Questions and Answers #28 Trigonometry

Questions and Answers Sheet 28 Trigonometry Question #1 Solving 𝐸= 1 √3 − sin 1 0∘ cos 1 0∘ Answer: We have cos 1 0∘ − √3 sin 1 0∘ 𝐸= sin 1 0∘ cos 1 0∘ (1/2) cos 1 0∘ − (√3/2) sin 1 0∘ =4 2 sin 1 0∘ cos 1 0∘ ∘ cos(60 + 10∘ ) =4 sin(20∘ ) cos 7 0∘ =4 sin 2 0∘ sin 2 0∘ =4 sin 2 0∘ =4 Question #2 Prove 3𝜋 3𝜋 cos ( + 𝑥) cos(2𝜋 + 𝑥) [cot ( − 𝑥) + cot(2𝜋 + 𝑥) = 1] 2 2 Answer: 𝜋 𝜋 𝜋 𝜋 + 𝑥) = cos (2𝜋 + 𝑥 − ) = cos (𝑥 − ) = cos ( − 𝑥) = sin 𝑥 2 2 2 2 cos(2𝜋 + 𝑥) = cos 𝑥 𝜋 𝜋 𝜋 cot (3 − 𝑥) = cot (2𝜋 − ( + 𝑥)) = − cot ( + 𝑥) 2 2 2 = −(− tan 𝑥) = tan 𝑥 cot(2𝜋 + 𝑥) = cot 𝑥 LHS = (sin 𝑥 cos 𝑥)(tan 𝑥 + cot 𝑥) sin 𝑥 cos 𝑥 = (sin 𝑥 cos 𝑥) ( + ) cos 𝑥 sin 𝑥 2 2 = sin 𝑥 + cos 𝑥 = 1 cos (3 Question #3 Prove this equality lim ( 𝑥→0 cos(𝜋𝑥) 𝜋𝑥 − 2 )=0 sin(𝜋𝑥) sin (𝜋𝑥) Answer: Using l’Hospital’s rule: cos 𝜋 𝑥 𝜋𝑥 sin 𝜋 𝑥 cos 𝜋 𝑥 − 𝜋𝑥 − ) = lim 𝑥→0 sin 𝜋 𝑥 sin2 𝜋𝑥 sin2 𝜋𝑥 sin 2 𝜋𝑥 − 2𝜋𝑥 = lim 𝑥→0 2 sin2 𝜋 𝑥 2𝜋 cos 2 𝜋𝑥 − 2𝜋 = lim 𝑥→0 4𝜋 sin 𝜋 𝑥 cos 𝜋 𝑥 cos 2 𝜋𝑥 − 1 = lim 𝑥→0 sin 2 𝜋𝑥 −2𝜋 sin 2 𝜋𝑥 = lim 𝑥→0 2𝜋 cos 𝜋 𝑥 =0 lim ( 𝑥→0 Question #4 𝜋 𝜋 6 5 Finding a for sin (4𝑎 + ) = sin (2𝑎 + ) Answer: 𝜋 𝜋 sin (4𝑎 + ) = sin (2𝑎 + ) 6 5 𝜋 𝜋 sin (4𝑎 + ) − sin (2𝑎 + ) = 0 6 5 Consider the formula: sin 𝑎 − sin 𝑏 = 2 cos 𝑎+𝑏 2 and 𝐴 ⋅ 𝐵 = 0 ⇒ 𝐴 = 0 ∨ 𝐵 = 0 11𝜋 𝜋 6𝑎 + 2𝑎 − 30 30 2 cos sin =0 2 2 11𝜋 𝜋 6𝑎 + 2𝑎 − 30 30 cos = 0 ∨ sin =0 2 2 11𝜋 6𝑎 + 30 = 0 ∨ cos 𝑚 = 0 ⇒ 𝑚 = 𝜋 + 𝑘𝜋 2 2 Hence we have: 11𝜋 6𝑎 + 30 = 𝜋 + 𝑘𝜋 ⇒ 𝑎 = 𝑘𝜋 + 19𝜋 , 𝑘 ∈ 𝑍 2 2 3 180 𝜋 30 2𝑎− 2 sin 𝑎−𝑏 2 = 𝑛 ⇒ sin 𝑛 = 0 ⇒ 𝑛 = 𝑘 𝜋 Hence we have: 𝜋 30 2𝑎− 2 = 𝑘𝜋 ⇒ 𝑎 = 𝑘𝜋 + 3, 𝑘 ∈ 𝑍 Question #5 Does the equation 𝑎 ⋅ cos(𝜃) − 𝑏 ⋅ sin(𝜃) = 𝑐 have a closed-form solution for 𝜃? Answer: Using the identity sin(𝑥)2 + cos(𝑥)2 = 1, we have, 𝑎 ⋅ cos(𝜃) − 𝑏 ⋅ sin(𝜃) = 𝑐 ⇒ 