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- 2. A.3. Determine whether an equation is an identity or conditional equation M.4. Model trigonometric identities to find other trigonometric values M.5 Analyze and solve situational problems involving trigonometric identities LEARNING COMPETENCIES:
- 4. MATH-CHIK
- 7. -2(x+4) + 3x = x – 8 5x – 4 = 11 0 = 0 x = 3
- 8. is an equation that is true and legitimate values of the variables An equation that is true only for certain values of the variable, and false for others
- 9. 5x - 11 = -2x – 4 3(2x - 1) = 2(3x - 2) + 1 x2 – 6x + 9 = (x + 3)2 x + 7(x – 1) = x + 3 Conditional Identity Identity Conditional 5x + 2x = 4 + 11 7x = 7 x = 1 3(2x - 1) = 2(3x - 2) + 1 6x – 3 = 6x – 4 + 1 6x – 3 = 6x – 3 0 = 0 x2 – 6x + 9 = (x + 3)2 (x + 3) (x + 3) = (x + 3) (x + 3) -x + 7(x – 1) = x + 3 -x + 7x – 7 = x + 3 6x - 7 = x + 3 6x – x = 3 + 7 5x = 10 x = 2
- 10. Which of the following equations are identities? a.3s + 7s = 10 s b.5c(c – 2s) = 5c2 – 10s c.2t – 1 = 3
- 11. Which of the following equations are not identities? a.(c – s) (c + s) = c2 – s2 b.3t2 = 1 c.(2c + 1) + (s – 3) = 2c + s – 2
- 13. 2x2 – x – 1 = (2x + 1) (x – 1) 0 = 0
- 14. 2x2 – x – 1 = (2x + 1) (x – 1) 0 = 0 2(cosθ)2 – cosθ – 1 = (2cosθ + 1) (θ – 1) 0 = 0
- 15. Which of the following are identities? a.3sin θ + 7 sin θ = 10 sin θ b.5cos θ (cos θ – 2 sin θ) = 5cos θ 2 – 10cos θ sin θ c.2tan θ – 1 = 3
- 16. Which of the following equations are identities? a.3s + 7s = 10 s b.5c(c – 2s) = 5c2 – 10s c.2t – 1 = 3
- 17. Which of the following are identities? a.3sin θ + 7 sin θ = 10 sin θ b.5cos θ (cos θ – 2 sin θ) = 5cos θ 2 – 10cos θ sin θ c.2tan θ – 1 = 3
- 18. Which of the following are identities? a.(cos θ – sin θ) (cos θ + sin θ) = cos2 θ – sin2 θ b.3tan2 θ = 1 c.2cos θ + 1 + (sin θ – 3) = 2cos θ + sin θ - 2
