Trigonometry Practice Questions Hard

Trigonometry can be challenging, especially when you're tackling advanced problems. We’ll focus on a series of hard trigonometry practice questions designed to test your understanding of key concepts.

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.

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).

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

sin(A + B) = 63/65.

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

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

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^∘.

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^∘.

x = 60^∘, 120^∘, 240^∘, 300^∘.

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

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°

x = 60°, 180°, 300°.

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

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

sin(10^∘) ⋅ sin(30^∘) ⋅ sin(50^∘) ⋅ sin(70^∘) = 1/16​.

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

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).

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)]

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

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.

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

Practice Questions (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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