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Chapter 5 Trigonometric Functions of Angles Section 4 The Other Trigonometric Functions Setting the Stage We will consider the relationship between the sides of a triangle here in terms of side lengths instead of the value sine and cosine. Setting the Stage We will consider the relationship between the sides of a triangle here in terms of side lengths instead of the value sine and cosine. (x, y) • r y θ x Setting the Stage We will consider the relationship between the sides of a triangle here in terms of side lengths instead of the value sine and cosine. (x, y) • r y θ x sin(θ ) = Setting the Stage We will consider the relationship between the sides of a triangle here in terms of side lengths instead of the value sine and cosine. (x, y) • r y θ x sin(θ ) = y r Setting the Stage We will consider the relationship between the sides of a triangle here in terms of side lengths instead of the value sine and cosine. (x, y) • r y θ x sin(θ ) = cos(θ ) = y r Setting the Stage We will consider the relationship between the sides of a triangle here in terms of side lengths instead of the value sine and cosine. (x, y) • r y θ x sin(θ ) = cos(θ ) = y r x r Other Trig Functions (x, y) • r y θ x Cosecant csc(θ ) = Other Trig Functions (x, y) • r y θ x Cosecant csc(θ ) = 1 = sin(θ ) Other Trig Functions (x, y) • r y θ x Cosecant csc(θ ) = 1 r = sin(θ ) y Other Trig Functions (x, y) • r y θ x Secant sec(θ ) = Other Trig Functions (x, y) • r y θ x Secant sec(θ ) = 1 = cos(θ ) Other Trig Functions (x, y) • r y θ x Secant sec(θ ) = 1 r = cos(θ ) x Other Trig Functions (x, y) • r y θ x Tangent tan(θ ) = Other Trig Functions (x, y) • r y θ x Tangent tan(θ ) = sin(θ ) = cos(θ ) Other Trig Functions (x, y) • r y θ x Tangent tan(θ ) = sin(θ ) y = cos(θ ) x Other Trig Functions (x, y) • r y θ x Cotangent cot(θ ) = Other Trig Functions (x, y) • r y θ x Cotangent cot(θ ) = cos(θ ) = sin(θ ) Other Trig Functions (x, y) • r y θ x Cotangent cot(θ ) = cos(θ ) x = sin(θ ) y Examples Example Evaluate tan(225◦ ). Examples Example Evaluate tan(225◦ ). The reference angle is Examples Example Evaluate tan(225◦ ). The reference angle is 45◦ . Examples Example Evaluate tan(225◦ ). The reference angle is 45◦ . In what quadrant is 225◦ ? Examples Example Evaluate tan(225◦ ). The reference angle is 45◦ . In what quadrant is 225◦ ? quadrant III Examples Example Evaluate tan(225◦ ). The reference angle is 45◦ . In what quadrant is 225◦ ? quadrant III tan(45◦ ) = Examples Example Evaluate tan(225◦ ). The reference angle is 45◦ . In what quadrant is 225◦ ? quadrant III tan(45◦ ) = sin(45◦ ) cos(45◦ ) = Examples Example Evaluate tan(225◦ ). The reference angle is 45◦ . In what quadrant is 225◦ ? quadrant III tan(45◦ ) = sin(45◦ ) cos(45◦ ) √ = 2 √2 2 2 = Examples Example Evaluate tan(225◦ ). The reference angle is 45◦ . In what quadrant is 225◦ ? quadrant III tan(45◦ ) = sin(45◦ ) cos(45◦ ) √ = 2 √2 2 2 =1 Examples Example Evaluate tan(225◦ ). The reference angle is 45◦ . In what quadrant is 225◦ ? quadrant III tan(45◦ ) = sin(45◦ ) cos(45◦ ) √ = 2 √2 2 2 =1 So, tan(225◦ ) = Examples Example Evaluate tan(225◦ ). The reference angle is 45◦ . In what quadrant is 225◦ ? quadrant III tan(45◦ ) = sin(45◦ ) cos(45◦ ) √ = 2 √2 2 2 =1 So, tan(225◦ ) = 1 Examples Example Evaluate csc 5π 6 Examples Example Evaluate csc 5π 6 The reference angle is Examples Example Evaluate csc 5π 6 The reference angle is π 6. Examples Example Evaluate csc 5π 6 The reference angle is In what quadrant is π 6. 