# Lesson HOW TO - Solve Trigonometric equations

Introduction The solution of trigonometric equations is one topic that students have particular problems with. There are a few reasons for this: 1. there is usually a simplify part first that requires use of some TRIG identities. 2. there is the use of RADIANS rather than degrees, for which some students are not at ease with. 3. there is the repetitive aspect of TRIG functions that students find bewildering. All in all, a potentially daunting topic. Solving the TRIG Equation Of the 3 topics listed above, I am concentrating on part 3, here in this Lesson. First thing, when solving a TRIG equation, is to understand or accept that each of SINE, COSINE and TANGENT have 2 angles that will satisfy the given equation within any 360 degree range. If we have multiples, such as cos(2x), then we also multiply up the possible number of solutions. sin(x) --> 2 solutions sin(2x) --> 4 solutions sin(3x) --> 6 solutions sin(4x) --> 8 solutions sin(5x) --> 10 solutions etc Look at the following lines: 1.

Amplitude and Period Amplitude and Period Learning Objective(s) · Understand amplitude and period. · Graph the sine function with changes in amplitude and period. · Graph the cosine function with changes in amplitude and period. · Match a sine or cosine function to its graph and vice versa. You know how to graph the functions and . or , where a and b are constants. We used the variable previously to show an angle in standard position, and we also referred to the sine and cosine functions as . for the input (as well as to label the horizontal axis). . You know that the graphs of the sine and cosine functions have a pattern of hills and valleys that repeat. . (or ) on the interval looks like the graph on the interval . The graph below shows four repetitions of a pattern of length . is on the interval is one cycle. You know from graphing quadratic functions of the form that as you changed the value of a you changed the “width” of the graph. and see how changes to b will affect the graph. periodic, and if so, what is the period?

SOLVING TRIGONOMETRIC EQUATIONS Note: If you would like a review of trigonometry, click on trigonometry. Example 1: Solve for x in the following equation. There are an infinite number of solutions to this problem. First isolate the cosine term. To solve for x, we have to isolate x. Let's restrict the domain so the function is one-to-one on the restricted domain while preserving the original range. If we restrict the domain of the cosine function to that interval , we can take the arccosine of both sides of each equation. The angle x is the reference angle. Therefore, if , then The period of equals and the period of , this means other solutions exists every units. where n is an integer. The approximate values of these solutions are You can check each solution algebraically by substituting each solution in the original equation. You can also check the solutions graphically by graphing the function formed by subtracting the right side of the original equation from the left side of the original equation. Algebraic Check: Left Side:

Sine, Cosine and Tangent in Four Quadrants Sine, Cosine and Tangent The three main functions in trigonometry are Sine, Cosine and Tangent. They are easy to calculate: Divide the length of one side of a right angled triangle by another side ... but we must know which sides! For an angle θ, the functions are calculated this way: Example: What is the sine of 35°? Cartesian Coordinates Using Cartesian Coordinates we mark a point on a graph by how far along and how far up it is: The point (12,5) is 12 units along, and 5 units up. Four Quadrants When we include negative values, the x and y axes divide the space up into 4 pieces: Quadrants I, II, III and IV (They are numbered in a counter-clockwise direction) In Quadrant I both x and y are positive, in Quadrant II x is negative (y is still positive), in Quadrant III both x and y are negative, and in Quadrant IV x is positive again, and y is negative. Like this: Example: The point "C" (−2,−1) is 2 units along in the negative direction, and 1 unit down (i.e. negative direction). There is a pattern!

Graphing Trigonometric Functions Graphing Trigonometric Functions (page 1 of 3) Sections: Introduction, Examples with amplitude and vertical shift, Example with phase shift You've already learned the basic trig graphs. But just as you could make the basic quadratic x2, more complicated, such as –(x + 5)2 – 3, so also trig graphs can be made more complicated. Let's start with the basic sine function, f(t) = sin(t). Now let's look at g(t) = 3sin(t): Do you see that the graph is three times as tall? Now let's look at h(t) = sin(2t): Copyright © Elizabeth Stapel 2010 All Rights Reserved Do you see that the graph is squished in from the sides? For sines and cosines (and their reciprocals), the "regular" period is 2π, so the formula is: For tangents and cotangents, the "regular" period is π, so the formula is: In the sine wave graphed above, the value of B was 2. (Note: Different books use different letters to stand for the period formula. Now let's looks at j(t) = sin(t – π/3): Now let's look at k(t) = sin(t) + 3:

