As the functions are periodic, we can calculate the sine, cosine, and tangent of angles outside of this range by adding multiples of 3 6 0 ∘ for s i n and c o s , or by adding multiples of 1 8 0 ∘ for t a n .
Next, we recall that the special angles are 3 0 ∘ , 4 5 ∘ , and 6 0 ∘ . The sine, cosine, and tangent of these angles are given below.
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While we will not consider the derivation or proof of these results in this explainer, it is worth recalling the identity s i n c o s t a n 𝑥 𝑥 = 𝑥 . This enables us to calculate the tangent of any angle if we are given the sine and cosine of that angle.
For example, since s i n 6 0 = √ 3 2 ∘ and c o s 6 0 = 1 2 ∘ , then t a n 6 0 = √ 3 2 ÷ 1 2 = √ 3 ∘ .
It will also be necessary for us to recall the definitions of the reciprocal trigonometric functions.
The reciprocal trigonometric functions cosecant 𝜃 , secant 𝜃 , and cotangent 𝜃 are the reciprocal of sine 𝜃 , cosine 𝜃 , and tangent 𝜃 such that c s c s i n s e c c o s c o t t a n 𝜃 = 1 𝜃 , 𝜃 = 1 𝜃 , 𝜃 = 1 𝜃 .
We can use these identities to calculate the cosecant, secant, and cotangent of 3 0 ∘ , 4 5 ∘ , and 6 0 ∘ .
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We will now recall the related angles of the trigonometric functions:
One way of recalling whether the sine, cosine, and tangent of any angle between 0 ∘ and 3 6 0 ∘ are positive or negative is using the CAST diagram. This is a memory device that we use to remember the signs of the trigonometric ratios in each of the four quadrants.
The Quadrant in Which the Terminal Side of the Angle Lies | The Interval in Which the Measure of the Angle Belongs | Signs of Trigonometric Functions | ||
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, | , | , | ||
First | + | + | + | |
Second | + | |||
Third | + | |||
Fourth | + |
We note that the angles are measured, from 0 ∘ to 3 6 0 ∘ or from 0 to 2 𝜋 radians in a counterclockwise direction, where the positive 𝑥 -axis is the initial side of the angle. The terminal side is where the angle stops. Any angle between 0 ∘ and 9 0 ∘ lies in the first quadrant. Any angle between 9 0 ∘ and 1 8 0 ∘ lies in the second quadrant. Any angle between 1 8 0 ∘ and 2 7 0 ∘ lies in the third quadrant. Any angle between 2 7 0 ∘ and 3 6 0 ∘ lies in the fourth quadrant.
Let us consider an example where we have to evaluate the cosine of an angle using the property of related angles to relate it to a special angle.
Find the value of c o s 1 1 𝜋 6 .
We begin by recalling that 𝜋 = 1 8 0 r a d i a n s ∘ .
So, 𝜋 6 = 3 0 1 1 𝜋 6 = 3 3 0 . r a d i a n s r a d i a n s ∘ ∘
We therefore need to calculate c o s 3 3 0 ∘ .
Let us recall the property of related angles c o s c o s ( 3 6 0 − 𝜃 ) = 𝜃 . ∘
If 𝜃 = 3 0 ∘ , then c o s c o s c o s c o s ( 3 6 0 − 3 0 ) = 3 0 3 3 0 = 3 0 . ∘ ∘ ∘ ∘ ∘
From our knowledge of special angles, we know that c o s 3 0 = √ 3 2 ∘ .
So, c o s 3 3 0 = √ 3 2 . ∘
We can therefore conclude that c o s 1 1 𝜋 6 = √ 3 2 .
Let us now consider a similar example where, this time, we need to evaluate the sine and tangent of given angles.
Evaluate 2 𝜋 6 − 8 4 𝜋 3 t a n s i n .
So, 𝜋 6 = 3 0 . r a d i a n s ∘
Also, 𝜋 3 = 6 0 4 𝜋 3 = 2 4 0 . r a d i a n s r a d i a n s ∘ ∘
We therefore need to calculate 2 3 0 − 8 2 4 0 t a n s i n ∘ ∘ .
