# Trigonometry—Double-Angle and Half-Angle Identities Double-Angle IdentitiesHalf-Angle Identities cos (2  ) = cos 2  – sin 2  = 2 cos 2  – 1 = 1 – 2.

## Presentation on theme: "Trigonometry—Double-Angle and Half-Angle Identities Double-Angle IdentitiesHalf-Angle Identities cos (2  ) = cos 2  – sin 2  = 2 cos 2  – 1 = 1 – 2."— Presentation transcript:

Trigonometry—Double-Angle and Half-Angle Identities Double-Angle IdentitiesHalf-Angle Identities cos (2  ) = cos 2  – sin 2  = 2 cos 2  – 1 = 1 – 2 sin 2  sin (2  ) = 2 sin  cos  tan (2  ) = The Proofs: sin (2  ) = tan (2  ) = cos (2  ) = Recall: Addition Identities cos (  +  ) = cos  cos  – sin  sin  sin (  +  ) = sin  cos  + cos  sin  tan (  +  ) = Page 33

Trigonometry—Double-Angle and Half-Angle Identities (cont’d) Double-Angle IdentitiesHalf-Angle Identities cos (2  ) = cos 2  – sin 2  = 2 cos 2  – 1 = 1 – 2 sin 2  sin (2  ) = 2 sin  cos  tan (2  ) = The Applications: If cos  = 3/5 ( 3  / 2 <  < 2  ) and tan  = 7/24 (  <  < 3  / 2 ), find a) sin (2  ) b) cos ( 1 / 2  ) c) cot (2  ) Page 34

Trigonometry—Word Problems (cont’d) 3. After having built a fence around a triangular plot determined by a pine tree (P), an oak tree (O) and a willow tree (W), a rancher hired a surveyor to determine the area of the plot. The surveyor asked, “Do you have some measurements?” The rancher replied that he had the measurements of all three angles and length of one of its sides. The surveyor then asked him for the measurements. “The angle at the pine tree formed by the fencing is 78 , the angle at the oak tree is 56 , and...” said the rancher. Feeling annoyed, the surveyor interrupted, “Just give the side you’ve measured.” So, the rancher said, “The fencing along the oak tree and the willow tree is 200 yards.” What is the area of the plot? 4. What is the perimeter? Page 35

Trigonometry—Equations When we are solving the word problems on the previous slides, we set up equations to solve them. Of course, there are trigonometric equations we don’t need to set up but we need to solve them. Most of the time, it is the angle (usually denoted by x or  ) we need to solve. Example: Solve for x: sin x = ½ Solution: x = sin –1 (½) = 30°. But wait, x could be in the 2 nd quadrant too since sine of a 2 nd quadrant angle is positive also. So what is it? But wait, there is more—since trigonometric function value of coterminal angles are equal. So what are they? Conclusion: Since there will be _______ many angles that satisfy the equation, we usually want _______ many of them, namely, those are between ____ and ____ (in degrees) or between ____ and ____ (in radians). Hence there will be always a stated/restricted domain for the equation we need to solve: Solve for x where 0°  x < 360°: sin x = ½ or Solve for x where x  [0, 2  ): sin x = ½ Page 36

Trigonometry—Equations (cont’d) Problems: Solve for x where 0°  x < 360°: 1. cos x = –½2.tan 2 x = 1 3.2sin 2 x + sin x – 1 = 04. sin 2x + cos x = 0 0°, 360° 90° 180° 270° In Degrees 0, 2   /2  3  /2 0, 6.28 1.57 3.14 4.71 In Radians In terms of  In terms of decimal If we know  ref and  is in the respective quadrants we can find  as follows: Q I:  = _______ Q II:  = _______ Q III:  = _______ Q IV:  = _______ Page 37

Trigonometry—Equations (cont’d) Problems: Solve for x where 0°  x < 360°: 1.cos 2 x = ¼ 2.cos 2x – cos x – 2 = 0 Solve for x where 0  x < 2  : 1. cos 2x – 3sin x + 4 = 02. 2tan 2 x + tan x – 3 = 0 3. 2sec 2 x + 3sec x + 1 = 0 90° 0°, 360°180° 270° In Degrees If we know  ref and  is in the respective quadrants we can find  as follows: Q I:  =  ref Q II:  = 180  –  ref Q III:  = 180  +  ref Q IV:  = 360  –  ref  /2 0, 2   3  /2 In Radians In terms of  1.57 0, 6.28 3.14 4.71 In terms of decimal SIN 00 –1 COS 1–1 0 Page 38

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