Trigonometric Functions Miscellaneous Exercise

Question:

Prove that: sin3x+sin2x−sinx=4sinxcos(x​/2)cos(3x/2)

Answer:

  1. sin3x+sin2x−sinx
  2. sin3x+2sinxcosx−sinx (Using the identity sin2x = 2sinxcosx)
  3. sin3x+2sinxcosx (Cancelling out the common term -sinx)
  4. 4sinxcosx/2 (Using the identity sin3x = 3sinxcosx - sinx)
  5. 4sinxcosx/2cos3x/2 (Dividing both sides by cos3x/2)
  6. Proved.

Question:

Prove that: 2cosπ​/13cos9π​/13+cos3π​/13+cos5π​/13=0

Answer:

  1. Expand the equation: 2cosπ/13 + cos3π/13 + cos5π/13 = 0

  2. Rewrite in terms of sines and cosines: 2sin7π/13cosπ/13 + sin11π/13cos3π/13 + sin15π/13cos5π/13 = 0

  3. Apply the sum-to-product identities: 2sin7π/13sin2π/13 + 2sin11π/13sin4π/13 + 2sin15π/13sin6π/13 = 0

  4. Simplify the equation: 2sin(7π/13 + 2π/13) + 2sin(11π/13 + 4π/13) + 2sin(15π/13 + 6π/13) = 0

  5. Use the double-angle identity: 2sin(9π/13) + 2sin(15π/13) + 2sin(21π/13) = 0

  6. Apply the sum-to-product identities again: 4sin9π/13cos6π/13 + 4sin15π/13cos12π/13 + 4sin21π/13cos18π/13 = 0

  7. Simplify the equation: 4sin(9π/13 + 6π/13) + 4sin(15π/13 + 12π/13) + 4sin(21π/13 + 18π/13) = 0

  8. Use the double-angle identity again: 4sin15π/13 + 4sin27π/13 + 4sin39π/13 = 0

  9. Apply the sum-to-product identities one last time: 8sin15π/13cos12π/13 + 8sin27π/13cos24π/13 + 8sin39π/13cos36π/13 = 0

  10. Simplify the equation: 8sin(15π/13 + 12π/13) + 8sin(27π/13 + 24π/13) + 8sin(39π/13 + 36π/13) = 0

  11. Use the double-angle identity one last time: 8sin27π/13 + 8sin51π/13 + 8sin75π/13 = 0

  12. Use the fact that sin(π + θ) = -sinθ: 8sin27π/13 - 8sin3π/13 - 8sin9π/13 = 0

  13. Rewrite the equation in terms of cosines: 2cosπ/13cos9π/13 + cos3π/13cos5π/13 - cos5π/13cos3π/13 = 0

  14. Apply the product-to-sum identities: 2cosπ/13cos9π/13 + cos3π/13cos5π/13 + cos3π/13cos5π/13 = 0

  15. Simplify the equation: 2cosπ/13cos9π/13 + 2cos3π/13cos5π/13 = 0

  16. Apply the product-to-sum identities again: 2cos(π/13 + 9π/13)cos(3π/13 + 5π/13) = 0

  17. Use the double-angle identity one last time: 2cos14π/13cos8π/13 = 0

  18. Apply the fact that cos(2θ) = cos2θ - 1: 2cos14π/13(cos8π/13 - 1) = 0

  19. Simplify the equation: 2cos14π/13(-1) = 0

  20. Solve for 0: 2cos14π/13 = 0

  21. Apply the fact that cosθ = 0 when θ = π/2 + kπ: 14π/13 = π/2 + kπ

  22. Solve for k: k = (14π/13 - π/2)/π = 13/6

Therefore, 2cosπ/13cos9π/13 + cos3π/13cos5π/13 + cos5π/13cos3π/13 = 0.

Question:

Prove that: (cosx+cosy)2+(sinx−siny)2=4cos2x+y/2

Answer:

  1. Expand the left side of the equation: (cosx+cosy)2+(sinx−siny)2 = cos2x + 2cosxcosy + cos2y + sin2x - 2sinxsiny + sin2y

  2. Simplify the left side: cos2x + 2cosxcosy + cos2y + sin2x - 2sinxsiny + sin2y = cos2x + cos2y + 2cosxcosy + sin2x + sin2y - 2sinxsiny

  3. Rewrite the left side using the trigonometric identity: cos2x + cos2y + 2cosxcosy + sin2x + sin2y - 2sinxsiny = 2cos2x + 2cosxcosy + 2sinxsiny

  4. Simplify the right side of the equation: 4cos2x + y/2 = 2cos2x + cosy + y/2

  5. Set the left side and the right side of the equation equal to each other: 2cos2x + 2cosxcosy + 2sinxsiny = 2cos2x + cosy + y/2

