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Pythagorean Theorem Literal equations Right trianglesHere you will learn about Trigonometry including how to use right triangle trigonometry(SOHCAHTOA), inverse trigonometric functions, and exact trigonometric values. You will also learn about the sine rule, the cosine rule, how to find the area of a triangle using \cfrac{1}{2} \, ab \sin(C) and lastly, verify trigonometric identities.
Students first learn about trigonometry in algebra II and expand their knowledge as they progress through precalculus and calculus.
Trigonometry is a branch of mathematics that focuses on triangles. Specifically, the relationship between the angles and sides of triangles. Trigonometry is derived from the Greek word βtrigononβ meaning triangle and βmetronβ meaning to measure.
There are three main functions of trigonometry.
However, in higher levels of mathematics, the use of trigonometry extends beyond triangles, in particular, when verifying trigonometric identities, the unit circle and the graphs of trigonometric functions.
Trigonometric relationships are crucial for higher levels of mathematics as well as physics, engineering, astronomy, and even in art fields such as music.
Use this quiz to check your grade 9 to 12 studentsβ understanding of Trigonometry. 15+ questions with answers covering a range of 9th to 12th grade trigonometry topics to identify areas of strength and support!
DOWNLOAD FREEUse this quiz to check your grade 9 to 12 studentsβ understanding of Trigonometry. 15+ questions with answers covering a range of 9th to 12th grade trigonometry topics to identify areas of strength and support!
DOWNLOAD FREESOHCAHTOA is a mnemonic device that gives you an easy way to remember the three main trigonometric ratios of the sides of a right triangle.
The trig ratios are used to find the missing sides of the triangle (right triangle) and the missing acute angles.
The names of the three ratios are:
When writing the trigonometric ratios, you must write them with respect to one of the acute angles of the right triangle. Letβs use Angle A to write the ratios.
The Sine ratio with respect to acute angle A=\sin A=\frac{ \text{opposite side length}} {\text{hypotenuse length}}
The Cosine ratio with respect to acute angle A=\cos A=\frac{\text{adjacent side length}}{\text{hypotenuse length}}
The Tangent ratio with respect to acute angle A=\tan A=\frac{\text{opposite side length}}{\text{adjacent side length}}
For example, letβs use SOHCAHTOA to find the missing side length of the triangle below.
Using the angle marked as 23^{\circ} and the hypotenuse, to find x, use the cosine function because cosine is \frac{\text{adjacent side length}}{\text{hypotenuse length}}.
The equation to solve for x is:
\cos 23=\cfrac{x}{14}Solve for x by multiplying both sides of the equation by 14.
14 \times \cos 23=xThen, use a calculator to find the value of x (round the answer to the nearest tenth).
x=12.9You can use SOHCAHTOA to calculate lengths and angles in 2D figures (like the example above) and 3D shapes by recognizing right-angle triangles.
For example, we can find the length AC of a parallelogram or the length AH in the cuboid below.
Step-by-step guide: SOHCAHTOA
The sine, cosine, and tangent are the main trigonometric functions. However, there are three other trigonometric functions that are used to find missing values.
These functions are directly related to the sine, cosine and tangent functions.
Letβs look at the triangle below to write the trigonometric ratios with respect to the angle marked as \theta.
Using SOHCAHTOA:
\sin \theta=\cfrac{4}{5} \cos \theta=\cfrac{3}{5} \tan \theta=\cfrac{4}{3}The cosecant, secant, and cotangent are reciprocals of the sine, cosine, and tangent, respectively.
\begin{aligned}& \sin \theta=\cfrac{4}{5} \rightarrow \csc \theta=\cfrac{5}{4} \\\\ & \cos \theta=\cfrac{3}{5} \rightarrow \sec \theta=\cfrac{5}{3} \\\\ & \tan \theta=\cfrac{4}{5} \rightarrow \cot \theta=\cfrac{5}{4} \end{aligned}Step-by-step guide: Trig functions
Inverse trigonometric functions allow us to calculate the size of an angle in a right-angle triangle. The inverse trig functions are expressed as this:
For example, letβs look at the triangle below.
Instead of solving for a side measurement, you need to solve for the angle, \theta. In order to do this, you need to use the inverse trig function. The given side values represent the opposite side and the adjacent side (with respect to \theta ).
The equation to find \theta is:
\tan \, \theta=\cfrac{5}{12}In order to solve for \theta, you must use the inverse tan function (\tan ^{-1})
\theta=\tan ^{-1}\left(\cfrac{5}{12}\right)Type this into a calculator to find the value of \theta.
\theta=\tan ^{-1}\left(\cfrac{5}{12}\right) \approx 22.6199Rounding the answer to the nearest tenth, \theta=22.6^{\circ}.
Step-by-step guide: Sin cos tan
Exact trigonometric values are found when the relationships between the sides and the angles in a triangle have exact values.
