Trig Identities. or (x + y)2 = x2 + 2xy + y 2. Dr. Ken W. Smith Other examples of identities are: (x + 3)2 = x2 + 6x + 9 and

Size: px

Start display at page:

Download "Trig Identities. or (x + y)2 = x2 + 2xy + y 2. Dr. Ken W. Smith Other examples of identities are: (x + 3)2 = x2 + 6x + 9 and"

Mervin Boone
6 years ago
Views:

Trig Identities An identity is an equation that is true for all values of the variables. Examples of identities might be obvious results like Part 4, Trigonometry Lecture 4.

1 Trig Identities An identity is an equation that is true for all values of the variables. Examples of identities might be obvious results like Part 4, Trigonometry Lecture 4.8a, Trig Identities and Equations x + x = 4x or (x + y) = x + xy + y. Dr. Ken W. Smith Other examples of identities are: Sam Houston State University 1 / 1 1 (x + ) = x + 6x + 9 and (a very important one!) A B = (A B)(A + B). / 1 Trig identities vs. trig equations Central Angles and Arcs What is a trig identity? A trig identity is an equation which is true for all inputs (such as angles, θ.) Some of our trig identities come from our definitions. For example, from the definition of the tangent function we know that For example, from the Pythagorean theorem on the unit circle, we know that the equation for the unit circle is x + y = 1 and so this turns into an identity for trig functions: sin θ cos θ We also have some identities given by symmetry. For example, since the sine function is odd then tan θ = sin( x) sin(x); (cos θ) + (sin θ) = 1 Since cosine is even then cos θ + sin θ = 1 cos( x) = cos x. This is true regardless of the choice of θ. By looking at the graphs of sine and cosine we observed that Other examples of trig identities are: sin θ 1 tan θ = cos θ sin( x) = sin x cos(z) = sin(z + π/). cos x = sin(x + π/). A trig equation, unlike an identity, may not necessarily be true for all angles θ. In general, with a trig equation, we wish to solve for θ. / 1 4 / 1

2 Finding all solutions to a trig equation Finding all solutions to a trig equation When we solve an equation, it is important that we find all solutions. For example, if we are solving the equation x x + = 0 then it is not sufficient to just list x = 1 as a solution. To find all solutions we might factor the quadratic x x + = (x )(x 1) and then set this equal to zero: (x )(x 1) = 0 This equation implies that either x = 0 (so x = ) or x 1 = 0 (so x = 1.) We have found two solutions. Another example: suppose we want to solve the equation x = 4. It is not sufficient to notice that x = is a solution, but we want to find both the solutions x = and x =. We often remind ourselves of the possibilities of two solutions by writing a plus-or-minus symbol (±) as in the computation x = 4 = x = ± 4 = ±. From our understanding about zeroes of a polynomial, we now know that we have found all the solutions. Once reason for the concept of factoring is that it aids us in finding all solutions! 5 / 1 Finding all solutions to a trig equation For many trig problems, we will find solutions within one revolution of the unit circle and then use the period of the trig functions to find an infinite number of solutions. For example, let s solve the equation sin θ =. Since the sine here is positive then the angle θ must be in quadrants I or II. Recall our discussion of theunit circle and triangles and recognize that sin(π/) =. So θ = π/ is a nice solution in the first quadrant for our equation. In the second quadrant, we note that π/ has reference angle π/ and so sin(π/) =. So θ = π/ is a nice solution in the second quadrant. 7 / 1 6 / 1 Finding all solutions to a trig equation sin θ = We found two solutions to this equation: θ = π/ and θ = π/. However, these two solutions are not all the solutions! Since the sine function is periodic with period π we can add π to any angle without changing the value of sine. So if θ = π/ is a solution then so are θ = π/ π, θ = π/ + π, θ = π/ + 4π, θ = π/ + 6π,... etc. We can write this infinite collection of solutions in the form θ = π/ + πk where we understand that k is a whole number, that is, k Z. Similarly, if θ = π/ is a solution then so are π/ + πk (k Z) We collect these together as our solution: 8 / 1

