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 Introduction  Recap  Different Trigonometric Identities › Pythagorean identities › Reciprocal Identities  How these work  Q and A.

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Presentation on theme: " Introduction  Recap  Different Trigonometric Identities › Pythagorean identities › Reciprocal Identities  How these work  Q and A."— Presentation transcript:

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2  Introduction  Recap  Different Trigonometric Identities › Pythagorean identities › Reciprocal Identities  How these work  Q and A

3  Equations and Equalities that involve trigonometric functions  Variables (a, b, x, y, etc.) in every identity › TRUE for any value of the variable

4 Hypotenuse Longest line opposite right angle Opposite Line opposite the angle that you are finding Adjacent Shorter line touching the angle you are finding

5  Basic Functions (SOH CAH TOA) › Sine (sin) – opposite / hypotenuse › Cosine (cos) – adjacent / hypotenuse › Tangent (tan) – opposite / adjacent  Inverse Functions › ArcSine (sin -1 ) › ArcCosine (cos -1 ) › ArcTangent (tan -1 )

6  Other Functions › Cosecant (cosec) – hypotenuse / opposite › Secant (sec) – hypotenuse / adjacent › Cotangent (cotan) – adjacent / opposite

7  (sinθ) 2 + (cosθ) 2 = 1  1 + (tanθ) 2 = (secθ) 2  1 + (cotanθ) 2 = (cosecθ) 2

8  (sinθ) 2 + (cosθ) 2 = 1  ( 3 / 5 ) 2 + ( 4 / 5 ) 2 = 1  9/25 + 16/5 = 1  25/25 = 1  Pythagoras Theorem  Adjacent 2 + Opposite 2  = Hypotenuse 2 37.9 o 3cm 5cm 4cm

9  1 + (tanθ)2 = (secθ)2  1 + ( 3 / 4 ) 2 = ( 5 / 4 ) 2  1 + 9/16 = 25/16  25/16 – 9/16 = 1  16/16 = 1 37.9 o 3cm 5cm 4cm

10  1 + (cotanθ) 2 = (cosecθ) 2  1 + ( 4 / 3 ) 2 = ( 5 / 3 ) 2  1 + 16/9 = 25/9  25/9 – 16/9 = 1  9/9 = 1 37.9 o 3cm 5cm 4cm

11  Identities, where: › A function being used on a particular angle = the reciprocal of another function used on the exact same number › Thus the name is pretty much largely self- explanatory ^^  Essentially, it was made with very close reference to the reciprocal functions

12  sinθ = 1 / cosecθ & cosecθ = 1 / sinθ  cosθ = 1 / secθ & secθ = 1 / cosθ  tanθ = 1 / cotanθ & cotanθ = 1 / tanθ

13  Any Questions?


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