Questions and Answers #29 Trigonometry

Questions and Answers Sheet 29 Trigonometry Question #1 How to solve the following equation? tan 𝑥 = tan(𝑥 + 10∘ ) tan(𝑥 + 20∘ ) tan(𝑥 + 30∘ ) Answer: sin 𝑥 cos(𝑥 + 10) sin(𝑥 + 20) sin(𝑥 + 30) = cos 𝑥 sin(𝑥 + 10) cos(𝑥 + 20) cos(𝑥 + 30) Applying componendo and dividendo, cos 𝑥 sin(𝑥 + 10) + sin 𝑥 cos(𝑥 + 10) cos 𝑥 sin(𝑥 + 10) − sin 𝑥 cos(𝑥 + 10) cos(𝑥 + 20) cos(𝑥 + 30) + sin(𝑥 + 20) cos(𝑥 + 30) = cos(𝑥 + 20) cos(𝑥 + 30) − sin(𝑥 + 20) cos(𝑥 + 30) sin(2𝑥 + 10) cos 1 0 = sin 1 0 cos(2𝑥 + 50) applying sin(𝐴 ± 𝐵) and cos(𝐴 ± 𝐵) So, sin(2𝑥 + 10) cos(2𝑥 + 50) = sin 1 0 cos 1 0 sin(4𝑥 + 60) − sin 4 0 = sin 2 0 (applying 2 𝑠𝑖𝑛 𝐴 𝑐𝑜𝑠 𝐴 = 𝑠𝑖𝑛 2𝐴 and 2 𝑠𝑖𝑛 𝐴 𝑐𝑜𝑠 𝐵 = 𝑠𝑖𝑛(𝐴 + 𝐵) + 𝑠𝑖𝑛(𝐴 − 𝐵) ) 20 + 40 40 − 20 sin(4𝑥 + 60) = sin 4 0 + sin 2 0 = 2 sin cos 2 2 (applying sin 2 𝐶 + sin 2 𝐷 = 2 sin(𝐶 + 𝐷) cos(𝐶 − 𝐷) and sin(90 ± 𝐴) = cos 𝐴) Now sin(4𝑥 + 60) = cos[90 − (4𝑥 + 60)] = cos(30 − 4𝑥) = 𝑐𝑜𝑠s (4𝑥 − 30) cos(−𝐴) = cos 𝐴 So, cos(4𝑥 − 30) = cos 1 0 or, 4𝑥 − 30 = 360𝑛 ± 10 where n is any integer. Find suitable n to keep 𝑥 ∈ [0,90] Question #2 Use the de Moivre theorem to evaluate the following: (1 + 𝑖)(√3 + 𝑖) 3 3 = 1−𝑖 (1 − √3𝑖) Answer: 𝜋 𝜋 1 + 𝑖 = √2 (cos ( ) + 𝑖 sin ( )) 4 4 𝜋 𝜋 √3 + 𝑖 = 2 (cos ( ) + 𝑖 sin ( )) 6 6 −𝜋 −𝜋 1 − 𝐼√3 = 2 (cos ( ) + 𝑖 sin ( )) 3 3 Then you can simply apply De Moivre's theorem: 9𝜋 9𝜋 3𝜋 3𝜋 12 12 4 4 The numerator becomes 8√2 (cos ( ) + 𝑖 sin ( )) = 8√2 (cos ( ) + 𝑖 sin ( )) The denominator becomes 8(cos(−𝜋) + 𝑖 sin(−𝜋)) = −8 3𝜋 3𝜋 4 4 So the fraction is equal to −√2 (cos ( ) + 𝑖 sin ( )) = 1 − 𝑖 Question #3 1 1 Assuming cos(36∘ ) = + √5 4 4 How to prove that tan2 (18∘ ) tan2 (54∘ ) is a rational number? Answer: Use the fact that 1 − cos 3 6∘ 2 tan2 1 8∘ = = 