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Trigonometry Practice Questions Hard

Last Updated : 23 Jul, 2025
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Trigonometry can be challenging, especially when you're tackling advanced problems. In this article, we’ll focus on a series of hard trigonometry practice questions designed to test your understanding of key concepts.

Whether you’re preparing for exams or looking to strengthen your problem-solving skills, these questions will help you improve your grasp of trigonometric identities, equations, and real-world applications. Solving these questions, ranging from easy to hard will help you better your problem-solving skills in trigonometry.

Check: Tricks to Remember Trigonometry Tables

Solved Questions on Trigonometry (Hard)

Question 1: Prove that cos⁡(3x) = 4cos⁡3(x) − 3cos⁡(x) using trigonometric identities.

Solution:

To Prove cos⁡(3x) = 4cos⁡3(x) − 3cos⁡(x).

Use the angle sum formula for cos⁡(3x) = cos⁡(2x + x)
cos⁡(3x) = cos⁡(2x) cos⁡(x) − sin⁡(2x) sin⁡(x)

Substitute cos⁡(2x) = 2cos⁡2(x) − 1 and sin⁡(2x) = 2sin⁡(x) cos⁡(x):
cos⁡(3x) = (2cos⁡2(x) − 1) cos (x) - (2sin⁡(x) cos⁡(x)) sin(x).
⇒ cos(3x) = 2cos3(x) − cos(x) − 2sin2(x) cos(x).

Use sin⁡2(x) = 1 − cos⁡2(x):
cos(3x) = 2cos3(x) − cos(x) − 2(1−cos2(x))cos(x).
⇒ cos⁡(3x) = 2cos⁡3(x) − cos⁡(x) − 2cos⁡(x) + 2cos⁡3(x).
⇒ cos(3x) = 4cos3(x) − 3cos(x)

Which is the required formula.

Question 2: In a triangle, two angles A and B satisfy sin⁡(A) = 3/5​ and cos⁡(B) = 5/13. Find the value of sin⁡(A + B).

Solution:

Use the formula sin⁡(A + B) = sin⁡(A) cos⁡(B) + cos⁡(A) sin⁡(B).

Calculate cos⁡(A) using cos⁡2(A) = 1 −sin⁡2(A):
\cos(A) = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{\frac{16}{25}} = \frac{4}{5}

Calculate sin⁡(B) using sin⁡2(B) = 1 −cos⁡2(B):
\sin(B) = \sqrt{1 - \left(\frac{5}{13}\right)^2} = \sqrt{\frac{144}{169}} = \frac{12}{13}.

Thus, sin⁡(A + B) = sin⁡(A) cos⁡(B) + cos⁡(A) sin⁡(B).
⇒ sin(A + B) = 3/5 × 5/13 + 4/5 × 12/13
⇒ sin(A + B) =15/65 + 48/65 = 63/65

Answer: sin(A + B) = 63/65.

Question 3: If sin⁡(x) = 3/5, find cos⁡(2x) and sin⁡(2x).

Solution:

Use cos2(x) = 1 - sin2(x):
\cos(x) = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{\frac{16}{25}} = \frac{4}{5}.

cos(2x) = cos2(x) - sin2(x)
⇒ cos(2x) = (4/5)2 - (3/5)2
⇒ cos(2x) = 16/25 - 9/25
⇒ cos(2x) = 7/25

Now, sin(2x) = 2 sin(x)cos(x):
sin(2x) = 2 × 3/5 × 4/5 = 24/25

Answer: cos⁡(2x )= 7/25, sin⁡(2x) = 24/25.

Question 4: Prove that 1 + tan⁡2(x) = sec⁡2(x), and then use this to solve sec⁡2(x) = 4 for 0≤ x ≤ 360.