𝑎√1 − sin2 (𝜃) = 𝑏 ⋅ sin(𝜃) + 𝑐 ⇒ 𝑎2 (1 − sin2 (𝜃)) = (𝑏 sin(𝜃) + 𝑐)2 (𝑎2 + 𝑏 2 ) sin(𝜃)2 + 2𝑏𝑐 sin(𝜃) + (𝑐 2 − 𝑎2 ) = 0 The above equation is quadratic in sin(𝜃). Solving the equation in sin(𝜃) yields, −2𝑏𝑐 ± √4𝑏 2 𝑐 2 − 4(𝑎2 + 𝑏 2 )(𝑐 2 − 𝑎2 ) sin(𝜃) = 2(𝑎2 + 𝑏 2 ) Question #6 Evaluate: 𝑚 1 ∑ arctan ( 2 ) 𝑛 +𝑛+1 𝑛=1 Answer: This telescopes, using the fact that arctan(𝑢) − arctan(𝑣) = arctan ( Specifically take u=n+1 and b=n . Then arctan ( This gives ∑𝑚 𝑛=1 arctan ( 1 𝑛2 +𝑛+1 1 𝑛2 +𝑛+1 𝑢−𝑣 1+𝑢𝑣 ) ) = arctan(𝑛 + 1) − arctan(𝑛) ) = arctan(𝑚 + 1) − 𝜋/4 Question #7 How does this fractions simplify: 1 + cos 2 𝜃 = (2 − √2)/2 Answer: 1 + cos 2 𝜃 = 2 − √2 2 2 √2 − 2 2 √2 =1− 2 (√2) 1 =1− √2 ⇒ 1 + cos 2 𝜃 1 =1− √2 ⇒ cos 2 𝜃 1 =− √2 = Question #8 How can the tangent addition theorem for complex numbers tan(𝑧 + 𝑤) = tan 𝑧+tan 𝑤 1−tan 𝑧 tan 𝑤 be proved? Answer: The proof is the same as for real numbers. You do need to convince yourself of the two addition formulas over 𝐶, namely sin(𝑧 + 𝑤) = sin 𝑧 cos 𝑤 + cos 𝑧 sin 𝑤 with cos(𝑧 + 𝑤) = cos 𝑧 cos 𝑤 − sin 𝑧 sin 𝑤 These follow from identities such as cos 𝑧 = 𝑒 𝑖𝑧 +𝑒 −𝑖𝑧 2 and sin 𝑧 = Then, same as for the reals, write out the fraction tan(𝑧 + 𝑤) = cos 𝑧 cos 𝑤 Question #9 Solve 2 + cot 𝜃 = 𝑐𝑠𝑐 𝜃 where 0 ≤ 𝜃 < 2𝜋 ? Answer: We know csc 2 𝜃 − cot 2 𝜃 = 1 ⇒ (csc 𝜃 + cot 𝜃)(csc 𝜃 − cot 𝜃) = 1 𝑒 𝑖𝑧 −𝑒 −𝑖𝑧 2𝑖 sin(𝑧+𝑤) cos(𝑧+𝑤) and divide numerator and denominator by Given csc 𝜃 − cot 𝜃 = 2 So, csc 𝜃 + cot 𝜃 = 5 1 2 4 So, 2 csc 𝜃 = ⇒ sin 𝜃 = > 0 2 5 1 1 5 2 2 4 cot 𝜃 = − csc 𝜃 = − < 0 𝜋 So, 𝜃 will lie in the 2nd quadrant, the principal value in ( , 𝜋) 2 Question #10 Please help me prove the identity: 1 1 2 2 cos 2 𝑎 − cos 4 𝑎 + sin4 𝑎 = − cos 2 𝑎 Answer: cos 2 𝑎 − cos 4 𝑎 + sin4 𝑎 = cos 2 𝑎 + (sin4 𝑎 − cos 4 𝑎) = cos 2 𝑎 + (sin2 𝑎 + cos 2 𝑎)(sin2 𝑎 − cos 2 𝑎) = cos 2 𝑎 + sin2 𝑎 − cos 2 𝑎 = sin2 𝑎 = 1/2 − 1/2 cos 2 𝑎 Question #11 I have an equation that I'm trying to solve: sin 𝑥 + √3 cos 𝑥 = 1 Answer: There's a nice trick: sin 𝑥 + √3 cos 𝑥 = 1 1 √3 = 2 ( sin 𝑥 + cos 𝑥) 2 2 𝜋 𝜋 = 2 (cos ( + 2𝑘𝜋) sin 𝑥 + sin ( + 2𝑘𝜋) cos 𝑥) 3 3 𝜋 = 2 sin (𝑥 + + 2𝑘𝜋) = 1 3 Then 𝜋 sin (𝑥 + + 2𝑘𝜋) = 1/2 3 Question #12 I have given this: 1 𝑡 𝑡 cot ( ) csc 2 ( ) csc(𝑡) 8 2 2 And I have to show that it is that: 1 𝑡 csc 4 ( ) 16 2 How do I do this? Answer: 1 t t cot( ) csc 2 ( ) csc(t ) 8 2 2 1 t t 1 = cot( ) csc 2 ( ) 8 2 2 sin t t cos( ) 1 1 2 csc 2 ( t ) = 8 sin( t ) 2 2sin( t cos( t ) 2 2 2 1 t = csc 4 ( ) 16 2 Question #13 How can I integrate the following: 1 ∞ −2exp (−𝑎𝑧) sin2 (𝑏𝑧) 𝑑𝑧 ∫ 𝑧 𝑏2 0 for 𝑎, 𝑏 > 0 ? Answer: Let I(a) denote the integral. Consider 1 ∞exp (−𝑎𝑧) sin2 (𝑏𝑧) 𝑑𝑧 𝐼 ′′ (𝑎) = 2 ∫ 𝑏 0 1 ∞exp (−𝑎𝑧)(1 − cos(2𝑏𝑧))𝑑𝑧 = 2∫ 2𝑏 0 ∞ 1 = 2 𝑅 ∫ (exp(−𝑎𝑧) − exp(−(𝑎 + 2𝑖𝑏)𝑧))𝑑𝑧 2𝑏 0 = 1 2𝑏 2 1 1 𝑎 𝑎+2𝑖𝑏 𝑅( − )= 2 1 𝑎 𝑎2 +4𝑏 2 Now integrate back, using 𝐼(∞) = 0 𝑎𝑛𝑑 𝐼 ′ (∞) = 0: ∞ 2 1 1 4𝑏 2 𝐼 ′ (𝑎) = ∫ 𝑑𝑡 = log (1 + ) 2 2 4𝑏 2 𝑎2 𝑎 𝑡 𝑡 + 4𝑏 ∞ 1 2𝑏 𝑎 4𝑏 2 𝐼(𝑎) = ∫ 𝐼 ′ (𝑡)𝑑𝑡 = arctan ( ) − 2 log (1 + 2 ) 𝑏 𝑎 4𝑏 𝑎 𝑎 Question #14 Find the limit lim (𝑥 2 −1) sin(3𝑥−3) 𝑥→1 cos(𝑥 3 −1) tan2 (𝑥 2 −𝑥) Answer: sin(3𝑥−3) tan2 (𝑥 2 −𝑥) = sin(3𝑥−3) 3𝑥−3 ×( 𝑥 2 −𝑥 tan(𝑥 2 −𝑥) 2 ) × 3(𝑥−1) 𝑥 2 (𝑥−1)2 Hence 2 (𝑥 2 − 1) sin(3𝑥 − 3) sin(3𝑥 − 3) 𝑥2 − 𝑥 3(𝑥 − 1)(𝑥 2 − 1) = × ( ) × cos(𝑥 3 − 1) tan2 (𝑥 2 − 𝑥) 3𝑥 − 3 tan(𝑥 2 − 𝑥) 𝑥 2 (𝑥 − 1)2 cos(𝑥 3 − 1) = sin(3𝑥−3) 3𝑥−3 ×( 𝑥 2 −𝑥 2 ) × tan(𝑥 2 −𝑥) 3(𝑥+1) 𝑥 2 cos(𝑥 3 −1) Now the first and second term on the right has limit 1 as 𝑥 → 1 . The last term limit can be obtained by plugging 𝑥 = 1 , to give the limit as 