From our knowledge of special angles, we know that t a n 3 0 = √ 3 3 ∘ and s i n 6 0 = √ 3 2 ∘ .
By considering the CAST diagram, as shown below, we see that 2 4 0 ∘ lies in the 3rd quadrant and the sine of any angle here is negative.
Since s i n s i n ( 1 8 0 + 𝜃 ) = − 𝜃 , ∘ then s i n s i n s i n s i n ( 1 8 0 + 6 0 ) = − 6 0 ( 2 4 0 ) = − 6 0 . ∘ ∘ ∘ ∘ ∘
So, s i n ( 2 4 0 ) = − √ 3 2 . ∘
Substituting the values of t a n 3 0 ∘ and s i n 2 4 0 ∘ into our expression, we have 2 3 0 − 8 2 4 0 = 2 √ 3 3 − 8 − √ 3 2 = 2 √ 3 3 + 4 √ 3 = 2 √ 3 3 + 1 2 √ 3 3 = 1 4 √ 3 3 . t a n s i n ∘ ∘
Therefore, 2 𝜋 6 − 8 4 𝜋 3 = 1 4 √ 3 3 t a n s i n .
In the remaining examples in this explainer, we will also need to use reciprocal trigonometric functions.
Find the value of c o s t a n c s c c o s 1 3 5 + 1 3 5 + 2 2 5 + 2 2 5 ∘ ∘ ∘ ∘ .
In order to answer this question, we need to recall reciprocal trigonometric identities and special angles.
From our knowledge of special angles, s i n 4 5 = √ 2 2 ∘ , c o s 4 5 = √ 2 2 ∘ , and t a n 4 5 = 1 ∘ .
Using the CAST diagram, we know that both the cosine and tangent of any angle in the second quadrant are negative. The sine and cosine of any angle in the third quadrant are also both negative.
The properties of related angles state that c o s c o s ( 1 8 0 − 𝜃 ) = − 𝜃 . ∘
So, c o s c o s c o s c o s ( 1 8 0 − 4 5 ) = − 4 5 1 3 5 = − 4 5 = − √ 2 2 . ∘ ∘ ∘ ∘ ∘
The properties of related angles also state that t a n t a n ( 1 8 0 − 𝜃 ) = − 𝜃 . ∘
So, t a n t a n ( 1 8 0 − 4 5 ) = − 4 5 . ∘ ∘ ∘
Therefore, t a n t a n 1 3 5 = − 4 5 = − 1 . ∘ ∘
They also state that s i n s i n ( 1 8 0 + 𝜃 ) = − 𝜃 . ∘
So, s i n s i n ( 1 8 0 + 4 5 ) = − 4 5 . ∘ ∘ ∘
Therefore, s i n s i n 2 2 5 = − 4 5 = − √ 2 2 . ∘ ∘
They also state that c o s c o s ( 1 8 0 + 𝜃 ) = − 𝜃 . ∘
So, c o s c o s ( 1 8 0 + 4 5 ) = − 4 5 . ∘ ∘ ∘
Therefore, c o s c o s 2 2 5 = − 4 5 = − √ 2 2 . ∘ ∘
Since c s c s i n 𝜃 = 1 𝜃 , then c s c s i n 2 2 5 = 1 2 2 5 = − √ 2 . ∘ ∘
Substituting these values into our expression, we have c o s t a n c s c c o s 1 3 5 + 1 3 5 + 2 2 5 + 2 2 5 = − √ 2 2 + ( − 1 ) + − √ 2 + − √ 2 2 = − √ 2 2 − 1 − √ 2 − √ 2 2 = − 2 √ 2 − 1 . ∘ ∘ ∘ ∘
So, our final answer is − 2 √ 2 − 1 .
For the next example, we will need to evaluate a second-degree trigonometric expression.
Evaluate 3 − 4 3 3 0 c o s ∘ .