5π 6 ? Examples Example Evaluate csc 5π 6 The reference angle is In what quadrant is π 6. 5π 6 ? quadrant II Examples Example Evaluate csc 5π 6 The reference angle is In what quadrant is π 6. 5π 6 ? csc quadrant II π 6 = Examples Example Evaluate csc 5π 6 The reference angle is In what quadrant is π 6. 5π 6 ? csc quadrant II π 6 = 1 sin π 6 = Examples Example Evaluate csc 5π 6 The reference angle is In what quadrant is π 6. 5π 6 ? csc quadrant II π 6 = 1 sin π 6 = = 1 1 2 Examples Example Evaluate csc 5π 6 The reference angle is In what quadrant is π 6. 5π 6 ? csc quadrant II π 6 = 1 sin π 6 = 1 1 2 =2 Examples Example Evaluate csc 5π 6 The reference angle is In what quadrant is π 6. 5π 6 ? csc quadrant II π 6 = 1 sin π 6 = 1 1 2 =2 So, csc 5π 6 = Examples Example Evaluate csc 5π 6 The reference angle is In what quadrant is π 6. 5π 6 ? csc quadrant II π 6 = 1 sin π 6 = 1 1 2 =2 So, csc 5π 6 =2 Examples Example Evaluate cot 11π 3 Examples Example Evaluate cot 11π 3 The reference angle is Examples Example Evaluate cot 11π 3 The reference angle is π 3. Examples Example Evaluate cot 11π 3 The reference angle is In what quadrant is π 3. 11π 3 ? Examples Example Evaluate cot 11π 3 The reference angle is In what quadrant is π 3. 11π 5π 3 ? 3 ? Examples Example Evaluate cot 11π 3 The reference angle is In what quadrant is π 3. 11π 5π 3 ? 3 ? quadrant IV Examples Example Evaluate cot 11π 3 The reference angle is In what quadrant is π 3. 11π 5π 3 ? 3 ? cot π 3 quadrant IV = Examples Example Evaluate cot 11π 3 The reference angle is In what quadrant is π 3. 11π 5π 3 ? 3 ? cot π 3 quadrant IV cos = sin = π 3 π 3 Examples Example Evaluate cot 11π 3 The reference angle is In what quadrant is π 3. 11π 5π 3 ? 3 ? cot π 3 quadrant IV cos = sin = = π 3 π 3 1 √2 3 2 Examples Example Evaluate cot 11π 3 The reference angle is In what quadrant is π 3. 11π 5π 3 ? 3 ? cot π 3 quadrant IV cos = sin = π 3 π 3 1 √2 3 2 1 =√ 3 Examples Example Evaluate cot 11π 3 The reference angle is In what quadrant is π 3. 11π 5π 3 ? 3 ? cot π 3 quadrant IV cos = sin = π 3 π 3 1 √2 3 2 1 =√ 3 So, cot 11π 3 = Examples Example Evaluate cot 11π 3 The reference angle is In what quadrant is π 3. 11π 5π 3 ? 