Questions and Answers #31 Trigonometry

Questions and Answers Sheet 31 Trigonometry Question #1 How do I prove the identity: 𝜋 1 − sin 2 𝑥 tan ( − 𝑥) = 4 cos 2 𝑥 Answer: 1 − sin 2 𝑥 cos 2 𝑥 (cos 𝑥 − sin 𝑥)2 = cos 2 𝑥 − sin2 𝑥 (cos 𝑥 − sin 𝑥)2 = (cos 𝑥 + sin 𝑥)(cos 𝑥 − sin 𝑥) cos 𝑥 − sin 𝑥 = cos 𝑥 + sin 𝑥 1 − tan 𝑥 = 1 + tan 𝑥 𝜋 = tan ( − 𝑥) 4 Question #2 Is there an easy way to prove the identity? 𝜋 2𝜋 𝜋 7 7 3 cos ( ) − cos ( ) + cos(3𝜋7) = cos ( ) Answer: 2𝜋 2𝜋 7 7 Let 𝑤 = cos ( ) + 𝑖 sin ( ) so that 𝑤 7 = 1. Thus 𝑤 7 − 1 = 0 (𝑤 − 1)(𝑤 6 + 𝑤 5 + 𝑤 4 + 𝑤 3 + 𝑤 2 + 𝑤 + 1) = 0 𝑤6 + 𝑤5 + 𝑤4 + 𝑤3 + 𝑤2 + 𝑤 + 1 = 0 (𝑤 3 + 𝑤 −3 ) + (𝑤 2 + 𝑤 −2 ) + (𝑤 + 𝑤 −1 ) = −1 2𝜋 2𝜋 2𝜋 Since 𝑤 + 𝑤 −1 = cos ( ) + 𝐼 sin ( ) + cos (− ) + 𝑖 sin (− 7 7 7 2𝜋 2𝜋 2𝜋 2 cos (3 × ) + 2 cos (2 × ) + 2 cos ( ) = −1 7 7 7 6𝜋 4𝜋 2𝜋 1 𝜋 cos ( ) + cos ( ) + cos ( ) = − = − cos ( ) 7 7 7 2 3 Using cos(𝜃) = − cos(𝜋 − 𝜃): 𝜋 3𝜋 2𝜋 𝜋 − cos ( ) − cos ( ) + cos ( ) = − cos ( ) 7 7 7 3 𝜋 2𝜋 3𝜋 𝜋 7 7 7 3 And hence cos ( ) − cos ( ) + cos ( ) = cos ( ) Question #3 Find all rational numbers p, q, r such that 𝑝 cos 𝜋 2𝜋 3𝜋 + 𝑞 cos + 𝑟 cos =1 7 7 7 2𝜋 7 2𝜋 ) = 2 cos ( ), using de Moivre's theroem: 7 Answer: 𝜋 At first we call 𝛼 = and we are having the expression 7 𝑝 ⋅ cos(𝛼) + 𝑞 cos(2𝛼) + 𝑟 cos(3𝛼) = 1 As cos(𝛼 + 𝛽) = cos(𝛼) cos(𝛽) − sin(𝛼) sin(𝛽) we know that cos(2𝛼) = cos 2 (𝛼) − sin2 (𝛼) and cos(3𝛼) = cos(𝛼) cos(2𝛼) − sin(𝛼) sin(2𝛼) As sin(2𝛼) = 2 cos(𝛼) sin(𝛼) we get cos(3𝛼) = cos(𝛼) (cos 2 (𝛼) − sin2 (𝛼)) − sin(𝛼) (2 cos(𝛼) sin(𝛼)) Expanding this gives us cos(3𝛼) = cos 3 (𝛼) − cos(𝛼) sin2 (𝛼) cos(𝛼) and simplifying lead to cos(3𝛼) = cos 3 (𝛼) − 3 cos(𝛼) sin2 (𝛼) So back to our original equation 𝑝 ⋅ cos(𝛼) + 𝑞 ⋅ cos(2𝛼) + 𝑟 ⋅ cos(3𝛼) = 1 Now we plug in the stuff above and get 𝑝 ⋅ cos(𝛼) + 𝑞 ⋅ (cos 2 (𝛼) − sin2 (𝛼)) + 𝑟 ⋅ (cos 3(𝛼) − cos(𝛼) sin2 (𝛼)) = 1 This is nearly fine but the sin2 (𝛼) terms are still annoying, we use that cos 2 (𝛼) + sin2 (𝛼) = 1 and hence sin2 (𝛼) = 1 − cos 2 (𝛼) so our equation is equal to p  cos( ) + q  (cos 2 ( ) − 1 + cos 2 ( )) + r  (cos3 ( ) − 3(1 − cos 2 ( )) cos( ) = 1 Expanding this gives us 𝑝 ⋅ cos(𝛼) + 𝑞 ⋅ (2 cos(𝛼) − 1) + 𝑟 ⋅ (4 cos 3(𝛼) − 3 cos(𝛼)) = 1 Now we call 𝛾 = cos(𝛼) and have 𝑝 ⋅ 𝛾 + 𝑞 ⋅ 2𝛾 2 − 𝑞 + 𝑟4𝛾 3 − 𝑟3𝑦 = 1 Now we sort it in powers of 𝛾 4𝑟𝛾 3 + 2 ⋅ 𝑞 ⋅ 𝛾 2 + 𝛾(𝑝 − 3𝑟) − 𝑞 = 1 Question #4 Solve the equation 3 sin2 𝑥 − 3 cos 𝑥 − 6 sin 𝑥 + 2 sin 2 𝑥 + 3 = 0 Answer: ⇒ 3(sin2 𝑥 − 2 sin 𝑥 + 1) + 2 sin 2 𝑥 − 3 cos 𝑥 = 0 ⇒ 3(1 − sin 𝑥)2 + cos 𝑥 (4 sin 𝑥 − 3) = 0 We know, sin 𝑥 = So, 1 − sin 𝑥 = 2𝑡 1+𝑡 2 (1−𝑡)2 1+𝑡 2 , cos 2 𝑥 = 1−𝑡 2 1+𝑡 2 , 4 sin 𝑥 − 3 = 4 where 𝑡 = tan 2𝑡 1+𝑡 2 −3 = − 𝑥 2 3𝑡 2 −8𝑡+3 1+𝑡 2 =0 If t is fine 3(1 − 𝑡)4 − (1 − 𝑡)(1 + 𝑡)(3𝑡 2 − 8𝑡 + 3) = 0 Or, (1 − 𝑡)[3(1 − 𝑡)3 − (1 + 𝑡)(3𝑡 3 − 8𝑡 + 3)] = 0 𝑥 𝑥 𝜋 𝜋 If 𝑡 = 1, tan = 1 ⇒ = 𝑛𝜋 + ⇒ 𝑥 = 2𝑛𝜋 + where n is any integer. 2 2 4 2 Else 2𝑡(3𝑡 2 − 7𝑡 + 2) = 0 If 𝑡 = 0, 𝑡 = 2𝑚𝜋 + 0 where m is any integer Else 3𝑡 2 − 7𝑡 + 2 = 0 𝑥 𝜋 Finally, if t is +∞, = 𝑟𝜋 + , 𝑥 = 2𝑟𝜋 + 𝜋 = (2𝑟 + 1) 𝜋 where r is any integer 𝑥 2 𝜋 2 if t is −∞, = 𝑠𝜋 − , 𝑥 = 2𝑠𝜋 − 𝜋 = (2𝑠 − 1) 𝜋 where s is any integer 2 2 In the last two case x is an odd multiple of 𝜋 which does not satisfy the given equation Question #5 Determine sin ( 𝜋 𝜋 𝜋 ) + 6 sin2 ( ) − 8 sin4 ( ) 14 14 14 Answer: A non-trivial result is that sin 2 3 𝜋 14 is a root to the equation 1 − 4𝑥 − 4𝑥 + 8𝑥 = 0 If you believe this, then you can simply apply polynomial division to obtain 1 1 𝑥 + 6𝑥 2 − 8𝑥 4 = − (𝑥 + ) (8𝑥 3 − 4𝑥 2 − 4𝑥 + 1) + 2 2 𝜋 So then, filling in 𝑥 = sin gives you the answer: 14 𝜋 𝜋 𝜋 1 2 4 sin + 6 sin − 8 sin = 14 14 