Substituting this value into our expression gives us 3 − 4 3 3 0 = 3 − 4 √ 3 2 = 3 − 4 3 4 = 3 − 3 = 0 . c o s ∘
So, our final answer is 0.
In the next example, we will investigate how different angles can satisfy a trigonometric equation involving multiple angles and the power of a trigonometric function.
Which of the following values of 𝑥 does not satisfy the equation s i n c o s 𝑥 + 6 ( 3 𝑥 ) − 1 2 = − 3 √ 2 ?
In this example, we have been given several values of 𝑥 and are being asked which does not satisfy the given equation. We note that since the given equation is a trigonometric one, it can possibly have an infinite number of solutions due to the properties of related angles.
The easiest way to check which value of 𝑥 does not satisfy the equation is to substitute the given values into the left-hand side one by one and see whether we get the right-hand side.
First, for 𝑥 = 4 5 ∘ , we have s i n c o s ∘ ∘ 4 5 + 6 ( 3 ⋅ 4 5 ) − 1 2 .
To evaluate this, we first recall that 4 5 ∘ is a special angle, and so we can look up the value of the first term directly to find that s i n 4 5 = √ 2 2 ∘ . For the second term, that is, 6 ( 3 ⋅ 4 5 ) = 6 1 3 5 c o s c o s ∘ ∘ , we can use the related angle property: c o s c o s ( 1 8 0 − 𝜃 ) = − 𝜃 . ∘
Taking 𝜃 = 4 5 ∘ , we have c o s c o s 1 3 5 = − 4 5 = − √ 2 2 . ∘ ∘
Putting this together, we have s i n c o s ∘ ∘ 4 5 + 6 ( 1 3 5 ) − 1 2 = √ 2 2 + 6 − √ 2 2 − 1 2 = 2 4 − 6 √ 2 2 − 1 2 = 1 2 − 3 √ 2 − 1 2 = − 3 √ 2 .
Thus, since this is equal to the right-hand side, 4 5 ∘ satisfies the equation.
Let us repeat this process for the other angles, continuing to use properties of related angles to help us. For 𝑥 = 1 3 5 ∘ , we have s i n c o s ∘ ∘ 1 3 5 + 6 4 0 5 − 1 2 .
We use the identities s i n s i n ( 1 8 0 − 𝜃 ) = 𝜃 ∘ and c o s c o s ( 𝜃 + 3 6 0 ) = 𝜃 ∘ , with 𝜃 = 4 5 ∘ , to get s i n c o s s i n c o s ∘ ∘ ∘ ∘ 1 3 5 + 6 4 0 5 − 1 2 = 4 5 + 6 4 5 − 1 2 = √ 2 2 + 6 √ 2 2 − 1 2 = 1 2 + 6 √ 2 2 − 1 2 = 3 √ 2 .
As this is not equal to the right-hand side, this shows that 1 3 5 ∘ does not satisfy the equation.
We have already found the answer, but for completeness, let us check the remaining options. For 𝑥 = 3 1 5 ∘ , we have s i n c o s ∘ ∘ 3 1 5 + 6 9 4 5 − 1 2 .
The first term can be simplified using s i n s i n ( 3 6 0 − 𝜃 ) = − 𝜃 ∘ (by taking 𝜃 = 4 5 ∘ ), to get s i n s i n 3 1 5 = − 4 5 ∘ ∘ . The second can be simplified by first using the periodicity of cosine, in other words, c o s c o s ( 𝜃 + 2 ⋅ 3 6 0 ) = 𝜃 ∘ , to get c o s c o s 9 4 5 = 2 2 5 ∘ ∘ , and then using c o s c o s ( 3 6 0 − 𝜃 ) = 𝜃 ∘ to get c o s c o s 2 2 5 = 1 3 5 ∘ ∘ . Recall that we have already found that c o s 1 3 5 = − √ 2 2 ∘ . Putting this together, we have s i n c o s s i n c o s ∘ ∘ ∘ ∘ ( 3 1 5 ) + 6 ( 9 4 5 ) − 1 2 = ( − 4 5 ) + 6 1 3 5 − 1 2 = − √ 2 2 + 6 − √ 2 2 − 1 2 = 1 2 − 3 √ 2 − 1 2 = − 3 √ 2 .
