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Trigonometric ratios

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This presentation illustrates the concepts of trigonometric ratios

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Trigonometric ratios

  1. 1. Trigonometric Ratios by SBR www.harekrishnahub.com
  2. 2. www.harekrishnahub.com Consider a circle with centre 𝑢 and radius 𝒓 units. Let 𝑷 (𝒙, π’š) be any point on the circumference of the circle. Join 𝑢𝑷. Let the radius vector 𝑢𝑷 make an angle 𝜽 with the positive 𝒙 βˆ’ π’‚π’™π’Šπ’”. ∴ βˆ π‘Ώπ‘Άπ‘· = 𝜽 and 𝑢𝑷 = 𝒓 Draw 𝑷𝑴 βŸ‚ the 𝒙 βˆ’ π’‚π’™π’Šπ’”. ∴ 𝑢𝑴 = 𝒙 and 𝑷𝑴 = π’š
  3. 3. www.harekrishnahub.com 𝑷𝑴𝑢 form a right angle triangle as shown below. Let us identify each of the sides of the triangle. 𝑢𝑷 = 𝒓 = 𝒙 𝟐 + π’š 𝟐 is the hypotenuse 𝑷𝑴 = π’š, the side opposite to 𝜽 is called the opposite side. 𝑢𝑴 = 𝒙, the side adjacent to 𝜽 is called the adjacent side.
  4. 4. www.harekrishnahub.com 6 trigonometric ratios Sine of angle ΞΈ 𝑠𝑖𝑛 πœƒ π‘œπ‘π‘π‘œπ‘ π‘–π‘‘π‘’ 𝑠𝑖𝑑𝑒 β„Žπ‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’ π’š 𝒓 = π’š 𝒙 𝟐 + π’š 𝟐 CoSine of angle ΞΈ π‘π‘œπ‘  πœƒ π‘Žπ‘‘π‘—π‘Žπ‘π‘’π‘›π‘‘ 𝑠𝑖𝑑𝑒 β„Žπ‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’ 𝒙 𝒓 = 𝒙 𝒙 𝟐 + π’š 𝟐 Tangent of angle ΞΈ π‘‘π‘Žπ‘› πœƒ π‘œπ‘π‘π‘œπ‘ π‘–π‘‘π‘’ 𝑠𝑖𝑑𝑒 π‘Žπ‘‘π‘—π‘Žπ‘π‘’π‘›π‘‘ 𝑠𝑖𝑑𝑒 π’š 𝒙 for x β‰  0 CoSecant of angle ΞΈ π‘π‘œπ‘ π‘’π‘ πœƒ β„Žπ‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’ π‘œπ‘π‘π‘œπ‘ π‘–π‘‘π‘’ 𝑠𝑖𝑑𝑒 𝒓 π’š = 𝒙 𝟐 + π’š 𝟐 π’š for y β‰  0 Secant of angle ΞΈ 𝑠𝑒𝑐 πœƒ β„Žπ‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’ π‘Žπ‘‘π‘—π‘Žπ‘π‘’π‘›π‘‘ 𝑠𝑖𝑑𝑒 𝒓 𝒙 = 𝒙 𝟐 + π’š 𝟐 𝒙 for x β‰  0 CoTangent of angle ΞΈ π‘π‘œπ‘‘ πœƒ π‘Žπ‘‘π‘—π‘Žπ‘π‘’π‘›π‘‘ 𝑠𝑖𝑑𝑒 π‘œπ‘π‘π‘œπ‘ π‘–π‘‘π‘’ 𝑠𝑖𝑑𝑒 𝒙 π’š for y β‰  0
  5. 5. www.harekrishnahub.com But 𝒕𝒂𝒏 𝜽 = π’π’‘π’‘π’π’”π’Šπ’•π’† π’”π’Šπ’…π’† 𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕 π’”π’Šπ’…π’† We have, 𝒕𝒂𝒏 𝜽 = π’π’‘π’‘π’π’”π’Šπ’•π’† π’”π’Šπ’…π’† π’‰π’šπ’‘π’π’•π’†π’π’–π’”π’† 𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕 π’”π’Šπ’…π’† π’‰π’šπ’‘π’π’•π’†π’π’–π’”π’† Dividing both the numerator and the denominator by hypotenuse, we get 𝒄𝒐𝒔 𝜽 = 𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕 π’”π’Šπ’…π’† π’‰π’šπ’‘π’π’•π’†π’π’–π’”π’† π’”π’Šπ’ 𝜽 = π’π’‘π’‘π’π’”π’Šπ’•π’† π’”π’Šπ’…π’† π’‰π’šπ’‘π’π’•π’†π’π’–π’”π’† ∴ 𝒕𝒂𝒏 𝜽 = π’”π’Šπ’ 𝜽 𝒄𝒐𝒔 𝜽
  6. 6. www.harekrishnahub.com Reciprocal relations π’”π’Šπ’ 𝜽 = 𝑢𝒑𝒑 π’”π’Šπ’…π’† π’‰π’šπ’‘ = 𝟏 π’‰π’šπ’‘ 𝑢𝒑𝒑 π’”π’Šπ’…π’† = 𝟏 𝒄𝒐𝒔𝒆𝒄 𝜽 𝒄𝒐𝒔 𝜽 = 𝒂𝒅𝒋 π’”π’Šπ’…π’† π’‰π’šπ’‘ = 𝟏 𝒂𝒅𝒋 𝑢𝒑𝒑 π’”π’Šπ’…π’† = 𝟏 𝒔𝒆𝒄 𝜽 𝒕𝒂𝒏 𝜽 = 𝑢𝒑𝒑 π’”π’Šπ’…π’† 𝒂𝒅𝒋 π’”π’Šπ’…π’† = 𝟏 𝒂𝒅𝒋 π’”π’Šπ’…π’† 𝑢𝒑𝒑 π’”π’Šπ’…π’† = 𝟏 𝒄𝒐𝒕 𝜽 Therefore, π’”π’Šπ’ 𝜽 and 𝒄𝒐𝒔𝒆𝒄 𝜽 are reciprocal to each other Therefore, 𝒄𝒐𝒔 𝜽 and 𝒔𝒆𝒄 𝜽 are reciprocal to each other Therefore, 𝒕𝒂𝒏 𝜽 and 𝒄𝒐𝒕 𝜽 are reciprocal to each other

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