14 2 Question #6 How is cos(arcsin(𝑥)) = sin(arccos(𝑥)) = √1 − 𝑥 2 Answer: sin(arcsin(𝑥)) = 𝑥 cos(arcsin(𝑥))2 = 1 − sin(arcsin(𝑥))2 cos(arcsin(𝑥)) = √1 − sin(arcsin(𝑥))2 cos(arcsin(𝑥)) = √1 − 𝑥 2 Question #7 Show that sin 𝑥 1 = (tan 3 𝑥 − tan 𝑥) cos 3 𝑥 2 Answer: sin 𝑥 = cos 3 𝑥 1 2 ⋅ sin 𝑥 ⋅ cos 𝑥 = ⋅ 2 cos 𝑥 ⋅ cos 3 𝑥 1 sin 2 𝑥 = ⋅ 2 cos 𝑥 ⋅ cos 3 𝑥 1 sin(3𝑥 − 𝑥) = ⋅ 2 cos 𝑥 ⋅ cos 3 𝑥 1 sin 3 𝑥 ⋅ cos 𝑥 − cos 3 𝑥 ⋅ sin 𝑥 = ⋅ 2 cos 𝑥 ⋅ cos 3 𝑥 1 sin 3 𝑥 ⋅ cos 𝑥 1 cos 3 𝑥 ⋅ sin 𝑥 = ⋅ − ⋅ 2 cos 𝑥 ⋅ cos 3 𝑥 2 cos 𝑥 ⋅ cos 3 𝑥 1 sin 3 𝑥 sin 𝑥 = ⋅( − ) 2 cos 3 𝑥 cos 𝑥 1 = ⋅ (tan 3 𝑥 − tan 𝑥) 2 Now successively replace x with 3x and 9x Question #8 Verify the following identity: tan 𝑥 1 + sec 𝑥 + = 2 csc 𝑥 1 + sec 𝑥 tan 𝑥 Answer: Method 1: tan 𝑥 As tan2 𝑥 = sec 𝑥 − 1 = (sec 𝑥 − 1)(sec 𝑥 + 1), sec 𝑥−1 sec 𝑥 + 1 = tan 𝑥 𝑡𝑎𝑛𝑥 1 + sec 𝑥 sec 𝑥 − 1 1 + sec 𝑥 ⇒ + = + 1 + sec 𝑥 tan 𝑥 tan 𝑥 tan 𝑥 2 2 sec 𝑥 cos 𝑥 2 = = = sin 𝑥 tan 𝑥 sin 𝑥 cos 𝑥 Method 2: sin 𝑥 tan 𝑥 sin 𝑥 (1 − cos 𝑥) cos 𝑥 = sin 𝑥 = = 1 + sec 𝑥 1 + 1 1 + cos 𝑥 (1 + cos 𝑥)(1 − cos 𝑥) cos 𝑥 1−cos 𝑥 = as sin2 𝑥 = 1 − cos 2 𝑥 sin 𝑥 and 1 1 + sec 𝑥 1 + cos 𝑥 1 + cos 𝑥 = = sin 𝑥 tan 𝑥 sin 𝑥 cos 𝑥 Question #9 Finding the limit of 1−cos(2𝑥) 1−cos(3𝑥) for 𝑥 → 0 Answer: Putting 𝑥 = 2𝑦 1 − cos(2𝑥) lim 𝑥→0 1 − cos(3𝑥) 1 − cos 4 𝑦 = lim 𝑦→0 1 − cos 6 𝑦 2 sin2 2 𝑦 = lim 𝑎𝑠 cos 2 𝜃 = 1 − 2 sin2 𝜃 𝑦→0 2 sin2 3 𝑦 sin 2 𝑦 2 4⋅( ) 2𝑦 = lim 𝑦→0 sin 3 𝑦 2 9⋅( ) 3𝑦 4 sin ℎ = 𝑎𝑠 lim =1 9 ℎ→0 ℎ Alternatively, 1−cos(2𝑥) 1−cos 3𝑥 =( sin 2𝑥 2 1+cos 3𝑥 sin 3𝑥 1+cos 2𝑥 ) ⋅( Now lim cos 𝑎 𝑥 = 1 for finite a 𝑥→0 1 + cos 3 𝑥 1+1 ⇒ lim ( )= =1 𝑥→0 1 + cos 2 𝑥 1+1 and lim sin 2 𝑥 2 𝑥→0sin2𝑥2𝑥 2 lim ( )= ⋅ = 𝑥→0 sin 3 𝑥 3 lim sin 3 𝑥 3 𝑥→0 3𝑥 ) Question #10 It is known that √9 − 8 sin 5 0∘ = 𝑎 + 𝑏 sin 𝑐 ∘ Answer: We have the following (working in