1 − √5 ∘ 1 + cos 3 6 5 Then use the fact that 1 tan2 54∘ = 2 tan 3 6∘ so that tan2 1 8∘ 1 1 tan2 1 8∘ tan2 5 4∘ = = (1 − tan2 1 8∘ )2 = tan2 3 6∘ 4 5 Question #4 Calculate the value of expresion: 𝐸(𝑥) = sin6 𝑥 + cos 6 𝑥 sin4 𝑥 + cos 4 𝑥 for tan(𝑥) = 2 Answer: sin6 𝑥 + cos 6 𝑥 sin6 𝑥 + cos 6 𝑥 tan6 𝑥 + 1 cos 6 𝑥 𝐸(𝑥) = = = 1 sin4 𝑥 + cos 4 𝑥 sin4 𝑥 + cos 4 𝑥 (tan4 𝑥 + 1) cos 2 𝑥 cos 6 𝑥 26 + 1 1 65 1 65 1 5 ⋅ 13 13 = cos 2 𝑥 ⋅ 4 = = 2 = ⋅ = 2 2 + 1 1 + tan 𝑥 17 2 + 1 17 5 17 17 6 sin 𝑥 6 6 6 6 (tan6 sin 𝑥 + cos 𝑥 = cos 𝑥 ( 6 ) cos 𝑥 𝑥 + 1) cos 𝑥 + 1 Question #5 Help me to prove that arcsin(√2 sin(𝑡)) + arcsin (√cos(2𝑡)) = 𝜋/2 Answer: Let arcsin(√2 sin 𝑡) = 𝑦 ⇒ sin 𝑦 = √2 sin 𝑡 sin 𝑦 2 cos 2 𝑡 = 1 − 2 sin2 𝑡 = 1 − 2 ( ) = 1 − sin2 𝑦 = cos 2 𝑦 √2 𝜋 So, arcsin √cos 2 𝑡 = arcsin(cos 𝑦) = arcsin (sin ( − 𝑦)) 2 𝜋 𝜋 2 2 Hence, arcsin √cos 2 𝑡 = − 𝑦 = − arcsin(√2 sin 𝑡) Question #6 How can I calculate sin(1010 100 − 10) ? Answer: Since 360 = 40 ⋅ 9,gcd(40,9) = 1 100 40|1010 100 by the Chinese Remainder Theorem, the residue of 1010 modulo 360 will be the unique residue congruent to 0 modulo 40 and 1 modulo 9. This is 280, so sin(1010 Question #7 How does one get the following equalities ? 𝜋 𝜋 sin ( + 𝑥) cos ( − 𝑥) 4 4 1 + tan 𝑥 = = √2 𝜋 cos 𝑥 cos cos 𝑥 4 Answer: 1 1 + tan 𝑥 = 1 + = cos x sin  4 cos sin 𝑥 cos 𝑥 = cos 𝑥+sin 𝑥 √2 1 cos 𝑥 √2 + sin x cos  4  4 = cos x Using the fact that cos sin( x + cos 𝜋 4 = sin 𝜋 4  4 =  4 cos x 1 √2 Question #8 Prove that if 𝜙 is not equal to 2𝑘𝜋 for any integer k, then n  sin( + t ) = t =0 sin(n + 1) n  )sin( + )sin( ) 2 2 2 Answer: Use the exponential representation of the sines and cosines: 1 cos(𝜃 + 𝑡𝜙) = (𝑒 𝑖(𝜃+𝑡𝜙) + 𝑒 −(𝜃+𝑡𝜙) ) = 𝑅[𝑒 𝑖(𝜃+𝑡𝜙) ] 2 1 sin(𝜃 + 𝑡𝜙) = (𝑒 𝑖(𝜃+𝑡𝜙) − 𝑒 −(𝜃+𝑡𝜙) ) = 