Solution:

To Prove: 1 + tan⁡2(x) = sec⁡2(x):

Proof:
Use sin⁡2(x) + cos⁡2(x) = 1 and divide through by cos⁡2(x):
⟹  sin⁡2(x)/cos⁡2(x) + 1 = 1/cos⁡2(x) 
⟹  tan⁡2(x) + 1 = sec⁡2(x).

Hence Proved

Solve sec⁡2(x) = 4:

sec2(x) = 4
⟹  tan⁡2(x) = 3
⟹  tan⁡(x) = ±√3
⇒ tan(x) = √3​ at x = 60, 240, and tan⁡(x) = −√3 at x = 120, 300.

Answer: x = 60, 120, 240, 300.

Question 5: Find all values of x such that cos⁡(2x) + cos⁡(x) = 0 for 0≤ x ≤ 360.

Solution:

Use the identity cos⁡(2x) = 2cos⁡2(x) − 1:
⇒ 2cos⁡2(x) − 1 + cos⁡(x) = 0

Let y = cos⁡(x):
2y2 + y − 1 = 0
y = \frac{-1 \pm \sqrt{1^2 - 4(2)(-1)}}{2(2)} = \frac{-1\pm\sqrt{9}}{4}
⇒ y = (-1 + 3)/4 = 1/2, or y = (-1 -3)/4 = -1

Now, solve fot x:
cos(x) = 1/2: x = 60°, 300°, and
cos(x) = -1: x = 180°

Answer: x = 60°, 180°, 300°.

Question 6: Prove that sin(10) ⋅ sin(30) ⋅ sin(50) ⋅ sin(70) = 1/16​.

Solution:

To prove: sin(10) ⋅ sin(30) ⋅ sin(50) ⋅ sin(70) = 1/16​.

Substitute: sin(30°) = 1/2.
1/2 ⋅ sin(10°) ⋅ sin(50) ⋅ sin(70)

Pair sin⁡(10) and sin⁡(70) using the product-to-sum formula: 2sin⁡(A) sin⁡(B) = cos⁡(A − B) − cos⁡(A + B)
Thus, sin(10°) ⋅ sin(70) = 1/2 ⋅ (cos(60°) -cos(80°)).

Substitute this back:
1/2 ⋅ 1/2 ⋅ (cos(60°) - cos(80°)) ⋅ sin(50°).

Simplify cos(60°) = 1/2
1/4 ⋅ (1/2 - cos(80°)) ⋅ sin(50°).

Expand sin⁡(50°) into terms
1/4 ⋅ 1/2 ⋅ sin(50°) - 1/4 ⋅ cos(80°) ⋅ sin(50°)

Simplify cos⁡(80) ⋅ sin⁡(50) using the product-to-sum formula: cos(A) sin(B) = 1/2 ​[sin(A + B) − sin(A − B)].

Substitute A = 80, B = 50:
cos(80) sin(50) = 1/2​[sin(130) − sin(30)].
cos(80) sin(50) = 1/2​[sin(50) − 1/2​].

Now substitute into the equation:
1/4​ ⋅ 1/2 ​⋅ sin(50) − 1/4​ ⋅ 1​/2 ⋅[sin(50) − 1/2​].
1/8 ⋅ sin(50) − 1/8​ ⋅ sin(50) + 1/16​.
The sin⁡(50) terms cancel out: 1/16

Answer: sin(10) ⋅ sin(30) ⋅ sin(50) ⋅ sin(70) = 1/16​.

Question 7: Prove the Identity \bold{\frac{sin(x) - sin(3x)}{cos(x)+cos(3x)} = tan(2x)}

Solution:

Applying sum-to-product formulas to both the numerator and denominator.

Numerator (sin⁡(x) − sin⁡(3x): sin(x) − sin(3x) = 2cos((x + 3x)/2​) sin ((x − 3x)/2​) = 2cos(2x) sin(−x).

Using sin⁡(−x) = −sin⁡(x), we have:
sin⁡(x) − sin⁡(3x) = −2cos⁡(2x) sin⁡(x).