3 ? cot π 3 quadrant IV cos = sin = π 3 π 3 1 √2 3 2 1 =√ 3 So, cot 11π 3 = − √13 Identities We all remember the Pythagorean Identity, right? Identities We all remember the Pythagorean Identity, right? The Pythagorean Identity sin2 θ + cos2 θ = 1 Identities We all remember the Pythagorean Identity, right? The Pythagorean Identity sin2 θ + cos2 θ = 1 We can use variations of this to prove different equations hold. Identities We all remember the Pythagorean Identity, right? The Pythagorean Identity sin2 θ + cos2 θ = 1 We can use variations of this to prove different equations hold. Variation 1 1 + cot2 θ = csc2 θ Identities We all remember the Pythagorean Identity, right? The Pythagorean Identity sin2 θ + cos2 θ = 1 We can use variations of this to prove different equations hold. Variation 1 1 + cot2 θ = csc2 θ Variation 2 tan2 θ + 1 = sec2 θ Identities Example Simplify secθcosθ − cos2 θ. Identities Example Simplify secθcosθ − cos2 θ. secθcosθ − cos2 θ Identities Example Simplify secθcosθ − cos2 θ. secθcosθ − cos2 θ = 1 cosθ − cos2 θ cosθ Identities Example Simplify secθcosθ − cos2 θ. secθcosθ − cos2 θ = 1 cosθ − cos2 θ cosθ = 1 − cos2 θ Identities Example Simplify secθcosθ − cos2 θ. secθcosθ − cos2 θ = 1 cosθ − cos2 θ cosθ = 1 − cos2 θ = sin2 θ Identities Example Show that cos2 θ + cos2 θtan2 θ = 1. Identities Example Show that cos2 θ + cos2 θtan2 θ = 1. cos2 θ + cos2 θtan2 θ Identities Example Show that cos2 θ + cos2 θtan2 θ = 1. cos2 θ + cos2 θtan2 θ = cos2 θ + cos2 θ sin2 θ cos2 θ Identities Example Show that cos2 θ + cos2 θtan2 θ = 1. sin2 θ cos2 θ 2 = cos θ + sin2 θ cos2 θ + cos2 θtan2 θ = cos2 θ + cos2 θ Identities Example Show that cos2 θ + cos2 θtan2 θ = 1. sin2 θ cos2 θ 2 = cos θ + sin2 θ cos2 θ + cos2 θtan2 θ = cos2 θ + cos2 θ =1 Identities Example Simplify sin4 θ − cos4 θ. Identities Example Simplify sin4 θ − cos4 θ. sin4 θ − cos4 θ Identities Example Simplify sin4 θ − cos4 θ. sin4 θ − cos4 θ = (sin2 θ + cos2 θ )(sin2 θ − cos2 θ ) Identities Example Simplify sin4 θ − cos4 θ. sin4 θ − cos4 θ = (sin2 θ + cos2 θ )(sin2 θ − cos2 θ ) = 1(sin2 θ − cos2 θ ) = Identities Example Simplify sin4 θ − cos4 θ. sin4 θ − cos4 θ = (sin2 θ + cos2 θ )(sin2 θ − cos2 θ ) = 1(sin2 θ − cos2 θ ) = sin2 θ − (1 − sin2 θ ) Identities Example Simplify sin4 θ − cos4 θ. sin4 θ − cos4 θ = (sin2 θ + cos2 θ )(sin2 θ − cos2 θ ) = 1(sin2 θ − cos2 θ ) = sin2 θ − (1 − sin2 θ ) = 2sin2 θ − 1 Identities Example Simplify 1−cos2 θ . sec2 θ −1 Identities Example Simplify 1−cos2 θ . sec2 θ −1 Whenever we see the square of a trig function ±1, we want to replace it with another square of a trig function when possible. Identities Example Simplify 1−cos2 θ . sec2 θ −1 Whenever we see the square of a trig function ±1, we want to replace it with another square of a trig function when possible. 1 − cos2 θ sec2 θ − 1 Identities Example Simplify 1−cos2 θ . sec2 θ −1 Whenever we see the square of a trig function ±1, we want to replace it with another square of a trig function when possible. sin2 θ 1 − cos2 θ = sec2 θ − 1 tan2 θ Identities Example Simplify 1−cos2 θ . sec2 θ −1 Whenever we see the square of a trig function ±1, we want to replace it with another square of a trig function when possible. sin2 θ 1 − cos2 θ = sec2 θ − 1 tan2 θ cos2 θ = sin2 θ · sin2 θ Identities Example Simplify 1−cos2 θ . sec2 θ −1 Whenever we see the square of a trig function ±1, we want to replace it with another square of a trig function when possible. sin2 θ 1 − cos2 θ = sec2 θ − 1 tan2 θ cos2 θ = sin2 θ · sin2 θ = cos2 θ