𝐼[𝑒 𝑖(𝜃+𝑡𝜙) ] 2𝑖 Then use a geometric series to sum. Specifically, for the sine series, write 𝑛 𝑛 ∑ sin(𝜃 + 𝑡𝜙) = 𝐼 [𝑒 𝑖𝜃 ∑𝑡=0 𝑒 𝑖𝑡𝜙 ] 𝑡=0 = 𝐼 [𝑒 𝑖𝜃 1 − 𝑒 𝑖(𝑛+1)𝜙 ] 1 − 𝑒 𝑖𝜙 𝜙 + 1) ] 2 =𝐼 𝜙 sin ( ) 2 𝜙 𝜙 sin [(𝑛 + 1) 2 ] = sin (𝜃 + 𝑛 ) 𝜙 2 sin ( ) 2 What is different for the cosine series? 𝜙 sin [(𝑛 𝑖(𝜃+𝑛 ) 2 ] [𝑒 100 ∘ − 10) = sin 2 70 = −1 For∑𝑛𝑡=0 cos 2 (2𝑡𝜃) write cos 2 (2𝑡𝜃) = 1/2 + (1/2) cos(4𝑡𝜃) and see if the work you did for the cosine series applies. Similar for the sin2 (2𝑡𝜃) series. Question #9 So far I got 𝑟 = 7/(4 − 2 cos 𝜃) (𝑟(4 − 2 𝑐𝑜𝑠 𝜃) = 7 𝑟(4 − 2(𝑥/𝑟)) = 7 Answer: You are on the right track. So we get 4𝑟 − 2𝑥 = 7 This implies that 4𝑟 = 7 + 2𝑥 Taking square on both sides, we get 16𝑟 2 = (7 + 2𝑥)2 Now note that 𝑥 2 + 𝑦 2 = 𝑟 2 . Put this into the last equation, we have 16(𝑥 2 + 𝑦 2 ) = (7 + 2𝑥)2 Question #10 Given 𝑥 ≥ 𝑦 ≥ 𝑧 ≥ 𝜋/12, 𝑥 + 𝑦 + 𝑧 = 𝜋/2 , find the maximum and minimum of cos 𝑥 sin 𝑦 cos 𝑧 Answer: Let 𝑃 = cos 𝑥 sin 𝑦 cos 𝑧 = = 2 [2 cos 𝑥 sin 𝑦] cos 𝑧 cos 𝑧 [sin(𝑥 + 𝑦) − sin(𝑥 − 𝑦)] ≤ ⋅ sin(𝑥 + 𝑦) 2 2 So 𝑃 ≤ = cos 𝑧 cos 𝑧⋅cos 𝑧 2 1 1 𝜋 4 4 12 = (1 + cos 2 𝑧) ≤ [1 + cos 2 ⋅ ] 2 + √3 8 𝜋 𝜋 2 12 Above equality hold when sin(𝑥 − 𝑦) = 0 ⇒ 𝑥 = 𝑦 and given 𝑥 + 𝑦 = − 𝑧 and 𝑥 ≥ 𝑦 ≥ 𝑧 ≥ And for max(𝑃), We must have 𝑧 = 𝜋 12 and 𝑥 = 𝑦 = 5𝜋 24 Question #11 Verify the following: 2 cot 𝑥 = csc 2 𝑥 − 2 tan 2 𝑥 Answer: 1 We put 𝑧 = 𝑒 𝑖𝑥 to obtain cot 𝑥 = 𝑖 2 −4 ( 1 𝑧− 1 𝑧 ) − 2 = −4 ( 𝑧 𝑧 2 −1 2 𝑧+𝑧 𝑧 2 +1 𝑧− 𝑧 2 −1 1 = 𝑖 𝑧 −4𝑧 2 ) − 2 = (𝑧 2 −1)2 1 and tan 2 𝑥 = 2 1 𝑧 −𝑧2 𝑖 𝑧 2 + 12 𝑧 𝑧 4 +1 = −2 (𝑧 2 −1)2 = 1 𝑧 4 −1 𝑖 𝑧 4 +1 2 and csc 2 𝑥 − 2 = ( 2𝑖 1) − 2 = 𝑧− 𝑧 But now = −2 (𝑧 2 2 cot 𝑥 tan 2𝑥 𝑧2 +1 = 𝑧 4 +1 −1)(𝑧 2 −1) 2𝑖 2 𝑧 −1 1𝑧4 −1 𝑖 𝑧4 +1 (𝑧 2 +1)(𝑧 4 +1) = −2 (𝑧 4 −1)(𝑧 2 −1) and we