Denominator (cos⁡(x) + cos⁡(3x): cos(x) + cos(3x) = 2cos((x + 3x)/2​) cos((x − 3x)​/2) = 2cos(2x) cos(−x).

Using cos⁡(−x) = cos⁡(x), we have:
cos(x) + cos(3x) = 2cos(2x) cos(x).

Simplify the fraction:
\frac{sin(x) - sin(3x)}{cos(x)+cos(3x)}=\frac{-cos(2x)sin(x)}{2cos(2x)cos(x)}

Cancel 2cos⁡(2x) (non-zero for valid x):
= -sin(x)/cos(x) = -tan(x).

Since LHS ≠ RHS, the given identity does not hold.

Question 8: Find the general solution of: sin(3x) + sin(5x) + sin(7x) = 0.

Solution:

Given: sin⁡(3x) + sin⁡(5x) + sin⁡(7x) = 0.

Use the sum-to-product formula for sin⁡(3x) + sin⁡(7x):
sin(A) + sin(B) = 2sin(A + B/2​) cos(A − B​/2).

Substituting A = 3x, B = 7x:
⇒ sin(3x) + sin(7x) = 2sin((3x + 7x)/2) cos((3x - 7x)/2).
⇒ sin(3x) + sin(7x) = 2sin(5x) cos(2x).

Subtitute back into the equation:
2sin(5x) cos(2x) + sin(5x) = 0.

Factorize sin(5x):
sin(5x)(2cos(2x) + 1) = 0.

Solve Each Factor:
Case 1: sin⁡(5x) = 0
⟹ 5x = nπ
⟹ x = nπ/5​, n ∈ Z.

Case 2: 2cos(2x) + 1 = 0
cos(2x)= −1/2​.
⟹ 2x = 2nπ ± 2π/3​
⟹ x = nπ ± π/3​, n ∈ Z.

Answer: The genral solutions are x = nπ/5​, n ∈ Z, or x = nπ ± π/3​, n ∈ Z.

Read More:

Practice with Quiz: - Trigonometry Quiz with Solutions

Practice Questions on Trigonometry (Hard)

Question 1: Prove that sin⁡(3x) = 3sin⁡(x) − 4sin⁡3(x) using trigonometric identities.

Question 2: Find all solutions for cos⁡(2x) + cos⁡(4x) + cos⁡(6x) = 0.

Question 3: Prove that tan⁡(3x) = (3tan⁡(x) − tan⁡3(x))/1 - 3tan⁡2(x).

Question 4: Solve sin⁡(2x) + sin⁡(4x) + sin⁡(8x) = 0.

Question 5: Prove that cos⁡(4x) − cos⁡(2x) = −2sin⁡(3x) sin⁡(x).

Question 6: Solve sin⁡(5x) + sin⁡(7x) + sin⁡(9x) = 0.

Question 7: Prove that (sin⁡(x) − sin⁡(3x))/(cos⁡(x) + cos⁡(3x)) = - tan⁡(x).

Question 8: Solve cos⁡(3x) + cos⁡(5x) + cos⁡(7x) = 0.

Answer Key:

  1. sin⁡(3x) = 3sin⁡(x) − 4sin⁡3(x).
  2. x = (2n + 1)π/8, x = nπ ± π/3, n ∈ Z
  3. (3x) = (3tan⁡(x) − tan⁡3(x))/1−3tan⁡2(x).
  4. x = nπ/5, x = 2nπ/3 ± π/9, n ∈ Z.
  5. cos⁡(4x) − cos⁡(2x) = −2sin⁡(3x)sin⁡(x).
  6. x = nπ/7, x = nπ ± π/6, n ∈ Z.
  7. (sin⁡(x) − sin⁡(3x))/( cos⁡(x)+cos⁡(3x)) = −tan⁡(2x).
  8. x = (2n + 1)π/6, x = nπ ± π/5, n ∈ Z.

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