Here are the exact trigonometric values for sine, cosine and tangent for angles in the first quadrant, 0^{\circ} \leqslant \theta \leqslant 90^{\circ}.
However, you can use those values to find the exact values of trig functions where the angles are not in the first quadrant by using the reference angle.
Also, the sign value of the angles can be found by the following:
Using the information above, letβs find the exact value of \cos 150^{\circ}
150^{\circ} is a second quadrant angle. So, to find the reference angle subtract 150 from 180.
180-150=30The reference angle to 150^{\circ} is 30^{\circ} , which means the \cos 150^{\circ} behaves similarly to \cos 30^{\circ}.
You know that the exact value of \cos 30^{\circ}=\cfrac{\sqrt{3}}{2}, you also know that the cosine is not positive in Quadrant II. So, the exact value of \cos 150^{\circ}=\cfrac{\sqrt{3}}{2}.
Step-by-step guide: How to find the exact value of a trig function
Step-by-step guide: Trig table
The ratios of the sides of a right triangle, SOHCAHTOA, are used to find missing angles and sides of only right triangles. So when the triangle is not a right triangle, you have to rely on other strategies to find missing sides and angles.
Through elementary school and middle school you learn how to find the area of a triangle using the formula, A=\cfrac{1}{2} \, b h.
Now that you know how to use trigonometric functions, there is another way to find the area of a triangle.
\begin{aligned}& A=\cfrac{1}{2} \, a b \sin (C) \\\\ & A=\cfrac{1}{2} \, a c \sin (B) \\\\ & A=\cfrac{1}{2} \, b c \sin (A) \end{aligned}Letβs use the new formula to calculate the area of triangle ABC.
Since Angle C is given, use this formula:
A=\cfrac{1}{2} \, ab \sin(C)Substitute in the given values.
\begin{aligned}A&=\cfrac{1}{2} \, ab\sin(C) \\\\ A&=\cfrac{1}{2} \, (12\times7\times\sin(77)) \\\\ A&=\cfrac{81.84708544}{2} \\\\ A&=40.92 \, cm^{2}\quad(2dp) \end{aligned}The area of triangle ABC rounded to the nearest hundredth is 40.92 \mathrm{~cm}^2.
Step-by-step guide: Trig formula for area of a triangle
Trigonometric identities are mathematical equations that involve trigonometric functions, like sine, cosine, and tangent and they are true for all values of the variables involved.
An identity is an equation that is always true.
Letβs verify that the following identity is true:
\cfrac{\sin (\theta)}{\cos (\theta)} \equiv \tan (\theta)When verifying an identity choose one of the equations to work with and leave the other side alone. There are multiple ways to verify a trig identity. In this case, letβs use the SOHCAHTOA relationships.
\begin{aligned}& \sin \theta=\cfrac{o}{h} \\\\ & \cos \theta=\cfrac{a}{h}\end{aligned}Rewriting the left hand side of the identity to be,
\begin{aligned}& \cfrac{\sin (\theta)}{\cos (\theta)} \equiv \tan (\theta) \\\\ & \cfrac{\cfrac{o}{h}}{\cfrac{a}{h}}=\tan \theta \\\\ & \cfrac{o}{h} \times \cfrac{h}{a}=\tan \theta \end{aligned}\cfrac{o}{a}=\tan \theta (This is true, the tangent ratio is equal to the opposite side divided by the adjacent side)
Step-by-step guide: Trig identities
Step-by-step guide: Trig formulas
How does this apply to high school math?
If you want more detailed steps and practice for solving for the sides and angles of right triangles and non-right triangles, finding the exact value of trig functions, and working with trig identities and formulas, check out the links highlighted above or take a look at the examples below.
Find the measure of side a.
Since you need to solve for the adjacent side and you are given the opposite side, use the tangent ratio.
\text { tangent ratio }=\cfrac{\text { opposite side }}{\text { adjacent side }}2Write an equation to find the missing part.
\tan 68=\cfrac{21}{a}3Solve for the missing part.
In this case, multiply both sides by a.
a \times \tan 68=21Then divide by \tan 68.
a=\cfrac{21}{\tan 68}Calculate the value of a with a calculator.
a=8.484551Round the answer to the nearest hundredth, 8.48 \, m.
Find the measure of angle \theta.
Identify the trig ratio to use.
You are given the opposite side and the hypotenuse, so use the sine ratio.
\text { sine ratio }=\cfrac{\text { opposite side }}{\text { hypotenuse }}
Write an equation to find the missing part.
Solve for the missing part.
Use the inverse sine function to find the value of \theta.
\begin{aligned}& \sin \theta=\cfrac{8}{20} \\\\
& \theta=\sin ^{-1}\left(\cfrac{8}{20}\right) \end{aligned}
Type into a calculator to get the value of \theta.
\theta=23.5782 (round the answer to the nearest hundredth)
\theta=23.58^{\circ}
Find the value of angle A.
Determine whether to use the sine rule or the cosine rule.