  6. Simplify the equation: 2cosxcosy + 2sinxsiny = cosy + y/2

  7. Rewrite the equation using the trigonometric identity: 2cosxcosy + 2sinxsiny = 2cos(x+y/2)cos(y/2)

  8. Simplify the equation: 2cos(x+y/2)cos(y/2) = 2cos(x+y/2)cos(y/2)

  9. Therefore, (cosx+cosy)2+(sinx−siny)2=4cos2x+y/2 is true.

Question:

Prove that: (sin7x+sin5x)+(sin9x+sin3x)​/(cos7x+cos5x)+(cos9x+cos3x)=tan6x

Answer:

  1. Start by expanding the left side of the equation: (sin7x + sin5x) + (sin9x + sin3x) / (cos7x + cos5x) + (cos9x + cos3x)

  2. Use the double angle formula for sine and cosine to simplify the left side of the equation: sin(7x + 5x) / cos(7x + 5x) + sin(9x + 3x) / cos(9x + 3x)

  3. Use the sum and difference formulas for sine and cosine to simplify the left side of the equation: sin(12x)cos(2x) / cos(12x)cos(2x) + sin(12x)sin(2x) / cos(12x)sin(2x)

  4. Use the double angle formula for sine and cosine to simplify the left side of the equation: tan(6x)

  5. Therefore, (sin7x+sin5x)+(sin9x+sin3x)​/(cos7x+cos5x)+(cos9x+cos3x)=tan6x

Question:

Prove that: (cosx−cosy)2+(sinx−siny)2=4sin2x−y/2

Answer:

Given: (cosx−cosy)2+(sinx−siny)2=4sin2x−y/2

Step 1: Expand the left side of the equation: (cosx−cosy)2+(sinx−siny)2 = cos2x + sin2x - 2cosxcosy + cos2y + sin2y - 2sinxsiny

Step 2: Simplify the left side of the equation: cos2x + sin2x - 2cosxcosy + cos2y + sin2y - 2sinxsiny = 1 - 2cosxcosy + 1 - 2sinxsiny = 2 - 2(cosxcosy + sinxsiny)

Step 3: Rewrite the right side of the equation: 4sin2x−y/2 = 2sin2x - y

Step 4: Set the left side of the equation equal to the right side: 2 - 2(cosxcosy + sinxsiny) = 2sin2x - y

Step 5: Isolate the terms containing cosxcosy and sinxsiny: 2(cosxcosy + sinxsiny) = 2sin2x - y + 2

Step 6: Divide both sides by 2: cosxcosy + sinxsiny = sin2x - y/2 + 1

Step 7: Prove the statement: (cosx−cosy)2+(sinx−siny)2=4sin2x−y/2

Proof: Starting with the given statement, we expanded the left side of the equation and simplified it to 2 - 2(cosxcosy + sinxsiny). We then rewrote the right side of the equation as 2sin2x - y. We set the left side equal to the right side and isolated the terms containing cosxcosy and sinxsiny. We then divided both sides by 2 to get cosxcosy + sinxsiny = sin2x - y/2 + 1. Therefore, (cosx−cosy)2+(sinx−siny)2=4sin2x−y/2 is true.

Question:

Prove that: sinx+sin3x+sin5x+sin7x=4cosxcos2xsin4x

Answer:

  1. Use the identity sinA + sinB = 2sin((A+B)/2)cos((A-B)/2)

  2. Rewrite the left side of the equation as: 2sin(2x)cos(0) + 2sin(4x)cos(2x) + 2sin(6x)cos(4x) + 2sin(8x)cos(6x)

  3. Simplify the right side of the equation using the identity cos2A = 2cos2A - 1

  4. Rewrite the right side of the equation as: 4cosx(2cos2(2x) - 1)sin4x

  5. Use the identity sin2A = 2sinAcosA

  6. Rewrite both sides of the equation as: 2sin(2x)cos(0) + 2sin(4x)cos(2x) + 2sin(6x)cos(4x) + 2sin(8x)cos(6x) = 8cos2xsin4x

  7. Use the identity sinA + sinB = 2sin((A+B)/2)cos((A-B)/2)

  8. Rewrite both sides of the equation as: 2sin(4x)cos(2x) = 2sin(4x)cos(2x)

  9. Therefore, sinx+sin3x+sin5x+sin7x=4cosxcos2xsin4x is true.

Question:

Prove that: (sin3x+sinx)sinx+(cos3x−cosx)cosx=0

Answer:

  1. Expand the equation on the left side: (sin3x + sinx)sin x + (cos3x - cosx)cos x

  2. Use the double angle identity for sine and cosine: (3sin²x - sinx)sin x + (3cos²x - cosx)cos x

  3. Simplify: 3sin³x - sin²xcosx + 3cos²xcosx - cosxsin x

  4. Use the Pythagorean identity: 3sin³x - sin²xcosx + 3cos³x - sin²xcosx

  5. Simplify: 3sin³x + 3cos³x

  6. Use the identity sin³x + cos³x = 0: 3(sin³x + cos³x)

  7. Simplify: 0