1×1× 3 × (1 + 1) =6 12 × cos(0) Question #15 I would like to try and prove 1+sin 𝑥 cos 𝑥 = 1+sin 𝑥+cos 𝑥 1−sin 𝑥+cos 𝑥 Answer: 1 + sin 𝑥 cos 𝑥 (1 + sin 𝑥)(1 − sin 𝑥 + cos 𝑥) = (cos 𝑥)(1 − sin 𝑥 + cos 𝑥) 1 − sin2 𝑥 + cos 𝑥 + cos 𝑥 sin 𝑥 = (cos 𝑥)(1 − sin 𝑥 + cos 𝑥) cos 2 𝑥 + cos 𝑥 + cos 𝑥 sin 𝑥 = (cos 𝑥)(1 − sin 𝑥 + cos 𝑥) (cos 𝑥)(1 + sin 𝑥 + cos 𝑥) = (cos 𝑥)(1 − sin 𝑥 + cos 𝑥) 1 + sin 𝑥 + cos 𝑥 = 1 − sin 𝑥 + cos 𝑥 Question #16 I am being asked to prove that 2𝑛 + 1 1 sin ( 2 𝑥) ∑ cos(𝑘𝑥) = + 2 2 sin(𝑥/2) 𝑛 𝑘=0 Answer: ∑ 𝑒 𝑖𝑘𝑥 = 0≤𝑟≤𝑛 = 𝑒 𝑖(𝑛+1)𝑥 2 𝑒 𝑖(𝑛+1)𝑥 − 1 𝑒 𝑖𝑥 − 1 (𝑒 𝑖𝑥 𝑖(𝑛+1)𝑥 2 −𝑒 𝑖𝑥 𝑖(𝑛+1)𝑥 2 ) 𝑖𝑥 (𝑒 2 − 𝑒 0 2 ) 𝑒2 (𝑛 + 1)𝑥 2 = 𝑥 2𝑖 sin 2 𝑎𝑠𝑒 𝑖𝑦 − 𝑒 −𝑖𝑦 = 2𝑖 sin 𝑦 𝑖𝑛𝑥 𝑒 2 = (cos 2𝑖 sin 𝑛𝑥 2 + 𝑖 sin 2 using Euler's identity. Its real part is cos (𝑛+1)𝑥 2 𝑥 sin 2 𝑛𝑥 sin 2 (𝑛+1)𝑥 2 𝑥 sin 2 𝑛𝑥 sin ) 𝑛𝑥 = (𝑛+1)𝑥 2 2 cos 2 sin 𝑥 2 sin 2 (2𝑛 + 1)𝑥 𝑥 sin + sin 2 2 = 𝑥 2 sin 2 using 2 sin 𝐴 cos 𝐵 = sin(𝐴 + 𝐵) + sin(𝐴 − 𝐵) Question #17 I need to prove by induction that: 𝑛 ∑ sin(𝑘𝑥) = sin ( 𝑘=1 𝑛+1 𝑛 𝑥) sin ( 𝑥) 2 2 𝑥 sin ( ) 2 Answer: You can actually use telescopy, which is just induction in disguise. We have that 𝑎+𝑏 𝑎−𝑏 cos 𝑏 − cos 𝑎 = 2 sin sin 2 2 Now let 1 𝑏 = (𝑘 + ) 𝑥 2 1 𝑏 = (𝑘 − ) 𝑥 2 Then 1 1 cos (𝑘 + ) 𝑥 − cos (𝑘 − ) 𝑥 = 2 sin 𝑘 𝑥 sin 2 2 1 1 cos (𝑘 + ) 𝑥 − cos (𝑘 − ) 𝑥 = 2 sin 𝑘 𝑥 sin 2 2 Now sum through 𝑘 = 1, … , 𝑛 𝑛 𝑥 2 𝑥 2 𝑛 1 1 𝑥 1 𝑥 ∑ cos (𝑘 + ) − cos (𝑘 − ) 𝑥 = 2 sin ∑ sin 𝑘 𝑥 cos (𝑛 + ) 𝑥 − cos 2 2 2 2 2 𝑘=1 𝑥 = 2 sin 2 whence 𝑘=1 ∑𝑘=1 sin 𝑘 𝑥 1 𝑥 𝑛 cos (𝑛 + ) 𝑥 − cos 2 2 = ∑ sin 𝑘 𝑥 𝑥 2 sin 𝑥=1 2 But using our first formula once more, we have 1 𝑥 2 2 cos (𝑛 + ) 𝑥 − cos so finally sin (𝑛+1)𝑥 𝑛𝑥 sin 2 2 𝑥 sin 2 = 2 sin (𝑛+1)𝑥 2 sin 𝑛𝑥 2 = ∑𝑛𝑘=1 sin 𝑘 𝑥 as desired. Question #18 The equation of motion given is 