Algebra/Trig Review - Solving Trig Equations Solve the following trig equations. For those without intervals listed find ALL possible solutions. For those with intervals listed find only the solutions that fall in those intervals. There’s not much to do with this one. Just divide both sides by 2 and then go to the unit circle. So, we are looking for all the values of t for which cosine will have the value of . From quick inspection we can see that is a solution. . To find this angle for this problem all we need to do is use a little geometry. with the positive x-axis, then so must the angle in the fourth quadrant. , but again, it’s more common to use positive angles so, we’ll use We aren’t done with this problem. that we want for the solution and sometimes we will want both (or neither) of the listed angles. This is very easy to do. to represent all the possible angles that can end at the same location on the unit circle, i.e. angles that end at . So, all together the complete solution to this problem is by using in the second solution.

Solving Trigonometric Equations Solving Trigonometric Equations (page 1 of 2) Solving trig equations use both the reference angles you've memorized and a lot of the algebra you've learned. Be prepared to need to think! Solve sin(x) + 2 = 3 for 0° < x < 360° Just as with linear equations, I'll first isolate the variable-containing term: sin(x) + 2 = 3 sin(x) = 1 Now I'll use the reference angles I've memorized: x = 90° Solve tan2(x) + 3 = 0 for 0° < x < 360° There's the temptation to quickly recall that the tangent of 60° involves the square root of 3 and slap down an answer, but this equation doesn't actually have a solution: tan2(x) = –3 How can the square of a trig function evaluate to a negative number? no solution Solve on 0° < x < 360° To solve this, I need to do some simple factoring: Now that I've done the algebra, I can do the trig. x = 30°, 90°, 270°, 330° Copyright © Elizabeth Stapel 2010-2011 All Rights Reserved Solve sin2(x) – sin(x) = 2 on 0° < x < 360° Only one of the factor solutions is sensible. x = 270° Hmm...

Amplitude, Period, Phase Shift and Frequency Some functions (like Sine and Cosine) repeat forever and are called Periodic Functions. The Period is the length from one peak to the next (or from any point to the next matching point): The Amplitude is the height from the center line to the peak (or to the trough). Or we can measure the height from highest to lowest points and divide that by 2. The Phase Shift is how far the function is horizontally to the right of the usual position. The Vertical Shift is how far the function is vertically from the usual position. All Together Now! We can have all of them in one equation: y = A sin(Bx + C) + D amplitude is A period is 2π/B phase shift is −C/B vertical shift is D Example: sin(x) This is the basic unchanged sine formula. So amplitude is 1, period is 2π, there is no phase shift or vertical shift: Example: 2 sin(4x − 2) + 3 amplitude A = 2 period 2π/B = 2π/4 = π/2 phase shift −C/B = −(−2)/4 = 1/2 vertical shift D = 3 In words: Note the Phase Shift formula −C/B has a minus sign: And we get: Frequency

Higher Bitesize Maths - Radians and equations : Revision, Page4 Chapter 5: Trigonometric Functions<BLURT> Chapter 5: Trigonometric Functions 1. Please solve 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. Copyright ©2001 The McGraw-Hill Companies. 4. Graphs of tan, cot, sec and csc by M. Bourne The graphs of tanx, cotx, secx and cscx are not as common as the sine and cosine curves that we met earlier in this chapter. However, they do occur in engineering and science problems. They are interesting curves because they have discontinuities. [For more on this topic, go to Continuous and Discontinuous Functions in an earlier chapter.] Recall from Trigonometric Functions, that tanx is defined as: tanx=​cos x​​sin x​​ For some values of x, the function cosx has value 0. When this happens, we have 0 in the denominator of the fraction and this means the fraction is undefined. The same thing happens with cotx, secx and cscx for different values of x. The Graph of y = tan x Sketch y = tan x. Solution As we saw above, tanx=​cosx​​sinx​​ This means the function will have a discontinuity where cos x = 0. x=…,−​2​​5π​​,−​2​​3π​​,−​2​​π​​,​2​​π​​,​2​​3π​​,​2​​5π​​,… It is very important to keep these values in mind when sketching this graph. Graph of y = tan x: Interactive Tangent Curve π 2π t

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