degrees): cos 2 0 − cos 8 0 = cos(50 − 30) − cos(50 + 30) = 2 sin 5 0 sin 3 0 = sin 5 0 Thus we have that 1 − 2 sin2 1 0 − sin 1 0 = sin 5 0 (using cos 2 0 = 1 − 2 sin2 1 0 and cos 8 0 = sin(90 − 80) = sin 1 0 ) And so 9 − 8 sin 5 0 = 9 − 8(1 − 2 sin2 1 0 − sin 1 0) = 1 + 8 sin 1 0 + 16 sin2 1 0 = (1 + 4 sin 1 0)2 Thus a=1,b=4, c=10 Question #11 Prove that tan−1 ( 𝑥 sin 𝛼 1−𝑥 cos 𝛼 ) − tan−1 ( 𝑥 cos 𝛼 sin 𝛼 ) is independent of x and is equal to 𝜋 2 −𝛼 Answer: The '-' sign is missing 𝑥 sin 𝛼 𝑥 − cos 𝛼 tan−1 ( ) − tan−1 ( ) 1 − 𝑥 cos 𝛼 sin 𝛼 𝑥 sin 𝛼 𝑥 cos 𝛼 − 1 − 𝑥 cos 𝛼 sin 𝛼 −1 = tan ( ) 𝑥 sin 𝛼 𝑥 − cos 𝛼 1+ ⋅ 1 − 𝑥 cos 𝛼 sin 𝛼 (sin (𝑥 𝑥 sin 𝛼 𝛼) − − cos 𝛼)(1 − 𝑥 cos 𝛼) = tan−1 ( ) (1 − 𝑥 cos 𝛼) sin 𝛼 + 𝑥 sin 𝛼 (𝑥 − cos 𝛼) assuming (1 − 𝑥 cos 𝛼) sin 𝛼 ≠ 0 = tan−1 ( 𝑥(sin2 𝛼−1−cos2 𝛼)+𝑥 2 cos 𝛼+cos 𝛼 = tan −1 ( sin 𝛼(1−2𝑥 cos 𝛼+𝑥 2 ) ) cos  (1 − 2 x cos  + x 2 ) ) using 𝑠𝑖𝑛2 𝛼 − 1 = −cos 2 𝛼 2 sin( (1 − 2 x cos  + x ) = tan−1 (cot 𝛼) assuming 1 − 2𝑥 cos 𝛼 + 𝑥 2 ≠ 0 𝜋 𝜋 2 2 = tan−1 (tan ( − 𝛼)) = − 𝛼 Question #12 How to deterine the exact value of sin(585∘ )? Answer: When calculating in degrees, 𝑠𝑖𝑛 is periodic with a period of 360 degrees. Hence sin(585∘ ) = sin(225∘ ) In particular, sin(𝑥 + 180∘ ) = − sin(𝑥). Hence sin(225∘ ) = sin(45∘ + 180∘ ) = − sin(45∘ ) 1 On the other hand, we know that sin(45∘ ) = cos(45∘ ) = . Hence √2 1 sin(585∘ ) = − √2 Question #13 Why is this silution to 2 sin2 𝑥(𝑥) + sin(𝑥) − 1 = 0 incorrect? Answer: Recall that sin(𝐴) = sin(𝐵) ⇒ 𝐴 = 𝑛𝜋 + (−1)𝑛 𝐵 Also, you could solve your problem in a slightly easier way. Note that 2 sin2 (𝑥) + sin(𝑥) − 1 = (2 sin(𝑥) − 1)(sin(𝑥) + 1) = 0 1 ⇒ sin(𝑥) = or -1 2 𝜋 𝜋 6 2 Hence, 𝑥 = 𝑛𝜋 + (−1)𝑛 , 2𝑛𝜋 − Question #14 In the equation: 𝑏 ((𝑎 − 𝑏) cot 𝜃) cos 2 𝜃 = − cot 2 Answer: From 𝑏 ((𝑎 − 𝑏 cot 𝜃) cos 2 𝜃 − 1) = cot 𝜃 cos 2 𝜃 2 With cot 𝜃 = cos 𝜃 sin 𝜃 this gives 𝑏 𝑎 sin 𝜃 2𝜃 = cos 2 𝜃 2 which finally results in 𝑎 𝑏 sin 2 𝜃 = cos 2 𝜃 2 2 and consequently 𝑏 tan 2 𝜃 = 𝑎 Question #15 I was asked to find the minimum and maximum values of the functions: 1. 