Trigonometry Practice Questions Medium

Trigonometry can be challenging, particularly for class 10 and 11 students dealing with trigonometric ratios and identities. In this article, we have gathered a set of medium-difficulty questions that cover key trigonometric concepts and formulas.

By solving these questions, ranging from easy to difficult, you can enhance your problem-solving skills in trigonometry.

Check: Tricks to Remember Trigonometry Tables

Solved Questions on Trigonometry (Medium)

Question 1: Find the exact value of tan(45° + 30°).

Solution:

Use the tangent sum formula: tan(A + B) = (tan(A) + tan(B)) / (1 - tan(A)tan(B)).
Here, A = 45° and B = 30°.

tan(45°) = 1 and tan(30°) = √3/3.
⇒ tan(45° + 30°) = (1 + √3/3) / (1 - 1×√3/3) = (1 + √3/3) / (1 - √3/3) =(√3 + 1) / (1 - √3).
(√3 + 1) / (1 - √3) × (1 + √3)/(1 + √3) = (1 + √3)²/[(1² - (√3)²) ] = (1 + √3)²/(1 - 3) = (1 + √3)²/(-2) = -(1 + √3)²/2

Thus, tan(45° + 30°) = -(1 + √3)²/2.

Question 2: If the angle of elevation from a point on the ground to the top of a tower is 30°, and the distance from the point to the base of the tower is 40 meters, find the height of the tower.

Solution:

The situation forms a right triangle with the tower as the height.
Use the tangent function: tan(θ) = opposite/adjacent.
Here, θ = 30°, adjacent side (distance to the tower) = 40 m.
So, tan(30°) = height / 40.
As tan(30°) is √3/3.

Therefore, height = 40 × √3/3 ≈ 23.09 meters.

Question 3: Solve for x: 2 sin(x) = √2, for 0° ≤ x ≤ 360°.

Solution:

First, isolate sin(x): sin(x) = √2 / 2.
The solutions to sin(x) = √2 / 2 are x = 45° and x = 135°.

So, the values of x that satisfy 2 sin(x) = √2 in the given range are x = 45° and x = 135°.

Read More about Trigonometric Equations.

Question 4: If the sides of a right triangle are in the ratio 3:4:5, and the shortest side is 9 cm, find the length of the hypotenuse.

Solution:

The sides of the triangle are in the ratio 3:4:5.
The shortest side (3 parts) is 9 cm.
Therefore, each part is 9 / 3 = 3 cm.
The hypotenuse (5 parts) is 5 × 3 cm = 15 cm.

Thus, the length of the hypotenuse is 15 cm.

Question 5: Prove that sin⁴x - cos⁴x = 1 - 2cos²x.

Solution:

Start with the left-hand side (LHS): sin⁴x - cos⁴x.

⇒ LHS= (sin²x)² - (cos²x)² = (sin²x + cos²x)(sin²x - cos²x).
⇒ LHS = 1 × (sin²x - cos²x) [As sin²x + cos²x = 1]
⇒ LHS = 1 - 2cos²x = RHS [As sin²x = 1 - cos²x]

Thus, sin⁴x - cos⁴x = 1 - 2cos²x,

Question 6: Find the exact value of sin(15°).

Solution:

Use the half-angle formula: sin(θ/2) = √((1 - cosθ)/2).
Here, θ = 30°, so sin(15°) = sin(30°/2).
As cos(30°) = √3/2.
Apply the formula: sin(15°) = √((1 - √3/2)/2).
⇒ sin(15°) = √((2 - √3)/4) = (√(2 - √3))/2.

Question 7: Solve for x: cos²x - sin²x = 1/2, for 0 ≤ x ≤ 2π.

Solution:

Use the Pythagorean identity: sin²x + cos²x = 1.
Substitute for sin²x: cos²x - (1 - cos²x) = 1/2.
Simplify: 2cos²x = 3/2.
cos²x = 3/4.
Find x: cos(x) = ±√(3/4) = ±√3/2.
Solve for x in the given range: x = π/6, 5π/6, 7π/6, 11π/6.

