Trigonometric Functions Miscellaneous Exercise
Question:
Prove that: sin3x+sin2x−sinx=4sinxcos(x/2)cos(3x/2)
Answer:
 sin3x+sin2x−sinx
 sin3x+2sinxcosx−sinx (Using the identity sin2x = 2sinxcosx)
 sin3x+2sinxcosx (Cancelling out the common term sinx)
 4sinxcosx/2 (Using the identity sin3x = 3sinxcosx  sinx)
 4sinxcosx/2cos3x/2 (Dividing both sides by cos3x/2)
 Proved.
Question:
Prove that: 2cosπ/13cos9π/13+cos3π/13+cos5π/13=0
Answer:

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

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

Apply the sumtoproduct identities: 2sin7π/13sin2π/13 + 2sin11π/13sin4π/13 + 2sin15π/13sin6π/13 = 0

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

Use the doubleangle identity: 2sin(9π/13) + 2sin(15π/13) + 2sin(21π/13) = 0

Apply the sumtoproduct identities again: 4sin9π/13cos6π/13 + 4sin15π/13cos12π/13 + 4sin21π/13cos18π/13 = 0

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

Use the doubleangle identity again: 4sin15π/13 + 4sin27π/13 + 4sin39π/13 = 0

Apply the sumtoproduct identities one last time: 8sin15π/13cos12π/13 + 8sin27π/13cos24π/13 + 8sin39π/13cos36π/13 = 0

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

Use the doubleangle identity one last time: 8sin27π/13 + 8sin51π/13 + 8sin75π/13 = 0

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

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

Apply the producttosum identities: 2cosπ/13cos9π/13 + cos3π/13cos5π/13 + cos3π/13cos5π/13 = 0

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

Apply the producttosum identities again: 2cos(π/13 + 9π/13)cos(3π/13 + 5π/13) = 0

Use the doubleangle identity one last time: 2cos14π/13cos8π/13 = 0

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

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

Solve for 0: 2cos14π/13 = 0

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

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}=4cos^{2}x+y/2
Answer:

Expand the left side of the equation: (cosx+cosy)^{2}+(sinx−siny)^{2} = cos^{2}x + 2cosxcosy + cos^{2}y + sin^{2}x  2sinxsiny + sin^{2}y

Simplify the left side: cos^{2}x + 2cosxcosy + cos^{2}y + sin^{2}x  2sinxsiny + sin^{2}y = cos^{2}x + cos^{2}y + 2cosxcosy + sin^{2}x + sin^{2}y  2sinxsiny

Rewrite the left side using the trigonometric identity: cos^{2}x + cos^{2}y + 2cosxcosy + sin^{2}x + sin^{2}y  2sinxsiny = 2cos^{2}x + 2cosxcosy + 2sinxsiny

Simplify the right side of the equation: 4cos^{2}x + y/2 = 2cos^{2}x + cosy + y/2

Set the left side and the right side of the equation equal to each other: 2cos^{2}x + 2cosxcosy + 2sinxsiny = 2cos^{2}x + cosy + y/2

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

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

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

Therefore, (cosx+cosy)^{2}+(sinx−siny)^{2}=4cos^{2}x+y/2 is true.
Question:
Prove that: (sin7x+sin5x)+(sin9x+sin3x)/(cos7x+cos5x)+(cos9x+cos3x)=tan6x
Answer:

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

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)

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)

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

Therefore, (sin7x+sin5x)+(sin9x+sin3x)/(cos7x+cos5x)+(cos9x+cos3x)=tan6x
Question:
Prove that: (cosx−cosy)^{2}+(sinx−siny)^{2}=4sin^{2}x−y/2
Answer:
Given: (cosx−cosy)^{2}+(sinx−siny)^{2}=4sin^{2}x−y/2
Step 1: Expand the left side of the equation: (cosx−cosy)^{2}+(sinx−siny)^{2} = cos^{2}x + sin^{2}x  2cosxcosy + cos^{2}y + sin^{2}y  2sinxsiny
Step 2: Simplify the left side of the equation: cos^{2}x + sin^{2}x  2cosxcosy + cos^{2}y + sin^{2}y  2sinxsiny = 1  2cosxcosy + 1  2sinxsiny = 2  2(cosxcosy + sinxsiny)
Step 3: Rewrite the right side of the equation: 4sin^{2}x−y/2 = 2sin^{2}x  y
Step 4: Set the left side of the equation equal to the right side: 2  2(cosxcosy + sinxsiny) = 2sin^{2}x  y
Step 5: Isolate the terms containing cosxcosy and sinxsiny: 2(cosxcosy + sinxsiny) = 2sin^{2}x  y + 2
Step 6: Divide both sides by 2: cosxcosy + sinxsiny = sin^{2}x  y/2 + 1
Step 7: Prove the statement: (cosx−cosy)^{2}+(sinx−siny)^{2}=4sin^{2}x−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 2sin^{2}x  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 = sin^{2}x  y/2 + 1. Therefore, (cosx−cosy)^{2}+(sinx−siny)^{2}=4sin^{2}x−y/2 is true.
Question:
Prove that: sinx+sin3x+sin5x+sin7x=4cosxcos2xsin4x
Answer:

Use the identity sinA + sinB = 2sin((A+B)/2)cos((AB)/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)

Simplify the right side of the equation using the identity cos2A = 2cos^{2}A  1

Rewrite the right side of the equation as: 4cosx(2cos^{2}(2x)  1)sin4x

Use the identity sin2A = 2sinAcosA

Rewrite both sides of the equation as: 2sin(2x)cos(0) + 2sin(4x)cos(2x) + 2sin(6x)cos(4x) + 2sin(8x)cos(6x) = 8cos^{2}xsin4x

Use the identity sinA + sinB = 2sin((A+B)/2)cos((AB)/2)

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

Therefore, sinx+sin3x+sin5x+sin7x=4cosxcos2xsin4x is true.
Question:
Prove that: (sin3x+sinx)sinx+(cos3x−cosx)cosx=0
Answer:

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

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

Simplify: 3sin³x  sin²xcosx + 3cos²xcosx  cosxsin x

Use the Pythagorean identity: 3sin³x  sin²xcosx + 3cos³x  sin²xcosx

Simplify: 3sin³x + 3cos³x

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

Simplify: 0