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Basic Trigonometric Identities

Basic Trigonometric Identities. Example. Verify the identity: sec x cot x = csc x.

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Basic Trigonometric Identities

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  1. Basic Trigonometric Identities

  2. Example Verify the identity: sec x cot x= csc x. Solution The left side of the equation contains the more complicated expression. Thus, we work with the left side. Let us express this side of the identity in terms of sines and cosines. Perhaps this strategy will enable us to transform the left side into csc x, the expression on the right. Apply a reciprocal identity: sec x = 1/cosx and a quotient identity: cot x = cosx/sinx. Divide both the numerator and the denominator by cos x, the common factor.

  3. Solution We start with the more complicated side, the left side. Factor out the greatest common factor, cos x, from each of the two terms. cos x- cos x sin2x= cos x(1 - sin2x) Factor cos x from the two terms. Use a variation of sin2 x + cos2 x = 1. Solving for cos2 x, we obtain cos2 x = 1 – sin2 x. = cos x ·cos2x = cos3x Multiply. Example Verify the identity: cosx - cosxsin2x = cos3x.. We worked with the left and arrived at the right side. Thus, the identity is verified.

  4. Guidelines for Verifying Trigonometric Identities • Work with each side of the equation independently of the other side. Start with the more complicated side and transform it in a step-by-step fashion until it looks exactly like the other side. • Analyze the identity and look for opportunities to apply the fundamental identities. Rewriting the more complicated side of the equation in terms of sines and cosines is often helpful. • If sums or differences of fractions appear on one side, use the least common denominator and combine the fractions. • Don't be afraid to stop and start over again if you are not getting anywhere. Creative puzzle solvers know that strategies leading to dead ends often provide good problem-solving ideas.

  5. Example • Verify the identity: csc(x) / cot (x) = sec (x) Solution:

  6. Example • Verify the identity: Solution:

  7. Example • Verify the following identity: Solution:

  8. 7.2 Trigonometric Equations

  9. Equations Involving a Single Trigonometric Function To solve an equation containing a single trigonometric function: • Isolate the function on one side of the equation. • Solve for the variable.

  10. x Trigonometric Equations y y = cos x 1 y = 0.5 x –4  –2  2  4  –1 cos x = 0.5 has infinitely many solutions for –< x <  y y = cos x 1 0.5 2  cos x = 0.5 has two solutions for 0 < x < 2 –1

  11. This is the given equation. 3 sin x- 2 = 5 sin x- 1 Subtract 5 sin x from both sides. 3 sin x- 5 sin x- 2 = 5 sin x- 5 sin x – 1 Simplify. -2 sin x- 2 =-1 Add 2 to both sides. -2 sin x= 1 Divide both sides by -2 and solve for sin x. sin x= -1/2 Example Solve the equation: 3 sin x- 2 = 5 sin x- 1, 0 ≤ x < 360° Solution The equation contains a single trigonometric function, sin x. Step 1Isolate the function on one side of the equation. We can solve for sin x by collecting all terms with sin x on the left side, and all the constant terms on the right side. x = 210° or x = 330°

  12. Solution The given equation is in quadratic form 2t2+t- 1 = 0 with t= cos x. Let us attempt to solve the equation using factoring. This is the given equation. 2 cos2x+ cos x- 1 = 0 Factor. Notice that 2t2 + t – 1 factors as (2t – 1)(2t + 1). (2 cos x- 1)(cos x+ 1) = 0 Set each factor equal to 0. 2 cos x- 1= 0 or cos x+ 1 = 0 Solve for cos x. 2 cos x= 1 cos x= -1 Example Solve the equation: 2 cos2 x+ cos x- 1 = 0, 0 £x< 2p. cos x= 1/2 x=px= 2pppx=p The solutions in the interval [0, 2p) are p/3, p, and 5p/3.

  13. Example • Solve the following equation for θ is any real number. Solution:

  14. Example • Solve the equation on the interval [0,2) Solution:

  15. Example • Solve the equation on the interval [0,2) Solution:

  16. 7.3 Sum and Difference Formulas

  17. The Cosine of the Difference of Two Angles The cosine of the difference of two angles equals the cosine of the first angle times the cosine of the second angle plus the sine of the first angle times the sine of the second angle.

  18. Solution We know exact values for trigonometric functions of 60° and 45°. Thus, we write 15° as 60° - 45° and use the difference formula for cosines. cos l5° = cos(60° - 45°) = cos 60° cos 45° + sin 60° sin 45° cos(-) = cos  cos + sin  sin  Example • Find the exact value of cos 15° Substitute exact values from memory or use special triangles. Multiply. Add.