3 Finding all solutions to a trig equation Solve the equation sin 4θ =. Sometimes inverse trig functions come in handy. For example, suppose we are solving the equation tan x =. One solution is x = arctan() which is approximately (There is no simple way to write this angle; we need the arctangent function.) Since tangent has period π and there is only one solution within an interval of length π, we know that the full set of solutions is {arctan() + πk : k Z}. 9 / 1 Solution. If sin 4θ = or So divide both sides by 4 to get or then from the notes above, we know that 4θ = π + πk 4θ = π + πk. θ = π 1 + π k θ = π 6 + π k (where k Z.) Note that since sin θ has period π then sin 4θ must have period π 4 = π. This is reflected in our solutions by our adding multiples of π to our first solutions, π/1 and π/6. 10 / 1 Solve sin x = 1. Solution. The equation sin x = 1 of the unit circle. They are x = π 6 and x = 5π 6. has two solutions within one revolution Solve sin(x π 4 ) + 1 =. Solution. Let s not worry about the expression x π 4 at the beginning of this problem. Think of x π 4 as an angle in its own right. First we subtract 1 from both sides and then divide both sides by. sin(x π 4 ) + 1 = = sin(θ π 4 ) = 1 = sin(θ π 4 ) = 1. Since sin x has period π we know that the collection of all solutions must be x = π 6 + πk and x = 5π 6 + πk. We know, from the previous problem, that sin( π 6 ) = 1 and sin( 5π 6 ) = 1 and so θ π 4 = π 6 + πk or θ π 4 = 5π 6 + πk. We can write this in set notation as { π 6 + πk : k Z} { 5π 6 + πk : k Z}. Now we can solve for θ by adding π/4 to both sides of these equations, obtaining solutions θ = π 4 + π 6 + πk or θ = π 4 + 5π 6 + πk. 11 / 1 The best answer is achieved by getting a common denominator for π/4 + Smith π/6: (SHSU) 1 / 1

4 Solve cos x =. Solve the equation tan θ =. Solution. The equation cos x = has two solutions within one revolution of the unit circle. Since the cosine function is even, we know that when we find one positive solution, the negative solution will also work. So our two solutions are x = π 6 and x = π6. Solution. tan θ = = tan θ = ±. If tan θ = then θ = π + πk (where k Z.) If tan θ = then θ = + πk. So the final answer is Since cos x has period π we know that the collection of all solutions must be { π6 + πk : k Z} { π6 + πk : k Z}. π 1 / 1 θ= π + πk or θ = π + πk. 14 / 1 16 / 1 Trig Equations Part 4, Trigonometry Lecture 4.8b, Trig Identities and Equations In the next presentation, we will look further at trig equations. (End) Dr. Ken W. Smith Sam Houston State University 15 / 1

5 Some worked problems. The standard rules of algebra are still important in solving trig equations. We will often use the algebra fact that if AB = 0 then either A = 0 or B = 0. If, for example, we need to solve the equation (tan x )( sin x 1) = 0 then we note that this implies that either tan x = 0 or sin x 1 = 0. These then imply tan x = or sin x = 1. In an earlier problem we solved the equation tan x = and got the solution set {arctan() + πk : k Z}. In a different problem we solved sin x = 1 and found the solution set In this case, we want all solutions to both equations. So the solution set to the equation (tan x )( sin x 1) = 0 is {arctan() + πk : k Z} { π 6 + πk : k Z} { 5π 6 + πk : k Z}. { π 6 + πk : k Z} {5π 6 + πk : k Z}. 17 / 1 Some worked problems. Solve sin x 5 sin x + 6 = 0. Solution. A standard algebra problem might have us solving z 5z + 6 = 0 by factoring the quadratic polynomial into (z )(z ) and solving (z )(z ) = 0. Here we can do something very similar; we factor sin x 5 sin x + 6 into (sin x )(sin x ). So sin x 5 sin x + 6 = 0 = (sin x )(sin x ) = 0 = sin x = 0 or sin x = 0. Now sin x = 0 = sin x = and since the sine function has range [ 1, 1], this equation has no solutions. Similarly the equation sin x = 0 has no solutions. So the quadratic equation sin x 5 sin x + 6 = 0 has NO solutions. 19 / 1 18 / 1 Some worked problems. Solve sin x sin x + 1 = 0. Solution. We factor sin x sin x + 1 = ( sin x 1)(sin x 1). So sin x sin x + 1 = 0 = ( sin x 1)(sin x 1) = 0 = sin x 1 = 0 or sin x 1 = 0 = sin x = 1 or sin x = 1. In the first case (sin x = 1 ) our solution set is { π 6 + πk : k Z} {5π + πk : k Z}. 6 In the second case (sin x = 1) our solution set is So our final solution is { π + πk : k Z}. { π 6 + πk : k Z} { 5π 6 + πk : k Z} { π + πk : k Z}. 0 / 1

6 Some worked problems In the next presentation, we will look further at trig identities and equations. (End) 1 / 1

Sum and difference formulae for sine and cosine. Elementary Functions. Consider angles α and β with α > β. These angles identify points on the

Sum and difference formulae for sine and cosine. Elementary Functions. Consider angles α and β with α > β. These angles identify points on the Consider angles α and β with α > β. These angles identify points on the unit circle, P (cos α, sin α) and Q(cos β, sin β). Part 5, Trigonometry Lecture 5.1a, Sum and Difference Formulas Dr. Ken W. Smith