𝑦 = sin2 𝑥/(1 + cos 2 𝑥) 2. 𝑦 = sin2 𝑥 − cos 4 𝑥 Answer: If you don't really need to use derivatives, 1. Clearly, 𝑦 = 2 sin2 𝑥 1+cos2 𝑥 ≥ 0 in fact =0 if sin 𝑥 = 0 sin 𝑥 sin2 𝑥 − (1 + cos 2 𝑥) − 1 = 1 + cos 2 𝑥 1 + cos 2 𝑥 cos2 𝑥 = −2 2 ≤ 0 in fact =0 if cos 𝑥 = 0 𝑦−1 = 1+cos 𝑥 ⇒𝑦−1≤0⇒𝑦 ≤1⇒0≤𝑦≤1 2. 𝑦 = sin2 𝑥 − cos 4 𝑥 = 1 − cos 2 𝑥 − cos 4 𝑥 = 1 − cos 2 𝑥 (1 + cos 2 𝑥) Now, 0  cos x  1  1  cos x + 1  2  0  cos x(cos x + 1  2  0  cos x(cos x + 1)  2 ⇒ 2 2 2 2 0 ≥ − cos 2 (cos 2 𝑥 + 1) ≥ −2 ⇒ 1 ≥ 1 − cos 2 𝑥 (cos 2 𝑥 + 1) ≥ −1 Question #16 Prove that 2 2 tan−1 𝑦𝑧 𝑧𝑥 𝑥𝑦 𝜋 + tan−1 + tan−1 = 𝑥𝑟 𝑦𝑟 𝑧𝑟 2 where 𝑥 2 + 𝑦 2 + 𝑧 2 = 𝑟 2 Answer: arctan 𝑎 + arctan 𝑏 + arctan 𝑐 will be 𝜋 2 𝜋 if arctan 𝑎 + arctan 𝑏 = − arctan 𝑐 2 𝜋 if tan(arctan 𝑎 + arctan 𝑏) = tan ( − arctan 𝑐) 2 if tan (𝑛𝜋 + arctan ( if 𝑎+𝑏 1−𝑎𝑏 = 𝑎+𝑏 1−𝑎𝑏 1 )) = cot(arctan 𝑐) = cot (arctan ) 𝑐 1 𝑐 if 𝑎𝑏 + 𝑏𝑐 + 𝑐𝑎 = 1 Question #17 How to find the solution x of 2 sin 1 1∘ sin 7 1∘ sin(𝑥 ∘ + 30∘ ) = sin 2 013∘ sin 2 10∘ Answer: You've over complicated the whole thing way too much. Don't forget, apart from sin(𝑥 + 30∘ ), the rest are just constants and so treat them like any other number. Therefore you can just do the whole thing simply and like normal: 2 sin(11∘ ) sin(71∘ ) sin(𝑥 + 30∘ ) = sin(2013∘ ) sin(210∘ ) = 2 sin(11∘ ) sin(71∘ ) sin(𝑥 ∘ + 30∘ ) = sin(33∘ ) sin(30∘ ) as they're all multiplied, divide through by the 2 sin(11) sin(71) to get sin(33∘ ) sin(30∘ ) sin(𝑥 + 30∘ ) = 2 sin(11∘ ) sin(71∘ ) Then 𝑎𝑟𝑐𝑠𝑖 𝑛 both sides to get sin(33∘ ) sin(30∘ ) 𝑥 + 30∘ = sin−1 ( ) = 