In this case, you are given angle C , side c, and side a. Since you need to solve for angle A, use the sine rule (Law of Sines).
Itβs also a SSA set up which lends itself to the law of sines.
Write the equation.
Solve the unknown part.
\cfrac{\sin A}{15}=\cfrac{\sin 72}{20} \quad (cross multiply)
\begin{aligned}& 20 \times \sin A=15 \times \sin 72 \\\\
& \sin A=\cfrac{15 \times \sin 72}{20} \\\\
& \sin A=0.713292 \end{aligned}
Use the inverse sine function to find angle A.
\text { Angle } A=45.50^{\circ}
Find the exact value of \sin 330^{\circ} .
Determine the quadrant and the sign of the value.
330^{\circ} is a 4 th quadrant angle and the sine has a negative value in the 4 th quadrant.
Find the reference angle.
To find the reference angle when the original angle is in the 4 th quadrant, subtract the angle from 360.
360-330=30
The reference angle is 30^{\circ}.
Find the value of the trig function of the reference angle.
Find the value of the trig function of the given angle.
Since the sine is negative in the 4 th quadrant and the \sin 30=\cfrac{1}{2}, then \sin 330=- \, \cfrac{1}{2}.
Find the area of the triangle below.
Label the sides of the triangle.
Substitute values into the formula.
Since you are given the values of side a, b, and angle C, use A=\cfrac{1}{2} \, b c \sin C
A=\cfrac{1}{2}(18)(12) \sin 79
Calculate the area.
Verify the identity.
\sin \theta \cot \theta \sec \theta=1Select which side of the identity to simplify.
In this case, use the left hand side of the identity to simplify.
Simplify one side of the identity.
In this case, use the simplifying strategy by changing all trig functions to sine and cosine by using the following:
1. Calculate the length of the hypotenuse.
Use SOHCAHTOA to find the missing side length. Label the sides with respect to angle B.
Use the cosine ratio since you have the adjacent side length and need to find the hypotenuse length.
\begin{aligned}& \cos \theta=\cfrac{\text { adjacent }}{\text { hypotenuse }} \\\\ & \cos 52=\cfrac{11}{H} \\\\ & H=\cfrac{11}{\cos (52)} \\\\ & H=17.87 \mathrm{~cm}\end{aligned}
2. Using your knowledge of exact trigonometric values, to find the value of \theta.
Use SOHCAHTOA to find the missing angle. Label the sides with respect to \theta .
Since you have the value of the opposite side and the adjacent side, use the tangent ratio.
\begin{aligned}\tan (\theta) & =\cfrac{\text { Opposite }}{\text { Adjacent }} \\\\ \tan (\theta) & =\cfrac{\sqrt{3}}{1} \\\\ \tan (\theta) & =\sqrt{3} \end{aligned}
Using the exact trigonometric values,
\tan (\theta)=\sqrt{3} when Β \theta=60^{\circ} \text {.}
3. Find angle \theta.
Since it is not a right triangle, you have to use either the law of sines or the law of cosines.
The values given are in a SSA set up, so you can use the law of sines.
\begin{aligned}\cfrac{\sin(A)}{a}&=\cfrac{\sin(B)}{b} \\\\ \cfrac{\sin(\theta)}{13}&=\cfrac{sin(70)}{15} \\\\ \sin(\theta)&= \cfrac{sin(70)}{15} \times 13 \\\\ \sin(\theta)&=0.8144002713 \\\\ \theta &=\sin^{-1}(0.8144002713) \\\\ \theta&=54.53^{\circ}\end{aligned}
4. Calculate the area of the following triangle:
Using trigonometry, find the area of the triangle using,
A=\cfrac{1}{2} \, ab \sin(C)
\begin{aligned}A&=\cfrac{1}{2} \times 12 \times 28 \times \sin(57) \\\\ A&= 140.90 \mathrm{~cm}^{2} \end{aligned}
5. Find the exact value of \cos 240^{\circ}.
240^{\circ} is in the third quadrant. The cosine value is always negative in the third quadrant.
To find the reference angle of an angle in the third quadrant, subtract 180 from the original angle.
240-180=60
The reference angle is 60^{\circ}.
\cos 60=\cfrac{1}{2}
So, since the cosine is negative in the third quadrant, \cos 240=- \, \cfrac{1}{2}.
6. Find the exact value of \csc 120^{\circ}.
120^{\circ} is an angle in the second quadrant. The csc is the reciprocal of the sin so it has a positive value in the second quadrant.
To find the reference angle in the second quadrant, subtract the original angle from 180.
180-120=60
The reference angle is 60^{\circ}.
\csc 60=\cfrac{2}{\sqrt{3}} which simplifies to be \cfrac{2 \sqrt{3}}{3}
Since the csc is positive in the second quadrant, \csc 120=\cfrac{2 \sqrt{3}}{3}.
Radians are a unit of measure of an angle.
The angle of elevation is an angle that rises up from the horizontal line.
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