Math 147 - Assignment 11 Solutions - Spring 2012 - BSU - Jaimos F Skriletz 1. Trigonometric Identities Prove the following trigonometric identities: (a) tan(x) − cot(x) = sin(x) cos(x) tan2 (x) − cot2 (x) sin(x) cos(x) − tan(x) − cot(x) cos(x) sin(x) = 2 2 2 tan (x) − cot (x) sin (x) cos2 (x) − 2 cos (x) sin2 (x) sin2 (x) − cos2 (x) sin(x) cos(x) = sin4 (x) − cos4 (x) sin2 (x) cos2 (x) 2 sin (x) − cos2 (x) sin2 (x) cos2 (x) = sin(x) cos(x) sin4 (x) − cos4 (x) (sin2 (x) − cos2 (x)) sin(x) cos(x) (sin2 (x) − cos2 (x))(sin2 (x) + cos2 (x)) sin(x) cos(x) = sin2 (x) + cos2 (x) = sin(x) cos(x) = (b) tan(x) + cot(x) = sec(x) csc(x) tan(x) + cot(x) = sin(x) cos(x) + cos(x) sin(x) sin2 (x) + cos2 (x) sin(x) cos(x) 1 = sin(x) cos(x) 1 1 = cos(x) sin(x) = = sec(x) csc(x) 1 Math 147 - Assignment 11 Solutions - Spring 2012 - BSU - Jaimos F Skriletz (c) sin(3x) + sin(7x) = cot(2x) cos(3x) − cos(7x) The following two sum to product identities will be used u+v u−v sin(u) + sin(v) = 2 sin cos 2 2 u+v u−v cos(u) − cos(v) = −2 sin sin 2 2 Thus we get the following: 3x + 7x 3x − 7x cos sin(3x) + sin(7x) 2 2 = 3x + 7x 3x − 7x cos(3x) − cos(7x) −2 sin sin 2 2 − sin(5x) cos(−2x) = sin(5x) sin(−2x) − cos(−2x) = = − cot(−2x) sin(−2x) = −(− cot(2x)) = cot(2x) 2 sin 2. Sum of Angle Identities Use the sum of angle identities to find the exact values of the following: (a) sin(7π/12) First 7π/12 = 3π/12 + 4π/12 = π/4 + π/3 so use the sum of angle identity sin(a + b) = sin(a) cos(b) + sin(b) cos(a) to get that sin(7π/12) = sin(π/4 + π/3) = sin(π/4) cos(π/3) + sin(π/3) cos(π/4) √ ! √ ! √ ! 2 1 3 2 = + 2 2 2 2 √ √ 2+ 6 = 4 (b) tan(15◦ ) First 15◦ = 45◦ − 30◦ so use the difference of angles identity tan(a − b) = tan(a) − tan(b) 1 + tan(a) tan(b) to get that √ √ tan(45◦ ) − tan(30◦ ) 1 − 3/3 3− 3 √ √ tan(15 ) = tan(45 − 30 ) = = = 1 + tan(45◦ ) tan(30◦ ) 1 + (1)( 3/3) 3+ 3 √ √ √ √ (3 − 3)2 9−6 3+3 12 − 6 3 √ √ = = = =2− 3 9 − 3 6 (3 + 3)(3 − 3) ◦ ◦ ◦ 2 Math 147 - Assignment 11 Solutions - Spring 2012 - BSU - Jaimos F Skriletz 3 3. Triple Angle Identities Prove the following triple angle identity: sin(3x) = 3 sin(x) − 4 sin3 (x) To prove this identity first use the sum of angle formulas to get sin(3x) = sin(2x + x) = sin(2x) cos(x) + sin(x) cos(2x) Next use the double angle identities sin(2x) = 2 sin(x) cos(x) and cos(2x) = 1 − 2 sin2 (x) to get sin(3x) = sin(2x) cos(x) + sin(x) cos(2x) = [2 sin(x) cos(x)] cos(x) + sin(x)[1 − 2 sin2 (x)] = 2 sin(x) cos2 (x) + sin(x) − 2 sin3 (x) = 2 sin(x)(1 − sin2 (x)) + sin(x) − 2 sin3 (x) = 2 sin(x) − 2 sin3 (x) + sin(x) − 2 sin3 (x) = 3 sin(x) − 4 sin3 (x) 4. Sums of Sine and Cosine Waves Consider the function √ 3 sin(x) − 3 3 cos(x) We can write√this as a single sine wave by first plotting the point (3, −3 3) and creating a right triangle as shown. By the Pythagorean Theorem we have that q √ √ r = (3)2 + (−3 3)2 = 36 = 6 And the reference angle for the triangle is π/3 so the actual angle in the fourth quadrant is φ= So sin(5π/3) = √ −3 3 6 5π 3 and cos(5π/3) = 63 . Putting it together along with the sum of angles formula we get that √ 3 sin(x) − 3 3 cos(x) = 6 ! √ 3 −3 3 sin(x) + cos(x) 6 6 = 6(cos(5π/3) sin(x) + sin(5π/3) cos(x)) = 6 sin(x + 5π/3) Is a sine wave with amplitude 6 and phase shift 5π/3. (note: Some prefer to use the angle −π/3 as opposed to 5π/3, but since both angles have the same terminal point they are equivalent.) Math 147 - Assignment 11 Solutions - Spring 2012 - BSU - Jaimos F Skriletz 5. Inverse Tangent Consider the tangent function, f (x) = tan(x) on the domain (−π/2, π/2). The graph of f (x) is as shown: (a) Domain: (−π/2, π/2). (b) Range: (−∞, ∞). (c) The graph passes the horizontal line test so it is a one-to-one function and has and inverse. (d) There are two vertical asymptotes at x= −π 2 and x = π 2 The inverse function is f −1 (x) = tan−1 (x) • Domain: (−∞, ∞) • Range: (−π/2, π/2) • Horizontal asymptotes at y= −π 2 and y = π 2 6. Inverse Trigonometric Functions (a) Find an algebraic expression for cos(tan−1 (x)) an where x is any real number. Let θ = tan−1 (x) so tan(θ) = x = x 1 Thus θ forms a right triangle as shown. Find the hypotenuse of the right triangle from the Pythagorean Theorem h2 = 11 + x2 p h = 1 + x2 Thus we have that cos(θ) = cos(tan−1 (x)) = √ 1 1 + x2 4 Math 147 - Assignment 11 Solutions - Spring 2012 - BSU - Jaimos F Skriletz (b) Find an algebraic expression for cot(sin−1 (x)) where −1 ≤ x ≤ 1. Let θ = sin−1 (x) so sin(θ) = x = x 1 Thus θ forms a right triangle as shown. Find the adjacent side of the right triangle from the Pythagorean Theorem 12 = a2 + x2 p a = 1 − x2 Thus we have that √ cot(θ) = cot(sin−1 (x)) = 1 − x2 x 5
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