Thus, 𝑥 = 3 1 5 ∘ works. The fourth option is 𝑥 = 4 0 5 ∘ . For this, we have s i n c o s ∘ ∘ 4 0 5 + 6 1 2 1 5 − 1 2 .
For both the first and second terms, we can use the periodicity of sine and cosine to simplify these terms. That is, s i n s i n ( 𝜃 + 3 6 0 ) = 𝜃 ∘ and c o s c o s ( 𝜃 + 3 ⋅ 3 6 0 ) = 𝜃 ∘ to get s i n c o s s i n c o s ∘ ∘ ∘ ∘ 4 0 5 + 6 1 2 1 5 − 1 2 = 4 5 + 6 1 3 5 − 1 2 .
This is the same left-hand side as we had for 𝑥 = 4 5 ∘ , so we can conclude that this value satisfies the equation.
Finally, we have 𝑥 = − 3 1 5 ∘ , which gives us a left-hand side of s i n c o s ∘ ∘ ( − 3 1 5 ) + 6 ( − 9 4 5 ) − 1 2 .
We can use the periodic identities s i n s i n 𝜃 = ( 𝜃 + 3 6 0 ) ∘ and c o s c o s 𝜃 = ( 𝜃 + 3 ⋅ 3 6 0 ) ∘ to get s i n c o s s i n c o s ∘ ∘ ∘ ∘ ( − 3 1 5 ) + 6 ( − 9 4 5 ) − 1 2 = 4 5 + 6 1 3 5 − 1 2 .
Once more, we have the same left-hand side as we had for 𝑥 = 4 5 ∘ and 𝑥 = 4 0 5 ∘ , so we can conclude that this value satisfies the equation.
In conclusion, option B does not satisfy the equation.
In the previous example, let us note that three of the possible solutions, namely, 𝑥 = 4 5 ∘ , 𝑥 = 4 0 5 ∘ , and 𝑥 = − 3 1 5 ∘ , ended up with the exact same expression on the left-hand side. This was a result of the periodicity of trigonometric functions, which can be seen by considering the positions of these angles on a CAST diagram.
That is to say, the values of c o s 𝑥 and s i n 𝑥 are equal for these values of 𝑥 because they share the same place on the CAST diagram. By extension, the values of c o s 𝑥 and s i n 𝑥 will also be the same.
It is important to be aware, however, that we cannot always use this approach of a CAST diagram. In particular, we should be careful since one of the terms in the equation (i.e., 6 ( 3 𝑥 ) c o s ) has a multiple angle. Recall that for trigonometric functions with multiple angles, the periodicity is different. This difference is illustrated in the graphs below.
Specifically, the period of c o s 𝑥 is 3 6 0 ∘ (or 2 𝜋 ), but the period of c o s 3 𝑥 is 1 2 0 ∘ (or 2 𝜋 3 ). Nevertheless, since the periods of c o s 𝑥 and c o s 3 𝑥 overlap, it turns out that angles that are 3 6 0 ∘ apart will still have the same value of c o s 3 𝑥 . We can see this by considering the periodic property of c o s 3 𝑥 , specifically that since it has a period of 1 2 0 ∘ , we have c o s c o s 3 𝑥 = ( 3 ( 𝑥 + 1 2 0 𝑛 ) ) , ∘ where 𝑛 is an integer. Since 3 6 0 ∘ is three times 1 2 0 ∘ , if we let 𝑛 = 3 𝑘 , where 𝑘 is also an integer, we have c o s c o s c o s 3 𝑥 = ( 3 ( 𝑥 + 1 2 0 ⋅ 3 𝑘 ) ) = ( 3 ( 𝑥 + 3 6 0 𝑘 ) ) . ∘ ∘
This shows us that the value of c o s 3 𝑥 will be the same for any angles that differ by 3 6 0 ∘ , so 4 5 ∘ , 4 0 5 ∘ , and − 3 1 5 ∘ will be the same too.