- 19. Which of the following equations are not identities? a.(c – s) (c + s) = c2 – s2 b.3t2 = 1 c.(2c + 1) + (s – 3) = 2c + s – 2
- 20. Which of the following are not identities? a.(cos θ – sin θ) (cos θ + sin θ) = cos2 θ – sin2 θ b.3tan2 θ = 1 c.2cos θ + 1 + (sin θ – 3) = 2cos θ + sin θ - 2
- 22. PYTHAGOREAN IDENTITIES RECIPROCAL IDENTITIES QUOTIENT IDENTITIES
- 24. Given that cos θ = 𝟑 𝟓 and 0 < θ < 𝜋 𝟐 , find sin θ. EXAMPLE (A)
- 26. Given that cos θ = 𝟑 𝟓 and 0 < θ < 𝜋 𝟐 , find sin θ. 𝒔𝒊𝒏𝟐𝜽 + 𝒄𝒐𝒔𝟐𝜽 = 1 𝒔𝒊𝒏𝟐𝜽 + ( 𝟑 𝟓 )𝟐 = 1 𝒔𝒊𝒏𝟐 𝜽 + 𝟗 𝟐𝟓 = 1 𝒔𝒊𝒏𝟐𝜽 = 1 - 𝟗 𝟐𝟓 𝒔𝒊𝒏𝟐 𝜽 = 𝟏𝟔 𝟐𝟓 𝒔𝒊𝒏 𝜽 = ± 𝟏𝟔 𝟐𝟓 𝒔𝒊𝒏 𝜽 = 𝟒 𝟓
- 27. CONFUSED??? 0 < θ < 𝜋 𝟐
- 28. QUADRANT II QUADRANT III QUADRANT IV QUADRANT I 𝜋 𝟐 < θ < 𝜋 0 < θ < 𝜋 𝟐 𝜋 < θ < 3𝜋 𝟐 Cos is positive Sin and Tan are negative All are positive Tan is positive Sin and Cos are negative Sin is positive Cos and Tan are negative 3𝜋 𝟐 < θ < 2 𝜋 C A S T
- 29. For restrictions: 0 < θ < 𝜋 𝟐 indicates 0 < θ < 90° (where θ is greater than 0 but less than 90°) 𝜋 𝟐 < θ < 𝜋 indicates 90 < θ < 𝟏𝟖𝟎° (where θ is greater than 90° but less than 180°) 𝜋 < θ < 3𝜋 𝟐 indicates 𝟏𝟖𝟎° < θ < 270° (where θ is greater than 180° but less than 270°) 3𝜋 𝟐 < θ < 2 𝜋 indicates 270° < θ < 360° (where θ is greater than 270° but less than 360°)
- 30. Given that cos θ = 𝟑 𝟓 and 0 < θ < 𝜋 𝟐 , find tan θ.
- 32. Given that cos θ = 𝟑 𝟓 and 0 < θ < 𝜋 𝟐 , find tan θ. 𝒕𝒂𝒏 𝜽 = 𝒔𝒊𝒏 𝜽 𝒄𝒐𝒔 𝜽 = 4 5 3 5 = 𝟒 𝟑
- 33. Find the other trigonometric functions 𝜽 from #1.
- 35. Find the other trigonometric functions 𝜽 from #1. 𝒄𝒔𝒄 𝜽 = 𝟏 𝒔𝒊𝒏 𝜽 𝒔𝒆𝒄 𝜽 = 𝟏 𝒄𝒐𝒔 𝜽 𝒄𝒐𝒕 𝜽 = 𝟏 𝒕𝒂𝒏 𝜽 = 𝟏 𝟒 𝟓 = 𝟓 𝟒 = 𝟏 𝟑 𝟓 = 𝟓 𝟑 = 𝟏 𝟒 𝟑 = 𝟑 𝟒
- 38. Given that sin θ = 𝟐 𝟑 and 𝜋 𝟐 < θ < 𝜋 find cos θ. EXAMPLE (B) 𝒔𝒊𝒏𝟐𝜽 + 𝒄𝒐𝒔𝟐𝜽 = 1 ( 𝟐 𝟑 )𝟐 + 𝒄𝒐𝒔𝟐𝜽 = 1 𝟒 𝟗 + 𝒄𝒐𝒔𝟐𝜽 = 1 𝒄𝒐𝒔𝟐 𝜽 = 1 - 𝟒 𝟗 𝒄𝒐𝒔𝟐𝜽 = 𝟓 𝟗 𝒄𝒐𝒔 𝜽 = ± 𝟓 𝟗 𝒄𝒐𝒔 𝜽 = ± 𝟓 𝟑
- 39. Find tan 𝜽 from the given. 𝒕𝒂𝒏 𝜽 = 𝒔𝒊𝒏 𝜽 𝒄𝒐𝒔 𝜽 = 2 3 − 5 3 = 𝟐(𝟑) − 𝟓(𝟑) = 𝟐 − 𝟓 = −𝟐 𝟓 𝟓