  19. cos(-) = cos  cos + sin  sin  Example Find the exact value of cos 80° cos 20° + sin 80° sin 20°. Solution The given expression is the right side of the formula for cos( - ) with = 80° and  = 20°. cos 80° cos 20° + sin 80° sin 20° = cos (80° - 20°) = cos 60° = 1/2

  20. Example • Find the exact value of cos(180º-30º) Solution

  21. Example • Verify the following identity: Solution

  22. Sum and Difference Formulas for Cosines and Sines

  23. Example • Find the exact value of sin(30º+45º) Solution

  24. Sum and Difference Formulas for Tangents The tangent of the sum of two angles equals the tangent of the first angle plus the tangent of the second angle divided by 1 minus their product. The tangent of the difference of two angles equals the tangent of the first angle minus the tangent of the second angle divided by 1 plus their product.

  25. Example • Find the exact value of tan(105º) Solution • tan(105º)=tan(60º+45º)

  26. Example • Write the following expression as the sine, cosine, or tangent of an angle. Then find the exact value of the expression. Solution

  27. Review Quiz Answers

  28. 7.4 Double-Angle and Half-Angle Formulas

  29. Double-Angle Identities

  30. Three Forms of the Double-Angle Formula for cos2

  31. Power-Reducing Formulas

  32. Example • If sin α = 4/5 and α is an acute angle, find the exact values of sin 2 α and cos 2 α. Solution If we regard α as an acute angle of a right triangle, we obtain cos α = 3/5. We next substitute in double-angle formulas: Sin 2 α = 2 sin α cos α = 2 (4/5)(3/5) = 24/25. Cos 2 α = cos2α – sin2α = (3/5)2 – (4/5)2 = 9/25 – 16/25 = -7/25.

  33. Half-Angle Identities

  34. Example Find the exact value of cos 112.5°. Solution Because 112.5° =225°/2, we use the half-angle formula for cos /2 with = 225°. What sign should we use when we apply the formula? Because 112.5° lies in quadrant II, where only the sine and cosecant are positive, cos 112.5° < 0. Thus, we use the - sign in the half-angle formula.

  35. Half-Angle Formulas for:

  36. Example • Verify the following identity: Solution

  37. 7.5 Product-to-Sum and Sum-to-Product Formulas

  38. Product-to-Sum Formulas

  39. Example • Express the following product as a sum or difference: Solution

  40. sin  sin  = 1/2 [cos( - ) - cos( + )] sin  cos  = 1/2[sin( + ) + sin( - )] Express each of the following products as a sum or difference. a. sin 8x sin 3xb. sin 4x cos x Text Example Solution The product-to-sum formula that we are using is shown in each of the voice balloons. a. sin 8x sin 3x= 1/2[cos (8x- 3x) - cos(8x+ 3x)] = 1/2(cos 5x- cos 11x) b. sin 4x cos x= 1/2[sin (4x+x) + sin(4x-x)] = 1/2(sin 5x+ sin 3x)

  41. Sum-to-Product Formulas

  42. Example • Express the difference as a product: Solution

  43. Example • Express the sum as a product: Solution

  44. Example • Verify the following identity: Solution

  45. 7.6 Inverse Trig Functions Objective: In this section, we will look at the definitions and properties of the inverse trigonometric functions. We will recall that to define an inverse function, it is essential that the function be one-to-one.

  46. 7.6 Inverse Trigonometric Functions and Trig Equations Domain: [–1, 1] Range: Domain: Range: Domain: [–1, 1] Range: [0, π]

  47. Let us begin with a simple question: What is the first pair of inverse functions that pop into YOUR mind? This may not be your pair but this is a famous pair. But something is not quite right with this pair. Do you know what is wrong? Congratulations if you guessed that the top function does not really have an inverse because it is not 1-1 and therefore, the graph will not pass the horizontal line test.

  48. Consider the graph of Note the two points on the graph and also on the line y=4. f(2) = 4 and f(-2) = 4 so what is an inverse function supposed to do with 4? By definition, a function cannot generate two different outputs for the same input, so the sad truth is that this function, as is, does not have an inverse.

  49. y=x 4 2 So how is it that we arrange for this function to have an inverse? We consider only one half of the graph: x > 0. The graph now passes the horizontal line test and we do have an inverse: Note how each graph reflects across the line y = x onto its inverse.

  50. A similar restriction on the domain is necessary to create an inverse function for each trig function. Consider the sine function. You can see right away that the sine function does not pass the horizontal line test. But we can come up with a valid inverse function if we restrict the domain as we did with the previous function. How would YOU restrict the domain?

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