For our final example, let us find the value of an expression involving the product of multiple trigonometric functions.
Find the value of 3 3 0 6 0 − 0 6 0 + 2 7 0 4 5 s i n s i n c o s s e c s i n c o s ∘ ∘ ∘ ∘ ∘ ∘ .
From our knowledge of special angles, we know that s i n 3 0 = 1 2 ∘ , s i n 6 0 = √ 3 2 ∘ , and c o s 4 5 = √ 2 2 ∘ .
From the sine and cosine graphs below, we see that c o s 0 = 1 ∘ and s i n 2 7 0 = − 1 ∘ .
Since s e c c o s 𝜃 = 1 𝜃 , then s e c c o s 6 0 = 1 6 0 = 1 ÷ 1 2 = 2 . ∘ ∘
We can now substitute all of these values into our expression: 3 3 0 6 0 − 0 6 0 + 2 7 0 4 5 = 3 1 2 √ 3 2 − ( 1 ) ( 2 ) + ( − 1 ) √ 2 2 = 3 √ 3 4 − 2 − 1 2 = 3 √ 3 4 − 1 0 4 = − 1 0 + 3 √ 3 4 . s i n s i n c o s s e c s i n c o s ∘ ∘ ∘ ∘ ∘ ∘
We will finish this explainer by recapping some of the key points.
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Problems on Trigonometric Ratios. Some trigonometric solutions based problems on trigonometric ratios are shown here with the step-by-step explanation. 1. If sin θ = 8/17, find other trigonometric ratios of <θ. Let us draw a ∆ OMP in which ∠M = 90°. Then sin θ = MP/OP = 8/17. Let MP = 8k and OP = 17k, where k is positive.
Solution to Problem 1: First we need to find the hypotenuse using Pythagora's theorem. (hypotenuse) 2 = 8 2 + 6 2 = 100. and hypotenuse = 10. We now use the definitions of the six trigonometric ratios given above to find sin A, cos A, tan A, sec A, csc A and cot A. sin A = side opposite angle A / hypotenuse = 8 / 10 = 4 / 5.
Correct answer: 23.81 meters. Explanation: To make sense of the problem, start by drawing a diagram. Label the angle of elevation as 25 o, the height between the ground and where the wire hits the flagpole as 10 meters, and our unknown, the length of the wire, as w. Now, we just need to solve for w using the information given in the diagram.
PROBLEMS ON TRIGONOMETRIC RATIOS. Problem 1 : For the measures in the figure shown below, compute sine, cosine and tangent ratios of the angle θ. Solution : In the given right angled triangle, note that for the given angle θ, PR is the 'opposite' side and PQ is the 'adjacent' side. Then,
How To Solve Trigonometry Problems Or Questions? Step 1: If no diagram is given, draw one yourself. Step 2: Mark the right angles in the diagram. Step 3: Show the sizes of the other angles and the lengths of any lines that are known. Step 4: Mark the angles or sides you have to calculate.
Thus, the other solution is 216.1°. 216.1 °. To solve simple equations involving a single trigonometric ratio (either sinθ,cosθ, sin θ, cos θ, or tanθ tan θ ), we can follow the steps below. 1. Isolate the trigonometric ratio. 2. Find one solution in 0° ≤ θ ≤ 360° 0 ° ≤ θ ≤ 360 °.
b) tan 41° = 1.9/x. c) tan θ = 11/8. Show Video Lesson. Applications of Trigonometric Ratios (Word Problems Involving Tangent, Sine and Cosine) Examples: Find the area of the parallelogram. A 70 foot ramp rises from the first floor to the second floor of a parking garage. The ramp makes an angle with the ground.