- 40. Given that tan θ = − 𝟓 𝟏𝟐 and 𝜋 𝟐 < θ < 𝜋 find sec θ. EXAMPLE (C)
- 42. Given that tan θ = − 𝟓 𝟏𝟐 and 𝜋 𝟐 < θ < 𝜋 find sec θ. EXAMPLE (C) 𝒔𝒆𝒄𝟐 𝜽 = 1 + 𝒕𝒂𝒏𝟐 𝜽 𝒔𝒆𝒄𝟐 𝜽 = 1 + −𝟓 𝟏𝟐 𝟐 𝒔𝒆𝒄𝟐 𝜽 = 1 + 𝟐𝟓 𝟏𝟒𝟒 𝒔𝒆𝒄𝟐 𝜽 = 𝟏𝟔𝟗 𝟏𝟒𝟒 𝒔𝒆𝒄𝟐 𝜽 = 𝟏𝟔𝟗 𝟏𝟒𝟒 𝒔𝒆𝒄 𝜽 = ± 𝟏𝟔𝟗 𝟏𝟒𝟒 𝒔𝒆𝒄 𝜽 = ± 𝟏𝟑 𝟏𝟐 𝒔𝒆𝒄 𝜽 = - 𝟏𝟑 𝟏𝟐
- 43. If 𝑠𝒆𝒄 𝜽 = - 𝟏𝟑 𝟏𝟐 𝒄𝒐𝒔 𝜽 = - 𝟏𝟐 𝟏𝟑 𝒔𝒊𝒏 𝜽 = 𝟓 𝟏𝟑 𝒄𝒔𝒄 𝜽 = 𝟏𝟑 𝟓 𝒄𝒐𝒕 𝜽 = - 𝟏𝟐 𝟓
- 45. EXAMPLE 1: Prove: tan θ = 𝐬𝐞𝐜 𝛉 𝐜𝐬𝐜 𝛉 Since the right-hand member is more complicated, we shall reduce it to the same as the left-handed member is using the fundamental identities, specifically, the identity for tangent function. Proof: tan θ = 𝐬𝐞𝐜 𝛉 𝐜𝐬𝐜 𝛉 Given tan θ = 𝐬𝐞𝐜 𝛉 𝐜𝐬𝐜 𝛉 ( 𝐬𝐢𝐧 𝛉 𝐜𝐨𝐬 𝛉 𝐬𝐢𝐧 𝛉 𝐜𝐨𝐬 𝛉 ) Multiply by sin θ cos θ tan θ = (𝐜𝐨𝐬 𝛉 𝐬𝐞𝐜 𝛉) (𝐬𝐢𝐧 𝛉 𝐜𝐬𝐜 𝛉) ( 𝐬𝐢𝐧𝛉 𝐜𝐨𝐬 𝛉 ) Commutative tan θ = 𝐬𝐢𝐧 𝛉 𝐜𝐨𝐬 𝛉 Reciprocal Identity tan θ = tan θ Quotient Identity
- 46. EXAMPLE 2: Prove: sec2 θ + csc2 θ = sec2 θ csc2 θ The left-hand side is more complicated than the left. We shall simplify it by converting secant and cosine and sine, respectively, the return it to secant and cosecant. Proof: sec2 θ + csc2 θ = sec2 θ csc2 θ Given 𝟏 𝒄𝒐𝒔𝟐𝜽 + 𝟏 𝒔𝒊𝒏𝟐𝜽 = sec2 θ csc2 θ Reciprocal 𝒔𝒊𝒏𝟐𝜽+𝒄𝒐𝒔𝟐𝜽 𝒄𝒐𝒔𝟐𝜽 𝒔𝒊𝒏𝟐𝜽 = sec2 θ csc2 θ LCD (Least Common Denominator) 𝟏 𝒄𝒐𝒔𝟐𝜽 𝒔𝒊𝒏𝟐𝜽 = sec2 θ csc2 θ Pythagorean sec2 θ csc2 θ = sec2 θ csc2 θ Reciprocal
- 47. REMEMBER: Each of the preceding examples represents only one way of proving the identity. The different ways which an identity can be proven provide the students like you with an opportunity to develop your ingenuity and originality.
- 48. EXAMPLE 3: Prove: (tan u + cot u)2 = sec2 u + csc2 u Proof: (tan u + cot u)2 = sec2 θ + csc2 θ Given tan2 u + 2tan u cot u + cot2 u= sec2 θ + csc2 θ Binomial Expansion sec2 u – 1 + 2 + csc2 u – 1 = sec2 u + csc2 u Pythagorean sec2 u + csc2 u = sec2 u + csc2 u Additive I
- 49. THAT WILL BE ALL!!! G O M A W O