are done. Question #12 Verify that 2 cos 2 𝑥 − 1 = 1−tan2 𝑥 1+tan2 𝑥 Answer: 2 cos 2 (𝑥) − 1 = 2 cos 2(𝑥) − sin2 (𝑥) − cos 2(𝑥) = cos 2 (𝑥) − sin2 (𝑥) = (cos(𝑥) − sin(𝑥)) ⋅ (cos(𝑥) + sin(𝑥)) = cos 2 (𝑥) (1 − tan2 (𝑥)) cos 2 (𝑥) = ⋅ (1 − tan2 (𝑥)) sin2 (𝑥) + cos 2 (𝑥) 1 − tan2 (𝑥) = 1 + tan2 (𝑥) Question #13 How would I solve the following trig equation? sin2 𝑥 = 1 − cos 𝑥 Answer: Recall sin2 𝑥 = 1 − cos 2 𝑥 So you can write the equation as a quadratic, in cos 𝑥 : sin2 𝑥 = 1 − cos(𝑥) ⇒ 1 − cos 2 𝑥 = 1 − cos 𝑥 ⇒ cos 2 𝑥 − cos 𝑥 = 0 ⇒ cos 𝑥 (cos 𝑥 − 1) = 0 Let 𝑢 = cos 𝑥, e.g. 𝑢(𝑢 − 1) = 0 ⇒ 𝑢 = 0 or 𝑢 = 1 𝑢 = cos 𝑥 = 0 ⇒ 𝑥 = (2𝑘+1)𝜋 2 ,𝑘 ∈ 𝑍 𝑢 = cos 𝑥 = 1 ⇒ 𝑥 = 2𝑘𝜋, 𝑘 ∈ 𝑍 Question #14 Simplify the expression √(1 − cos 𝛼 cos 𝛽)2 − sin2 𝛼 sin2 𝛽 Answer: Let 𝑎 = cos 𝛼 , 𝑏 = 𝑐𝑜𝑠 𝜃 So, the the given expression = √(1 − 𝑎𝑏)2 − (1 − 𝑎2 )(1 − 𝑏 2 ) = √{1 − 2𝑎𝑏 + 𝑎^2𝑏^2 − (1 − 𝑎^2 − 𝑏^2 + 𝑎^2𝑏^2} = √𝑎2 + 𝑏 2 − 2𝑎𝑏 = |𝑎 − 𝑏| = |cos 𝛼 − cos 𝛽| Question #15 𝜋 𝜋 2 2 Given that − ≤ arcsin 𝑥 ≤ 1 , find the exact of cos [arcsin (− )] 3 Answer: −1 2 Use, 𝑐𝑜𝑠 2 𝜃 + sin2 𝜃 and also sin(arcsin(𝑦)) = 𝑦 to find that cos 2 (arcsin(−1/3)) = 1 − ( ) = 3 Now observe that −𝜋/2 ≤ 𝑎𝑟𝑐𝑠𝑖𝑛 (−1/3)) ≤ 𝜋/2, so cos(arcsin(−1/3)) ≥ 0 So it is the positive root of the above quadratic equation: 8 2√2 cos(arcsin(−1/3)) = √ = 9 3 Question #16 𝑋 = (1 + tan 1∘ )(1 + tan 2∘ )(1 + tan 3∘ ) … (1 + tan 4 5∘ ) 1 tan 𝜃 tan(90 − 𝜃) = cot 𝜃 = Answer: Let 𝜃 = 1∘ = 45 = ∏ √2 𝑘=1 𝜋 180 45 rad. Then 𝑋 = ∏45 𝑘=1(1 + tan(𝑘𝜃)) = ∏𝑘=1 cos(𝑘𝜃) sin((𝑘 + 45)𝜃) cos(𝑘𝜃) Furthemore ∏45 𝑘=1 √2 = sin(𝑘𝜃)+cos(𝑘𝜃) sin((𝑘+45)𝜃) cos(𝑘𝜃) 44 ∘ ∏ sin(90 − 𝑛𝜃) 𝑛=0 245/2 ∏45 𝑘=1 cos(𝑘𝜃) ∘ = 245/2 ∏45 𝑘=1 sin((𝑘+45)𝜃) ∏45 𝑘=1 cos(𝑘𝜃) Now, using sin(90 − 𝛼) = cos(𝛼) ∘ ∏44 ∏44 cos(𝑛𝜃) 𝑛=0 sin(90 − 𝑛𝜃) 45/2 𝑛=0 245/2 = 2 ∏45 ∏45 𝑘=1 cos(𝑘𝜃) 𝑘=1 cos(𝑘𝜃) = 245/2 cos(0) cos(45∘ ) = 223 Question #17 Finding all solutions to tan5 𝑥 − 9 tan 𝑥 = 0 Answer: 1 − tan2 𝑥 1 − 3 1 2𝜋 = = − = cos 1 + tan2 𝑥 1 + 3 2 3 2𝜋 2𝜋 (3𝑛 2𝑥 = 2𝑛𝜋 ± = ± 1) where n is any integer. 3 3 𝜋 𝑥 = (3𝑛 ± 1) 3 𝜋 Now, we need 0 ≤ (3𝑛 + 1) < 2𝜋 ⇒ 0 ≤ 3𝑛 + 1 < 6 ⇒ 𝑛 = 0,1 tan2 𝑥 = 3 ⇒ cos 2 𝑥 = 3 𝜋 2𝜋 3 3 For 𝑛 = 1, 𝑥 = (3𝑛 − 1) = Question #18 Prove the identity 𝑥 1 + sin 𝑥 = 2 cos 2 (45∘ − ) 2 𝜋 5𝜋 3 3 ; 𝑛 = 2 ⇒ 𝑥 = (3𝑛 − 1) = 8 9 Answer: 𝑥 1 + sin 𝑥 = 2 cos 2 (45∘ − ) 2 cos 2 𝛼 = 2 cos 2 𝛼 − 1 𝑥 cos 2 ⋅ (45∘ − ) = cos(90∘ − 𝑥) = cos 9 0∘ cos 𝑥 + sin 9 0∘ sin 𝑥 = 0 + 1 ⋅ sin 𝑥 = sin 𝑥 2 Question #19 Show that : 31 + 8√15 = 16(1 + cos 6∘ )(1 + cos 4 2∘ )(1 + cos 6 6∘ )(1 − cos 7 8∘ ) Answer: (1 + cos 6∘ )(1 + cos 4 2∘ )(1 + cos 6 6∘ )(1 − cos 7 8∘ ) = (1 + cos 6∘ )(1 + cos 6 6∘ )(1 − cos 7 8∘ )(1 + cos 4 2∘ ) = (1 + cos 6∘ + cos 6 6∘ + cos 6∘ cos 6 6∘ )(1 + cos 4 2∘ − cos 7 8∘ − cos 7 8∘ ) 1 1 = [1 + 2 cos 3 0∘ cos 3 6∘ + (cos 6 0∘ + cos 7 2∘ )] [1 + 2 sin 1 8∘ sin 6 0∘ − (cos 3 6∘ + cos 1 20∘ )] 2 2 (Applying 2 cos 𝐴 cos 𝐵 = cos(𝐴 − 𝐵) + cos(𝐴 + 𝐵) , cos 2 𝐶 + cos 2 𝐷 = 2 cos(𝐶 − 𝐷) cos(𝐶 + 𝐷) and cos 2 𝐶 − cos 2 𝐷 = −2 sin(𝐶 − 𝐷) sin(𝐶 + 𝐷) ) Now sin 6 0∘ = cos 3 0∘ = So, cos 7 2∘ = sin 1 8∘ = √3 , 𝑐𝑜𝑠120∘ = cos(180 − 60)∘ = − cos 6 0∘ = − 2 √5−1 4 and cos 3 6∘ = 1 2 √5+1 4 Question #20 Simplify: sec 𝑥 tan 𝑥 (1 − sin2 𝑥) Answer: Assuming you mean sec(𝑥) tan(𝑥) (1 − sin2 (𝑥)) Remember that sin2 (𝑥) + cos 2(𝑥) = 1 that sec(𝑥) = 1 cos(𝑥) and also that tan(𝑥) = Now substituting these identities into your expression we get 1 sin(𝑥) sec(𝑥) tan(𝑥) (1 − sin2 (𝑥)) = ⋅ ⋅ cos 2 (𝑥) cos(𝑥) cos(𝑥) cos 2 (𝑥) = ⋅ sin(𝑥) cos 2 (𝑥) = sin(𝑥) sin(𝑥) cos(𝑥)