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Unit 8.2 Relationships Between Sine, Cosine and Tangent Sine and Cosine Consider triangle ABC: ο· Since the sum of the interior angles of a triangle must add up to 180°, the two acute angles of a right triangle must add to 90°. Angle A + Angle B = 90° ο· Angle A = 90° - Angle B Angle B = 90° - Angle A Using the trigonometric ratios for right triangles, the following ratios from β²ABC can be obtained: π π ο§ Sin A = = Cos B Cos A = = Sin B π π So, Sin ΞΈ = cos(90° - ΞΈ) and Cos ΞΈ = sin(90° - ΞΈ) sin 30° = cos 60° and cos 30° = sin 60° sin 55° = cos 35° and cos 55° = sin 35° For Example: Tangent Using the trigonometric ratios for right triangles, the following ratios from β²ABC can be obtained: Tan B = π π sin B = π π cos B = π π So, sin π΅ cos π΅ = π π π π = π π = tan B Therefore, tan ΞΈ = sin π cos π Sine2 and Cosine2 Consider Pythagorasβ Theorem: ο· ο· a2 + b2 = c2 Again we can use our trigonometric ratios for right triangles The following ratios can be obtained: sin B = cos B = π π π π b = c(sin B) b2 = c2(sin2B) a = c(cos B) a2 = c2(cos2B) * Note: (sin B)2 is usually written as sin2B, (cos B)2 is usually written as cos2 B By substitution: a2 + b2 = c2 c2sin2B + c2cos2B = c2 sin2B + cos2B = 1 Soβ¦. Sin2ΞΈ + Cos2ΞΈ = 1 Sine, Cosine and Tangent If sin ΞΈ + cos ΞΈ = 1, 2 Soβ¦ 2 tan2ΞΈ + 1 = then 1 cos2 ΞΈ sin2 ΞΈ cos2 ΞΈ + cos2 ΞΈ cos2 ΞΈ = 1 cos2 ΞΈ Summary 1. SinΞΈ = cos(90° - ΞΈ) 2. SinΞΈ = cos(90° - ΞΈ) 3. TanΞΈ = π πππ πππ π 4. Sin2ΞΈ + cos2ΞΈ = 1 5. Tan2ΞΈ + 1 = 1 cos2 ΞΈ * 3 Example 1: Given sinΞΈ = Solution 1: Draw a right triangle with acute angle ΞΈ opposite side of length 3, and hypotenuse of length 5. SOH CAH TOA 5 is an acute angle, find the exact value of cosΞΈ and tanΞΈ. 5 3 ΞΈ a Method 1: Method 2: By PYTHAGORUSβ Theorem sin2ΞΈ + cos2 ΞΈ = 1 a2 + b2 = c2 3 ( )2 + cos2ΞΈ = 1 5 a2 + 32 = 52 3 cos2ΞΈ = 1 β ( )2 5 a = 25 β 9 2 a2 = 16 cos2ΞΈ = 1 - 9 25 = 25 - 9 25 25 a=4 cosΞΈ = β So, ππ ππππππ cosΞΈ = ππππππππππ = ππππππππ tanΞΈ = ππ ππππππ = π π 16 25 = π π π π 3 π ππΞΈ 5 π tanΞΈ = = = πππ ΞΈ 4 π 5 = 16 25 2 , find the exact value of sinΞΈ and tanΞΈ. Example 2: Given cosΞΈ = Solution 2: Draw a right triangle with acute angle ΞΈ adjacent side of length 2, and hypotenuse of length 3. SOH CAH TOA 3 3 b ΞΈ 2 Method 1: Method 2: By PYTHAGORUSβ Theorem sin2ΞΈ + cos2 ΞΈ = 1 a2 + b2 = c2 2 sin2ΞΈ + ( )2 = 1 3 22 + b2 = 32 2 sin2ΞΈ = 1 β ( )2 3 a =9β4 2 a2 = 5 sin2ΞΈ = 1 - 4 9 = 9 4 5 - = 9 9 9 = βπ π a = β5 sinΞΈ = β So, ππππππππ sinΞΈ = ππππππππππ = ππππππππ tanΞΈ = ππ ππππππ = βπ π βπ π 5 9 β5 π ππΞΈ 3 βπ tanΞΈ = = 2 = πππ ΞΈ π 3