49∘ 2 sin(11∘ ) sin(71∘ ) Subtracting 30∘ then gives you 𝑥 = 19∘ and as you want it in the interval 90∘ < 𝑥 < 180∘ , you simply do 90∘ + 19∘ = 109∘ Question #18 How do you determine the local extrema points for 𝑦 = √3 cos(3𝑥) + sin(3𝑥) ? Answer: At 𝑥 = 𝜋/2, 𝑦 ′ ≠ 0; … 𝑦 ′ (𝜋/2) = 3√3 You need to solve for 𝑦 ′ = −\3√3 sin(3𝑥) + 3 cos(3𝑥) = 0 ⇒ 3√3 sin(3𝑥) = cos(3𝑥) sin(3𝑥) ⇒ √3 sin(3𝑥) = cos(3𝑥) ⇒ √3 = √3 tan(3𝑥) = 1 cos(3𝑥) 1 ⇒ tan(3𝑥) = √3 Question #19 Solve: 𝑥3 𝑥→0 tan3 (2𝑥) lim Answer: I'll assume that you know that lim 𝑥→0 sin(𝑥) 𝑥 =1 by knowing this you can show by doing a simple substitution u=ax that sin 𝑎 𝑥 𝑎 lim = 𝑥→0 𝑏𝑥 𝑏 this will also give you 𝑏𝑥 1 𝑏 lim = = 𝑥→0 sin(𝑎𝑥) sin(𝑎𝑥) 𝑎 lim 𝑏𝑥 𝑥→0 sin(𝑥) tan(𝑥) knowning this and since tan(𝑥) = you can show that lim =1 cos(𝑥) 𝑥→0 𝑥 and tan(𝑎𝑥) 𝑎 lim = 𝑥→0 𝑏𝑥 𝑏 which implies that 𝑏𝑥 𝑏 lim = 𝑥→0 tan(𝑎𝑥) 𝑎 that's all you need to solve your question and similar questions since lim lim 𝑥3 𝑥→0 tan3 (2𝑥) 𝑥 = lim 𝑥 𝑥→0 tan(2𝑥) × lim 𝑥 𝑥→0 tan(2𝑥) × 𝑥→0 tan(2𝑥) 1 1 1 × × 2 2 2 1 = 8 = Question #20 How am I supposed to prove this inequality? 1 2) 𝑘 tan(1) ≤ ∑ 1 2 ) 𝑘=1 cos ( 𝑘+1 ∞ sin ( Answer: 1 If 𝑘 ∈ 𝑁, 𝑘 ≥ 2 and 𝑥 ∈ [0, √ ] then 2 𝑑 𝑥 1 𝑥 𝑥 1 𝑥2 sin ( ) cos(𝑥) = cos ( ) cos(𝑥) − sin ( ) sin(𝑥) ≥ − ≥0 𝑑𝑥 𝑘 𝑘 𝑘 𝑘 2𝑘 𝑘 1 1 sin( 2 ) k (k + 1) k  for all 𝑘 ≥ 1 1 cos( ) cos( ) k k +1 sin( 𝑥 so sin ( ) cos(𝑥) is non-decreasing on this interval. In particular 𝑘 2. This inequality also holds in fact for k=1 since sin(1) < sin(𝜋 − 2) = sin(2) Now we get the following inequalities for all 𝑘 ≥ 1 : 1 1 1 1 sin ( ) cos ( ) − cos ( ) sin ( ) 1 1 𝑘 𝑘 + 1 𝑘 𝑘 + 1 tan ( ) − tan ( )= 1 1 𝑘 𝑘+1 cos ( ) cos ( ) 𝑘 𝑘+1 = 1 ) 𝑘(𝑘+1) 1 1 cos( ) cos( ) 𝑘 𝑘+1 sin( 1 ≤ sin( 2 ) 𝑘 cos2 ( 1 ) 𝑘+1 and summing from 1 to ∞ gives the required result.