Trig Identities An identity is an equation that is true for all - PowerPoint PPT Presentation

Trig Identities An identity is an equation that is true for all values of the variables. Examples of identities might be “obvious” results like Elementary Functions Part 4, Trigonometry 2 x + 2 x = 4 x Lecture 4.8a, Trig Identities and Equations or ( x + y ) 2 = x 2 + 2 xy + y 2 . Dr. Ken W. Smith Other examples of identities are: 1 ( x + 3) 2 = x 2 + 6 x + 9 Sam Houston State University and 2013 2 (a very important one!) A 2 − B 2 = ( A − B )( A + B ) . Smith (SHSU) Elementary Functions 2013 1 / 21 Smith (SHSU) Elementary Functions 2013 2 / 21 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 Some of our trig identities come from our definitions. For example, from inputs (such as angles, θ .) the definition of the tangent function we know that tan θ = sin θ For example, from the Pythagorean theorem on the unit circle, we know that the equation for the unit circle is x 2 + y 2 = 1 and so this turns into cos θ an identity for trig functions: We also have some identities given by symmetry. For example, since the sine function is odd then sin( − x ) − sin( x ); (cos θ ) 2 + (sin θ ) 2 = 1 Since cosine is even then cos 2 θ + sin 2 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: cos x = sin( x + π/ 2) . 1 tan θ = sin θ cos θ 2 sin( − x ) = − sin x A trig equation, unlike an identity, may not necessarily be true for all 3 cos( z ) = sin( z + π/ 2) . angles θ . In general, with a trig equation, we wish to solve for θ. Smith (SHSU) Elementary Functions 2013 3 / 21 Smith (SHSU) Elementary Functions 2013 4 / 21

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 2 − 3 x + 2 = 0 Another example: suppose we want to solve the equation x 2 = 4 . then it is not sufficient to just list x = 1 as a solution. To find all solutions It is not sufficient to notice that x = 2 is a solution, but we want to find we might factor the quadratic x 2 − 3 x + 2 = ( x − 2)( x − 1) and then set both the solutions x = 2 and x = − 2 . this equal to zero: ( x − 2)( x − 1) = 0 We often remind ourselves of the possibilities of two solutions by writing a plus-or-minus symbol ( ± ) as in the computation This equation implies that either x − 2 = 0 (so x = 2 ) or x − 1 = 0 (so √ x 2 = 4 = ⇒ x = ± 4 = ± 2 . x = 1 . ) We have found two solutions. 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! Smith (SHSU) Elementary Functions 2013 5 / 21 Smith (SHSU) Elementary Functions 2013 6 / 21 Finding all solutions to a trig equation Finding all solutions to a trig equation √ For many trig problems, we will find solutions within one revolution of the 3 sin θ = 2 unit circle and then use the period of the trig functions to find an infinite We found two solutions to this equation: θ = π/ 3 and θ = 2 π/ 3 . number of solutions. However, these two solutions are not all the solutions! Since the sine function is periodic with period 2 π we can add 2 π to any angle without For example, let’s solve the equation changing the value of sine. So if θ = π/ 3 is a solution then so are √ 3 θ = π/ 3 − 2 π, θ = π/ 3 + 2 π, θ = π/ 3 + 4 π, θ = π/ 3 + 6 π, ... etc. sin θ = 2 . We can write this infinite collection of solutions in the form Since the sine here is positive then the angle θ must be in quadrants I or II. θ = π/ 3 + 2 πk Recall our discussion of the unit circle and 30-60-90 triangles and where we understand that k is a whole number, that is, k ∈ Z . √ 3 recognize that sin( π/ 3) = 2 . So θ = π/ 3 is a nice solution in the first Similarly, if θ = 2 π/ 3 is a solution then so are quadrant for our equation. 2 π/ 3 + 2 πk ( k ∈ Z ) In the second quadrant, we note that 2 π/ 3 has reference angle π/ 3 and so We collect these together as our solution: √ 3 sin(2 π/ 3) = 2 . So θ = 2 π/ 3 is a nice solution in the second quadrant. Smith (SHSU) Elementary Functions 2013 7 / 21 Smith (SHSU) Elementary Functions 2013 8 / 21