The ratios of the sides of a right triangle are called trigonometric ratios. Three common trigonometric ratios are the sine (sin), cosine (cos), and tangent (tan). These are defined for acute angle A below: In these definitions, the terms opposite, adjacent, and hypotenuse refer to the lengths of the sides.
Welcome to our comprehensive video on solving problems using trigonometric ratios! In this tutorial, we delve into the practical application of trigonometry ...
Trigonometric Expressions. Solve and manipulate expressions involving trigonometric functions. Explore the steps for solving trigonometric equations: solve sin (x) = 1/2. solutions to sin (2 (x-pi/2)) = 1. Expand trigonometric expressions with the steps provided: simplify sin (x + pi/4) expand cos (2x)^2. simplify cos (4x)^2 + sin (2x)
Lesson 5: Introduction to the trigonometric ratios. Triangle similarity & the trigonometric ratios. Trigonometric ratios in right triangles. Trigonometric ratios in right triangles. Trigonometric ratios in right triangles. Math >. High school geometry >. Right triangles & trigonometry >. Introduction to the trigonometric ratios.
Trigonometric Ratios. The six trigonometric ratios are sine (sin), cosine (cos), tangent (tan), cotangent (cot), cosecant (cosec), and secant (sec). In geometry, trigonometry is a branch of mathematics that deals with the sides and angles of a right-angled triangle. Therefore, trig ratios are evaluated with respect to sides and angles.
@MathTeacherGon will solve problems involving right triangles. The main focus of this is to use trigonometric ratios in solving real life examples of right t...
Solution. We can solve this equation using only algebra. Isolate the expression tanx on the left side of the equals sign. 2(tanx) + 2(3) = 5 + tanx 2tanx + 6 = 5 + tanx 2tanx − tanx = 5 − 6 tanx = − 1. There are two angles on the unit circle that have a tangent value of − 1: θ = 3π 4 and θ = 7π 4.
Step 4: Set up the equation that states the trig ratio you found in step 3 equals the ratio of the two sides you identified in step 2. tan B = O p p o s i t e A d j a c e n t = 9 5. Step 5: Find ...
Problem 1 : The angle of elevation of the top of the building at a distance of 50 m from its foot on a horizontal plane is found to be 60 °. Find the height of the building. Solution : Draw a sketch. Here, AB represents height of the building, BC represents distance of the building from the point of observation.
To solve a problem involving two right triangles using trigonometry, draw and label a diagram showing the given information, and the length or angle measure to be found. identify the two triangles that can be used to solve the problem, and plan how to use each triangle. solve the problem and show each step in your solution.
Trigonometric ratios are frequently expressed as decimal approximations. Example 2 : Find the sine, the cosine, and the tangent of the indicated angle. a. ∠S b. ∠R. Solution (a) : The length of the hypotenuse is 13. For ∠ S, the length of the opposite side is 5, and the length of the adjacent side is 12.
1-Step; 2-Step; 3-Step; Mixed; Variable on One Side: Non-Calculator. ... Solving Real-Life Problems Using Trigonometry. Worksheets. Worksheet. PPT. A4. PDF. A5. PDF. Worksheet. PPT. ... Method. PPT. Introduction to Trigonometry Choosing a Trigonometric Ratio to Use Calculating Angles & Lengths Using Trigonometry. Angles of Elevation ...
Read the problem and make sure all the words and ideas are understood. Draw the right triangle and label the given parts. Identify what we are looking for.; Label what we are looking for by choosing a variable to represent it.; Find the required trigonometric ratio.; Solve the ratio using good algebra techniques.; Check the answer by substituting it back into the ratio in step 4 and by making ...
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Radian Measure. To use trigonometric functions, we first must understand how to measure the angles. Although we can use both radians and degrees, radians are a more natural measurement because they are related directly to the unit circle, a circle with radius 1.The radian measure of an angle is defined as follows.
Try This. In this explainer, we will learn how to evaluate the trigonometric functions with special angles and how to use them to evaluate trigonometric expressions. We will begin by recalling special angles, together with the sine, cosine, and tangent values of these angles. Let us consider a unit circle.