Trigonometric Problems

Questions

• 1. Prove x is an integer and state it's value in: (3sinx + cosx)^2 + (sinx-3cosx)^2
• 2. Show 6sin^2x = 4 + cosx is the same as 6cos^2x + cosx - 2 = 0
• 3. Show 7sin^2θ +sinθcosθ = 6 is the same as tan^2θ + tanθ - 6 = 0
• 4. Prove: sinA/1+cosA + 1+cosA/sinA ≡ 2/sinA
• 5. Show that 3+sin^2θ / cosθ-2 = 3cosθ
• 6. Solve cos(3x+50) = (root(3))/2 between -180≤x≤360
• 7. Solve 6tanxsinx=5
• 8. Solve: 5cos^2x - 13sinx + 1 = 0 (hence solve 5cos^2(x+30) - 13sin(x+30) + 1 = 0) range= 0<x<360

1....

• 1. Prove x is an integer and state it's value in: (3sinx + cosx)^2 + (sinx-3cosx)^2
• Expand: 9sin^2x + 6sinxcosx + cos^2x + sin^2x - 6sinxcosx + 9cos^2x
• Collect together: 10sin^2x + 10cos^2x
• Factor out 10: 10(sin^2x + cos^2x)
• Use identity that sin^2x + cos^2x ≡ 1 : 10(1) = 10

2....

• 2. Show 6sin^2x = 4 + cosx is the same as 6cos^2x + cosx - 2 = 0
• sin^2x + cos^2x≡ 1 , so sin^2x≡1-cos^2x.
• Here we can say: 6(1-cos^2x) = 4+cosx
• Expand and rearrange into quadratic: 6cos^2x + cosx - 2 = 0
• Solve for x=1/2 or -2/3

3....

• 3. Show 7sin^2θ +sinθcosθ = 6 is the same as tan^2θ + tanθ - 6 = 0
• Start at answer: Use identity tanθ = sinθ/cosθ ; tan^2θ + tanθ - 6 = 0 is same as sin^2θ / cos^2θ + sinθ/cosθ - 6 = 0
• Now remove denominators: sin^2θ + sinθcosθ - 6cosθ = 0
• We use identity sin^2θ + cos^2θ ≡ 1 and rearrange to cos^2θ ≡ 1- sinθ and plug in : sin^2θ + sinθcosθ - 6(1-sin^2θ)
• Same as: sin^2θ + sinθcosθ - 6 + 6sin^2θ = 0 , collect together for: 7sin^2θ+sinθcosθ-6=0
• Same as: 7sin^2θ +sinθcosθ = 6

4....

• 4. Prove: sinA/1+cosA + 1+cosA/sinA ≡ 2/sinA
• Left side: sinA/1+cosA + 1+cosA/sinA , remove denominator: (sinA(sinA) + (1+cosA)(1+cosA))/sinA(1+cosA)
• Expand top: (sin^2A+1+2cosA+cos^2A)/sinA(1+cosA) , same as: ((sin^2A + cos^2A) + 1 + 2cosA)/sinA(1+cosA)
• Use identity sin^2θ + cos^2θ ≡ 1to say: (1+1+2cosA)/sinA(1+cosA) , same as: (2(1+cosA)/sinA(1+cosA))
• Cancel (1+cosA): 2/sinA

5....

• 5. Show that 3+sin^2θ / cosθ-2 = 3cosθ
• Remove denominator: 3+sin^2θ=3cosθ(cosθ-2)
• Use identity sin^2θ + cos^2θ ≡ 1 and rearrange to sin^2θ ≡ 1- cosθ to say 3+(1-cos^2θ) = 3cosθ(cosθ-2)
• Same as: 4-cos^2θ = 3cos^2θ - 6cosθ
• Move to 1 side and collect: 4cos^2θ - 6cosθ - 4 = 0
• Solve quadratic: cos=-1/2 or 2
• Cos must be -1/2 as -1≤cos≤1

6....

• 6. Solve cos(3x+50) = (root(3))/2 between -180≤x≤360
• cos-1(root(3))/2 = 30 , other value therefore = -30 , +/-360 = 390,330,750,690,1110,1050
• -50 (3x) : -20 , -80 , 340 , 280 , 700 , 640 , 1060 , 1000
• Divided by 3 (x) : -20/3 , -80/3 , 340/2 , 280/3 , 700/3 , 640/3 , 1060/3 , 1000/3 , -380/3 , -440/3

7....

• 7. Solve 6tanxsinx=5
• First show that 6tanxsinx=5 is the same as 6cos^2x+5cosx-6=0 : 6(sin/cos)sin=5 , 6sin^2x/cosx =5, 6sin^2x=5cosx, 6(1-cos^2x)=5cos, 6cos^2x+5cosx-6=0
• Now solve: solve quadratic for cosx=2/3 or -3/2 , but the it must be 2/3 as -1≤cos≤1 and find all other values which right now i cba to do so sry future self

8.....

• 5cos^2x - 13sinx + 1 = 0 (range= 0<x<360)
• cos^2x + sin^2x ≡ 1
• 5(1 - sin^2x) - 13sinx + 1 = 0 , 5 - 5sin^2x - 13sin + 1 , 5sin^2x + 13sinx - 6 = 0
• Solve 5sin^2x + 13sinx - 6 = 0 as a quadratic to get sin = 2/5 or -3.
• Sin-1(2/5) = 23.6 , 180-23.6 = 156.4. +/- 360 but gives values not in range. Sin-1(-3) is invalid as sin is from -1 to 1.
• Hence 5cos^2(x+30) - 13sin(x+30) + 1 = 0
• x+30 = 23.6 , 156.4
• x= -6.4 , 126.4 +/-360 = 353.6
• Range= 0<x<360 , so answers = 126.4 , 353.6