Finding all solutions to a trig equation More worked problems. √ 3 Solve the equation sin 4 θ = 2 . √ 3 Solution. If sin 4 θ = 2 then from the notes above, we know that Sometimes inverse trig functions come in handy. For example, suppose we 4 θ = π 3 + 2 πk are solving the equation tan x = 2 . or 4 θ = 2 π 3 + 2 πk. One solution is x = arctan(2) which is approximately 1 . 10715 . (There is no simple way to write this angle; we need the arctangent function.) Since So divide both sides by 4 to get tangent has period π and there is only one solution within an interval of θ = π 12 + π 2 k length π , we know that the full set of solutions is or { arctan(2) + πk : k ∈ Z } . θ = π 6 + π 2 k (where k ∈ Z . ) Note that since sin θ has period 2 π then sin 4 θ must have period 2 π 4 = π 2 . This is reflected in our solutions by our adding multiples of π 2 to our first solutions, π/ 12 and π/ 6 . Smith (SHSU) Elementary Functions 2013 9 / 21 Smith (SHSU) Elementary Functions 2013 10 / 21 More worked problems. More worked problems. Solve 2 sin( x − π 4 ) + 1 = 2 . Solve sin x = 1 2 . Solution. Let’s not worry about the expression x − π 4 at the beginning of Solution. The equation sin x = 1 2 has two solutions within one revolution this problem. Think of x − π 4 as an angle in its own right. of the unit circle. First we subtract 1 from both sides and then divide both sides by 2. 2 sin( x − π ⇒ 2 sin( θ − π ⇒ sin( θ − π 4 ) = 1 They are x = π 6 and x = 5 π 4 ) + 1 = 2 = 4 ) = 1 = 2 . 6 . We know, from the previous problem, that sin( π 2 and sin( 5 π 6 ) = 1 6 ) = 1 Since sin x has period 2 π we know that the collection of all solutions must 2 and so be θ − π 4 = π 6 + 2 πk or θ − π 4 = 5 π x = π 6 + 2 πk and x = 5 π 6 + 2 πk. 6 + 2 πk. Now we can solve for θ by adding π/ 4 to both sides of these equations, We can write this in set notation as obtaining solutions θ = π 4 + π 6 + 2 πk or θ = π 4 + 5 π { π 6 + 2 πk : k ∈ Z } ∪ { 5 π 6 + 2 πk : k ∈ Z } . 6 + 2 πk. The best answer is achieved by getting a common denominator for π/ 4 + π/ 6 : Smith (SHSU) Elementary Functions 2013 11 / 21 Smith (SHSU) Elementary Functions 2013 12 / 21 5 π 13 π

More worked problems. More worked problems. √ 3 Solve cos x = 2 . Solve the equation tan 2 θ = 3 . √ √ √ 3 Solution. The equation cos x = 2 has two solutions within one Solution. tan 2 θ = 3 = ⇒ tan θ = ± 3 . If tan θ = 3 then revolution of the unit circle. Since the cosine function is even, we know θ = π 3 + 2 πk (where k ∈ Z . ) that when we find one positive solution, the negative solution will also work. √ 3 then θ = 2 π If tan θ = − 3 + 2 πk . So our two solutions are x = π 6 and x = − π 6 . So the final answer is Since cos x has period 2 π we know that the collection of all solutions must θ = π 3 + 2 πk or θ = 2 π 3 + 2 πk . be { π 6 + 2 πk : k ∈ Z } ∪ {− π 6 + 2 πk : k ∈ Z } . Smith (SHSU) Elementary Functions 2013 13 / 21 Smith (SHSU) Elementary Functions 2013 14 / 21 Trig Equations Elementary Functions Part 4, Trigonometry In the next presentation, we will look further at trig equations. Lecture 4.8b, Trig Identities and Equations (End) Dr. Ken W. Smith Sam Houston State University 2013 Smith (SHSU) Elementary Functions 2013 15 / 21 Smith (SHSU) Elementary Functions 2013 16 / 21