = cos (\ alpha) \ end {align *}
This explains the other equations in the same way. So these two formulas also apply:
\ begin {align *} cos (90 ° - \ alpha) = sin (\ alpha) \
tan (90 ° - \ alpha) = \ dfrac {1} {tan (\ alpha)} \ end {align *}
These relationships are called complement relationships. However, there are also the supplement relationships. These are easy to see if we look at the functions in a coordinate system.

 

Sinecosinetangent
sin (180 ° + α) = - sin (α)cos (180 ° + α) = - cos (α)tan (180 ° + α) = tan (α)
sin (180 ° −α) = sin (α)cos (180 ° −α) = - cos (α)tan (180 ° −α) = - tan (α)
sin (360 ° −α) = - sin (α)cos (360 ° −α) = cos (α)tan (360 ° −α) = - tan (α)

 

Sine law

The law of sines and cosines, which we will discuss next, reflect relationships between side lengths and angles in arbitrary triangles. The formulas that we just got to know form the basis for this.

In the picture above we see that the drawn height \ (h_ {c} \) divides the triangle into two right-angled triangles. Now we can apply the sine to both triangles and get:
\ begin {align *}
sin (\ alpha) = \ dfrac {h_ {c}} {b} \
sin (\ beta) = \ dfrac {h_ {c}} {a} \
\ end {align *}

 

If we now convert both equations to \ (h_ {c} \), we can equate them.
\ begin {align *}
h_ {c} = b \ cdot sin (\ alpha) \
h_ {c} = a \ cdot sin (\ beta) \
a \ cdot sin (\ beta) = b \ cdot sin (\ alpha) \
\ dfrac {a} {sin (\ alpha)} = \ dfrac {b} {sin (\ beta)} \ \ end {align *}

 

The law of sines thus states that the ratio between one angle and the opposite side is equal to the ratio of the other angle and the opposite side. Since the derivation behaves the same on each side of the triangle, we can summarize that
\ begin {align *} \ dfrac {a} {sin (\ alpha)} = \ dfrac {b} {sin (\ beta)} = \ dfrac {c} {sin (\ gamma)} \ end {align *}
applies. Let's take a look at a quick example of this. The sides are given with the lengths \ (a = 3cm \), \ (b = 4cm \) and the angle \ (\ alpha = 30 ° \). Now we want to determine the angle \ (\ beta \).

solution

All we have to do is plug the following quantities into the formula. But you should always make sure that the angle and the respective side are really opposite each other, otherwise the statement would be wrong.

\ begin {align *}
&& \ frac {a} {sin (\ alpha)} & = && \ frac {b} {sin (\ beta)} \
& \ Longleftrightarrow & sin (\ beta) & = && \ frac {b} {a} \ cdot sin (\ alpha) \
& \ Longleftrightarrow & sin (\ beta) & = && \ frac {4} {3} \ cdot sin (30 °) \
& \ Longleftrightarrow & sin (\ beta) & = && 0.6667 \
& \ Longleftrightarrow & \ beta & = && arcsin (0.6667) = 41.81 °
\ end {align *}

 

Cosine law

Unlike the sine law, the cosine law expresses a relationship between the three sides and an angle in the triangle. The basic consideration here is again the division into two right triangles.

Now we need the Pythagorean theorem. If we consider the triangle \ (\ triangle BCD \), the following applies to its hypotenuse \ (a \):
\ begin {align *} a ^ {2} = h ^ {2} + (cq) ^ {2} = h ^ {2} + c ^ {2} -2cq + q ^ {2} \ end {align * }
For the third term we used the second binomial formula. The same applies to the second triangle \ (\ triangle ADC \):
\ begin {align *} b ^ {2} = h ^ {2} + q ^ {2} \
h ^ {2} = b ^ {2} -q ^ {2} \ end {align *}
With the last change, we can now insert the term into our first equation.
\ begin {align *} a ^ {2} = b ^ {2} + c ^ {2} -2cq \ end {align *}
Now we still have to use the cosine that we learned at the beginning: \ (cos (angle) = \ dfrac {adjacent} {hypotenuse} \). This results in our angle \ (cos (\ alpha) = \ dfrac {q} {b} \).

If we change that, we get \ (q = b \ cdot cos (\ alpha) \). Now we can replace \ (q \) in the previous formula with this term and get
\ begin {align *} a ^ {2} = b ^ {2} + c ^ {2} -2bc \ cdot cos (\ alpha) \ end {align *}
This is our first formula of the cosine theorem. We get the other two in a similar way:
\ begin {align *} b ^ {2} = a ^ {2} + c ^ {2} -2ac \ cdot cos (\ beta) \
c ^ {2} = a ^ {2} + b ^ {2} -2ab \ cdot cos (\ gamma) \ end {align *}

You may have noticed a certain similarity to the Pythagorean Theorem. This is a special case of the law of cosines, namely when it is a right triangle. Let \ (\ gamma = 90 ° \), then \ (cos (90 °) = 0 \). This results in the Pythagorean theorem with \ (c ^ {2} = a ^ {2} + b ^ {2} \).

Let's look at an example here again. An arbitrary triangle with the side lengths \ (a = 10 \), \ (b = 11 \) and \ (c = 12 \) is given. How do we find out the size of the angle \ (\ alpha \)?

solution

First we have to choose the formula that contains \ (\ alpha \). We then try to convert this to \ (cos (\ alpha) \).
\ begin {align *} a ^ {2} = b ^ {2} + c ^ {2} -2bc \ cdot cos (\ alpha) \
a ^ {2} -b ^ {2} -c ^ {2} = - 2bc \ cdot cos (\ alpha) \
cos (\ alpha) = \ dfrac {a ^ {2} -b ^ {2} -c ^ {2}} {- 2bc} \ \ end {align *}

 

Then only the values ​​have to be inserted and the arccos applied.
\ begin {align *} cos (\ alpha) = \ dfrac {10 ^ {2} -11 ^ {2} -12 ^ {2}} {- 2 \ cdot11 \ cdot12} \
cos (\ alpha) = \ dfrac {-165} {- 264} = \ dfrac {5} {8} \
\ alpha = arccos (\ dfrac {5} {8}) = 51.32 ° \ end {align *}