Explanation:
For cos(x) = √3/2, x = π/6 and 11π/6 in the range 0 ≤ x ≤ 2π.
For cos(x) = -√3/2, x = 5π/6 and 7π/6 in the range 0 ≤ x ≤ 2π.

Question 8: Prove that tan(θ)tan(60° - θ)tan(60° + θ) = tan3θ.

Solution:

Start with the left-hand side (LHS): tan(θ)tan(60° - θ)tan(60° + θ).

Use the tan subtraction and addition formulas:
tan(60° - θ) = (tan(60°) - tan(θ))/(1 + tan(60°)tan(θ)).
tan(60° + θ) = (tan(60°) + tan(θ))/(1 - tan(60°)tan(θ)).

Multiply these expressions with tan(θ).

\tan \theta \cdot \tan(60^\circ - \theta) \cdot \tan(60^\circ + \theta) = \tan \theta \cdot \left( \frac{\sqrt{3} - \tan(\theta)}{1 + \sqrt{3} \tan(\theta)} \right) \cdot \left( \frac{\sqrt{3} + \tan(\theta)}{1 - \sqrt{3} \tan(\theta)} \right)
\Rightarrow \tan \theta \cdot \tan(60^\circ - \theta) \cdot \tan(60^\circ + \theta) = \frac{\tan \theta \cdot (\sqrt{3} - \tan \theta) (\sqrt{3} + \tan \theta)}{(1+\sqrt{3} \tan \theta)(1-\sqrt{3} \tan \theta) } = \frac{\tan \theta \cdot (\sqrt{3}^2 - \tan^2 \theta)}{1^2 - (\sqrt3 \tan \theta)^2} = \frac{\tan \theta \cdot (3 - \tan^2\theta)}{1 - 3 \tan^2 \theta}
\Rightarrow \tan \theta \cdot \tan(60^\circ - \theta) \cdot \tan(60^\circ + \theta) = \frac{\tan \theta \cdot (\sqrt{3} - \tan \theta) (\sqrt{3} + \tan \theta)}{(1+\sqrt{3} \tan \theta)(1-\sqrt{3} \tan \theta) } = \frac{\tan \theta \cdot (\sqrt{3}^2 - \tan^2 \theta)}{1^2 - (\sqrt3 \tan \theta)^2} = \frac{\tan \theta \cdot (3 - \tan^2\theta)}{1 - 3 \tan^2 \theta} = \text{RHS}

Also Read,

• Trigonometric Functions
• Trigonometric Ratios
• Trigonometric Identities

Practice with Quiz: - Trigonometry Quiz with Solutions

Practice Problems on Trigonometry (Medium)

Question 1: Solve for x in the equation cos(x) = 1/2 for 0^∘ ≤ x ≤ 360^∘.

Question 2: In a right triangle, the sides are in the ratio 5:12:13, and the shortest side is 10 units. Find the length of the hypotenuse.

Question 3: Prove that sin(2x) = 2 sin(x) cos(x).

Question 4: Solve for x in the equation sin²(x) + cos²(x) = 1 for 0 <= x<= 2π.

Question 5: Prove that cos(2x) = cos²(x) - sin²(x).

Question 6: Solve 2cos⁡²(x) - 1 = 0 for 0^∘ ≤ x ≤ 360^∘.

Question 7: In a right triangle, the hypotenuse is 10 units, and sin⁡(A) = 0.6. Find the lengths of the opposite and adjacent sides.

Question 8: Solve for x in the equation 2sin⁡(x) - 1 = 0 for 0^∘ ≤ x ≤ 360^∘_.

Answer Key

1. x = 60^∘,300^∘_.
2. Hypotenuse = 26 units.
3. 2sin(x)cos(x).
4. The equation holds for all x.
5. cos(2x) = cos2(x) - sin2(x).
6. x = 45^∘, 135^∘, 225^∘, 315^∘_.
7. Opposite side = 6 units, Adjacent side = 8 units.
8. x = 30^∘, 150^∘_.

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