Concept

Find the value of cos 2 if sin = and  is between 0° and 90°. Double-Angle Identities Find the value of cos 2 if sin = and  is between 0° and 90°. cos 2 = 1 – 2 sin2  Double-angle identity Simplify. Example 1

Step 1 Use the identity sin2 = 1 – cos2 to find the value of cos. Double-Angle Identities A. Find the exact value of tan 2 if cos = and  is between 0° and 90°. Step 1 Use the identity sin2 = 1 – cos2 to find the value of cos. cos2 + sin2 = 1 Subtract. Example 2

Take the square root of each side. Double-Angle Identities Take the square root of each side. Step 2 Find tan to use the double-angle identity for tan 2. Definition of tangent Example 2

Double-angle identity Double-Angle Identities Simplify. Step 3 Find tan 2. Double-angle identity Example 2

Square the denominator and simplify. Double-Angle Identities Square the denominator and simplify. Simplify. Answer: Example 2

Double-angle identity Double-Angle Identities B. Find the exact value of sin 2 if cos = and  is between 0° and 90°. Double-angle identity Simplify. Answer: Example 2

A. Find the exact value of cos2 if sin = and  is between 0° and 90°. Example 2

B. Find the exact value of tan2 if sin = and  is between 0° and 90°. Example 2

Concept

A. Find and  is the second quadrant. Half-Angle Identities A. Find and  is the second quadrant. Since we must find cos first. cos2 = 1 – sin2 sin2 + cos2 = 1 Simplify. Example 3A

Take the square root of each side. Half-Angle Identities Take the square root of each side. Since  is in the second quadrant, Half-angle identity Example 3

Rationalize the denominator. Half-Angle Identities Simplify the radicand. Rationalize the denominator. Multiply. Example 3

Half-Angle Identities Answer: Example 3

B. Find the exact value of sin165. Half-Angle Identities B. Find the exact value of sin165. 165 is in Quadrant II; the value is positive. Example 3B

Half-Angle Identities Simplify. Simplify. Answer: Example 3

A. Find and  is in the fourth quadrant. B. C. D. Example 3A

B. Find the exact value of cos157.5. Example 3B

Simplify Using Double-Angle Identities FOUNTAIN Chicago’s Buckingham Fountain contains jets placed at specific angles that shoot water into the air to create arcs. When a stream of water shoots into the air with velocity v at an angle of  with the horizontal, the model predicts that the water will travel a horizontal distance of D = sin 2 and reach a maximum height of H = sin2. The ratio of H to D helps determine the total height and width of the fountain. Find Example 4

Simplify the numerator and the denominator. Simplify Using Double-Angle Identities Original equation Simplify the numerator and the denominator. Example 4

Simplify. sin2 = 2sin cos Simplify. Simplify Using Double-Angle Identities Simplify. sin2 = 2sin cos Simplify. Example 4

Simplify Using Double-Angle Identities Answer: Example 4

Verify that sin θ (cos2 θ – cos 2θ) = sin3 θ is an identity. Verify Identities Verify that sin θ (cos2 θ – cos 2θ) = sin3 θ is an identity. Answer: sin θ (cos2 θ – cos 2θ) = sin3 θ Original equation ? sin θ [cos2 θ – (cos2 θ – sin2 θ)] = sin3 θ cos 2θ = cos2θ – sin2θ ? sin θ (cos2 θ – cos2 θ + sin2 θ) = sin3 θ Distributive Property ? sin θ (sin2 θ) = sin3 θ Simplify. ? sin3 θ = sin3 θ Multiply. Example 5

C. yes; cos2 θ + 2sin θ cos θ – sin2 θ = cos 2θ Determine whether cos θ (cos θ + sin θ) – sin θ (cos θ + sin θ) = cos 2θ is an identity. A. yes; cos2 θ – sin2 θ = cos 2θ B. yes; cos2 θ – sin2 θ = 1 C. yes; cos2 θ + 2sin θ cos θ – sin2 θ = cos 2θ D. no Example 5