Mathematics

Trigonometric Identities and Equations

223 Questions

Solve a variety of questions based on trigonometric identities and mathematical equations. Key areas covered include double angle formulas, the law of sines, and quadrant angles. This material helps students preparing for advanced mathematics exams, UPSC, and state public service commissions.

Double angle formulasLaw of sinesTrigonometric quadrantsLaplace transformSecant functionTaylor series

Trigonometric Identities and Equations Questions

Multiple choice

What is the period of the sine and cosine functions?

  1. $$2π$$
  2. $$π$$
  3. $$1$$
  4. $$0$$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The period of the sine and cosine functions is $$2π$$, meaning that they repeat their values every $$2π$$ units along the x-axis.

Multiple choice

What is the law of sines?

  1. $$ rac{sin(A)}{a} = rac{sin(B)}{b} = rac{sin(C)}{c}$$
  2. $$sin(A) + sin(B) + sin(C) = 1$$
  3. $$sin(A) = cos(B) = tan(C)$$
  4. $$sin(A) / cos(A) = tan(A)$$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The law of sines states that in a triangle, the ratio of the sine of an angle to the length of the opposite side is the same for all angles.

Multiple choice

Solve the equation (2\sin^2\theta + \sqrt{3}\sin\theta - 1 = 0) for (0 \le \theta \le 2\pi).

  1. \(\theta = \frac{\pi}{3}, \frac{5\pi}{3}\)
  2. \(\theta = \frac{\pi}{4}, \frac{3\pi}{4}\)
  3. \(\theta = \frac{\pi}{6}, \frac{5\pi}{6}\)
  4. \(\theta = \frac{\pi}{2}, \frac{3\pi}{2}\)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Factoring the equation, we get ((2\sin\theta - 1)(\sin\theta + 1) = 0). Solving each factor separately, we find (\sin\theta = \frac{1}{2}) or (\sin\theta = -1). Using the unit circle or reference angles, we find the solutions (\theta = \frac{\pi}{3}, \frac{5\pi}{3}).

Multiple choice

Find all solutions of the equation (\tan^2\theta - \tan\theta - 2 = 0) in the interval ([0, 2\pi)).

  1. \(\theta = \frac{\pi}{4}, \frac{3\pi}{4}\)
  2. \(\theta = \frac{\pi}{3}, \frac{2\pi}{3}\)
  3. \(\theta = \frac{\pi}{6}, \frac{5\pi}{6}\)
  4. \(\theta = 0, \pi\)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Factoring the equation, we get ((\tan\theta - 2)(\tan\theta + 1) = 0). Solving each factor separately, we find (\tan\theta = 2) or (\tan\theta = -1). Using the unit circle or reference angles, we find the solutions (\theta = \frac{\pi}{4}, \frac{3\pi}{4}).

Multiple choice

Solve the equation (2\cos^2\theta + 3\sin\theta - 5 = 0) for (0 \le \theta \le 2\pi).

  1. \(\theta = \frac{\pi}{6}, \frac{5\pi}{6}\)
  2. \(\theta = \frac{\pi}{3}, \frac{2\pi}{3}\)
  3. \(\theta = \frac{\pi}{4}, \frac{3\pi}{4}\)
  4. \(\theta = 0, \pi\)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the identity (\cos^2\theta = 1 - \sin^2\theta), we can rewrite the equation as (2(1 - \sin^2\theta) + 3\sin\theta - 5 = 0). Expanding and rearranging, we get (-2\sin^2\theta + 3\sin\theta - 3 = 0). Factoring, we find ((2\sin\theta - 3)(\sin\theta - 1) = 0). Solving each factor separately, we find (\sin\theta = \frac{3}{2}) or (\sin\theta = 1). Since (\sin\theta) cannot be greater than 1, we discard the first solution. Using the unit circle or reference angles, we find the solution (\theta = \frac{\pi}{6}, \frac{5\pi}{6}).

Multiple choice

Solve the equation (\sin^2\theta + \cos^2\theta - 2\sin\theta\cos\theta = 1) for (0 \le \theta \le 2\pi).

  1. \(\theta = \frac{\pi}{4}, \frac{3\pi}{4}\)
  2. \(\theta = \frac{\pi}{3}, \frac{2\pi}{3}\)
  3. \(\theta = \frac{\pi}{6}, \frac{5\pi}{6}\)
  4. \(\theta = 0, \pi\)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the identity (\sin^2\theta + \cos^2\theta = 1), we can simplify the equation to (1 - 2\sin\theta\cos\theta = 1). Rearranging, we get (\sin\theta\cos\theta = 0). This means either (\sin\theta = 0) or (\cos\theta = 0). Using the unit circle or reference angles, we find the solutions (\theta = \frac{\pi}{4}, \frac{3\pi}{4}).

Multiple choice

Solve the equation (\tan^2\theta + 2\tan\theta + 1 = 0) for (0 \le \theta \le 2\pi).

  1. \(\theta = \frac{\pi}{4}, \frac{3\pi}{4}\)
  2. \(\theta = \frac{\pi}{3}, \frac{2\pi}{3}\)
  3. \(\theta = \frac{\pi}{6}, \frac{5\pi}{6}\)
  4. \(\theta = 0, \pi\)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Factoring the equation, we get ((\tan\theta + 1)^2 = 0). Solving for (\tan\theta), we find (\tan\theta = -1). Using the unit circle or reference angles, we find the solutions (\theta = \frac{\pi}{4}, \frac{3\pi}{4}).

Multiple choice

Find all solutions of the equation (\sec^2\theta - \tan^2\theta = 1) in the interval ([0, 2\pi)).