𝑥(𝑡) = −2 1 cos(10𝑡) + sin(10𝑡) 3 2 Answer: So, 2 3 1 cos 1 0𝑡 = sin 1 0𝑡 2 so, cos 1 0𝑡 sin 1 0𝑡 = 3 4 4 4 So, tan 1 0𝑡 = ⇒ 10𝑡 = 𝑛𝜋 + arctan , where n is any integer. So, 𝑡 = 3 𝑛𝜋+arctan 10 For 𝑛 = 0, 𝑡 = Also, 3 4 3 arctan 10 4 3 cos 10𝑡 3 = sin 10𝑡 4 =± √cos2 10𝑡+sin2 10𝑡 √32 +4 2 =± 1 5 3 4 ⇒ cos 1 0𝑡 = ± , sin 1 0𝑡 = ± 5 5 4 cos 1 0𝑡 ⋅ sin 1 0𝑡 = cos 2 1 0𝑡 tan 1 0𝑡 = cos 2 1 0𝑡 ⋅ > 0 3 So, the sign of cos 1 0𝑡, sin 1 0𝑡 will be same. Question #19 How to integrate? ∫ arctan √𝑥 2 − 1 √𝑥 2 + 𝑥 𝑑𝑥 Answer: Consider the integral arccos 𝑥 ∫ 𝑥√𝑥+1 𝑑𝑥 𝜋 which was derived by Mike. Using the property arccos 𝑥 = − arcsin 𝑥 , we have 2 𝜋 𝑑𝑥 = ∫ −∫ 2 𝑥√𝑥+1 arcsin 𝑥 𝑥√𝑥+1 𝑑𝑥 The leftmost integral is evaluated trivially and so we consider the rightmost integral. We express 1/√𝑥 + 1 in terms of its MacLaurin series, valid for |𝑥| < 1 ∞ ∫ arcsin 𝑥 1 ∑ ((− ) , (𝑘)) 𝑥 𝑘 𝑑𝑥 𝑥 2 𝑘=0 1 𝑘−1 = ∑∞ arcsin 𝑥 𝑑𝑥 𝑘=0 ((− ) , (𝑘)) ∫ 𝑥 2 Then, upon consideration of the integral ∫ 𝑥 𝑘−1 arcsin 𝑥 𝑑𝑥 and an application of integration by parts, we find that 𝑥𝑘 1 𝑥 𝑘 𝑑𝑥 ∫ 𝑥 𝑘−1 arcsin 𝑥 𝑑𝑥 = arcsin 𝑥 − ∫ 𝑘 𝑘 √1 − 𝑥 2 According to Mathematica, the remaining integral may be expressed in terms of the hypergeometric function. This procedure may or may not provide an evaluation for |𝑥| < 1 . Question #20 Prove that cot 2 (𝜋/7) + cot 2 (2𝜋/7) + cot 2 (3𝜋/7) = 5 Answer: So: the roots of 𝑧 3 + 𝑧 2 − 3𝑧 − 1 = 0 are 2 cos If ⇒ 𝑟𝜋 cot 2 7 2(𝑢−1) 𝑢+1 2𝑟𝜋 = 𝑢, cos = 2 cos 3 7 2𝑟𝜋 7 𝑟𝜋 = 1−tan2 7 𝑟𝜋 1+tan2 2𝜋 7 𝑟𝜋 = 7 cot2 7 −1 𝑟𝜋 cot2 +1 7 , 2 cos = 4𝜋 7 , 2 cos 6𝜋 7 𝑢−1 𝑢+1 will satisfy 3 2(𝑢 − 1) 2(𝑢 − 1) 2(𝑢 − 1) ) +( ) − 3( )−1=0 𝑢+1 𝑢+1 𝑢+1 𝑟𝜋 On simplification, 7𝑢3 − 35𝑢2 + 21𝑢 − 1 = 0 whose roots are cot 2 ⇒( 7 Now, use Vieta's formulas, to find ∑ cot 2 𝑟𝜋 7 = 35 7