  1. \(\theta = \frac{\pi}{4}, \frac{3\pi}{4}\)
  2. \(\theta = \frac{\pi}{3}, \frac{2\pi}{3}\)
  3. \(\theta = \frac{\pi}{6}, \frac{5\pi}{6}\)
  4. \(\theta = 0, \pi\)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the identity (\sec^2\theta = 1 + \tan^2\theta), we can simplify the equation to (1 + \tan^2\theta - \tan^2\theta = 1). This simplifies to (1 = 1), which is true for all values of (\theta). Therefore, the equation has infinitely many solutions in the interval ([0, 2\pi)).

Multiple choice

Solve the equation (\sin^2\theta - \cos^2\theta = \frac{1}{2}) for (0 \le \theta \le 2\pi).

  1. \(\theta = \frac{\pi}{4}, \frac{3\pi}{4}\)
  2. \(\theta = \frac{\pi}{3}, \frac{2\pi}{3}\)
  3. \(\theta = \frac{\pi}{6}, \frac{5\pi}{6}\)
  4. \(\theta = 0, \pi\)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the identity (\sin^2\theta + \cos^2\theta = 1), we can rewrite the equation as (\sin^2\theta - (1 - \sin^2\theta) = \frac{1}{2}). Simplifying, we get (2\sin^2\theta - 1 = \frac{1}{2}). Solving for (\sin\theta), we find (\sin\theta = \pm\frac{\sqrt{6}}{4}). Using the unit circle or reference angles, we find the solutions (\theta = \frac{\pi}{4}, \frac{3\pi}{4}).

Multiple choice

Solve the equation (2\cos^2\theta + \sin\theta - 1 = 0) for (0 \le \theta \le 2\pi).

  1. \(\theta = \frac{\pi}{6}, \frac{5\pi}{6}\)
  2. \(\theta = \frac{\pi}{3}, \frac{2\pi}{3}\)
  3. \(\theta = \frac{\pi}{4}, \frac{3\pi}{4}\)
  4. \(\theta = 0, \pi\)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the identity (\cos^2\theta = 1 - \sin^2\theta), we can rewrite the equation as (2(1 - \sin^2\theta) + \sin\theta - 1 = 0). Expanding and rearranging, we get (-2\sin^2\theta + \sin\theta - 1 = 0). Factoring, we find ((2\sin\theta - 1)(\sin\theta - 1) = 0). Solving each factor separately, we find (\sin\theta = \frac{1}{2}) or (\sin\theta = 1). Using the unit circle or reference angles, we find the solutions (\theta = \frac{\pi}{6}, \frac{5\pi}{6}).

Multiple choice

Find all solutions of the equation (\tan^2\theta + \sec\theta - 1 = 0) in the interval ([0, 2\pi)).

  1. \(\theta = \frac{\pi}{4}, \frac{3\pi}{4}\)
  2. \(\theta = \frac{\pi}{3}, \frac{2\pi}{3}\)
  3. \(\theta = \frac{\pi}{6}, \frac{5\pi}{6}\)
  4. \(\theta = 0, \pi\)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the identity (\sec^2\theta = 1 + \tan^2\theta), we can rewrite the equation as (1 + \tan^2\theta + \tan\theta - 1 = 0). Simplifying, we get (\tan^2\theta + \tan\theta = 0). Factoring, we find (\tan\theta(\tan\theta + 1) = 0). Solving each factor separately, we find (\tan\theta = 0) or (\tan\theta = -1). Using the unit circle or reference angles, we find the solutions (\theta = \frac{\pi}{4}, \frac{3\pi}{4}).

Multiple choice

Solve the equation (\sin^2\theta + \cos^2\theta - \sin\theta - \cos\theta = 0) for (0 \le \theta \le 2\pi).

  1. \(\theta = \frac{\pi}{4}, \frac{3\pi}{4}\)
  2. \(\theta = \frac{\pi}{3}, \frac{2\pi}{3}\)
  3. \(\theta = \frac{\pi}{6}, \frac{5\pi}{6}\)
  4. \(\theta = 0, \pi\)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the identity (\sin^2\theta + \cos^2\theta = 1), we can simplify the equation to (1 - \sin\theta - \cos\theta = 0). Rearranging, we get (\sin\theta + \cos\theta = 1). This suggests that (\sin\theta) and (\cos\theta) have the same sign. The only way this can happen is if both (\sin\theta) and (\cos\theta) are positive. Using the unit circle or reference angles, we find the solutions (\theta = \frac{\pi}{4}, \frac{3\pi}{4}).

Multiple choice

Solve the equation (2\sin^2\theta - 3\sin\theta + 1 = 0) for (0 \le \theta \le 2\pi).

  1. \(\theta = \frac{\pi}{6}, \frac{5\pi}{6}\)
  2. \(\theta = \frac{\pi}{3}, \frac{2\pi}{3}\)
  3. \(\theta = \frac{\pi}{4}, \frac{3\pi}{4}\)
  4. \(\theta = 0, \pi\)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Factoring the equation, we get ((2\sin\theta - 1)(\sin\theta - 1) = 0). Solving each factor separately, we find (\sin\theta = \frac{1}{2}) or (\sin\theta = 1). Using the unit circle or reference angles, we find the solutions (\theta = \frac{\pi}{6}, \frac{5\pi}{6}).

Multiple choice

What is the value of (\sin 90^\circ) in Indian trigonometry?

  1. 0

  2. 1
  3. \(\frac{1}{2}\)
  4. \(\frac{\sqrt{2}}{2}\)
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In Indian trigonometry, the value of (\sin 90^\circ) is 1.

Multiple choice

What is the value of (\cos 0^\circ) in Indian trigonometry?

  1. 0

  2. 1
  3. \(\frac{1}{2}\)
  4. \(\frac{\sqrt{2}}{2}\)
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In Indian trigonometry, the value of (\cos 0^\circ) is 1.