6.1 MCQs-Trigonometry

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Trigonometry

Basic Concepts and Angles

1. In a right triangle, the sine of an acute angle is defined as:

1. Opposite side / Adjacent side

2. Adjacent side / Hypotenuse

3. Opposite side / Hypotenuse

4. Hypotenuse / Opposite side

Show me the answer

Answer: 3. Opposite side / Hypotenuse

Explanation:

• For an acute angle $\theta$ in a right triangle:

  • $\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}}$

  • $\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}$

  • $\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}}$

• This is the basic SOH-CAH-TOA mnemonic.

2. Which of the following trigonometric ratios is positive in the third quadrant?

1. Sine

2. Cosine

3. Tangent

4. Cosecant

Show me the answer

Answer: 3. Tangent

Explanation:

• The "ASTC" rule (All Students Take Calculus) helps remember sign conventions:

  • Quadrant I (0° to 90°): All ratios are positive.

  • Quadrant II (90° to 180°): Sine and cosecant are positive.

  • Quadrant III (180° to 270°): Tangent and cotangent are positive.

  • Quadrant IV (270° to 360°): Cosine and secant are positive.

• In QIII, only tangent and its reciprocal cotangent are positive.

3. The radian measure of 180° is:

1. $\pi$

2. $\frac{\pi}{2}$

3. $2\pi$

4. $\frac{\pi}{4}$

Show me the answer

Answer: 1. $\pi$

Explanation:

• The conversion between degrees and radians is: $180^\circ = \pi \text{ radians}$.

• Therefore:

  • $90^\circ = \frac{\pi}{2}$ radians

  • $360^\circ = 2\pi$ radians

  • $1^\circ = \frac{\pi}{180}$ radians

Trigonometric Identities

4. The fundamental Pythagorean identity is:

1. $\sin^2 \theta + \cos^2 \theta = 1$

2. $\tan^2 \theta + 1 = \sec^2 \theta$

3. $1 + \cot^2 \theta = \csc^2 \theta$

4. All of the above

Show me the answer

Answer: 4. All of the above

Explanation:

• All three are Pythagorean identities derived from $\sin^2 \theta + \cos^2 \theta = 1$:

  1. Divide identity (1) by $\cos^2 \theta$: $\tan^2 \theta + 1 = \sec^2 \theta$.

  2. Divide identity (1) by $\sin^2 \theta$: $1 + \cot^2 \theta = \csc^2 \theta$.

• These identities are fundamental for simplifying trigonometric expressions.

5. Which of the following is the double-angle formula for sine?

1. $\sin 2\theta = 2 \sin \theta \cos \theta$

2. $\sin 2\theta = \sin^2 \theta - \cos^2 \theta$

3. $\sin 2\theta = 1 - 2\sin^2 \theta$

4. $\sin 2\theta = 2\cos^2 \theta - 1$

Show me the answer

Answer: 1. $\sin 2\theta = 2 \sin \theta \cos \theta$

Explanation:

• The double-angle formulas are:

  • $\sin 2\theta = 2 \sin \theta \cos \theta$

  • $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$

    • Also equals $2\cos^2 \theta - 1$

    • Also equals $1 - 2\sin^2 \theta$

  • $\tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta}$

6. The value of $\sin(90^\circ - \theta)$ is equal to:

1. $\sin \theta$

2. $\cos \theta$

3. $-\cos \theta$

4. $-\sin \theta$

Show me the answer

Answer: 2. $\cos \theta$

Explanation:

• This is a co-function identity.

• For complementary angles (sum = 90°):

  • $\sin(90^\circ - \theta) = \cos \theta$

  • $\cos(90^\circ - \theta) = \sin \theta$

  • $\tan(90^\circ - \theta) = \cot \theta$

• These identities show the relationship between trigonometric functions of complementary angles.

Trigonometric Functions of Special Angles

7. The value of $\sin 30^\circ$ is:

1. $\frac{1}{2}$

2. $\frac{\sqrt{3}}{2}$

3. $\frac{1}{\sqrt{2}}$

4. 1

Show me the answer

Answer: 1. $\frac{1}{2}$

Explanation:

• Important exact values:

  • $\sin 0^\circ = 0$, $\sin 30^\circ = \frac{1}{2}$, $\sin 45^\circ = \frac{1}{\sqrt{2}}$, $\sin 60^\circ = \frac{\sqrt{3}}{2}$, $\sin 90^\circ = 1$

  • $\cos 0^\circ = 1$, $\cos 30^\circ = \frac{\sqrt{3}}{2}$, $\cos 45^\circ = \frac{1}{\sqrt{2}}$, $\cos 60^\circ = \frac{1}{2}$, $\cos 90^\circ = 0$

8. The value of $\tan 45^\circ$ is:

1. 0

2. 1

3. $\frac{1}{\sqrt{3}}$

4. $\sqrt{3}$

Show me the answer

Answer: 2. 1

Explanation:

• Since $\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}$,

• $\tan 45^\circ = \frac{\sin 45^\circ}{\cos 45^\circ} = 1$.

• Other important tangent values:

  • $\tan 30^\circ = \frac{1}{\sqrt{3}}$

  • $\tan 60^\circ = \sqrt{3}$

Graphs of Trigonometric Functions

9. The period of the sine function $y = \sin x$ is:

1. $\pi$

2. $2\pi$

3. $\frac{\pi}{2}$

4. $4\pi$

Show me the answer

Answer: 2. $2\pi$

Explanation:

• The period of a trigonometric function is the horizontal distance over which the graph completes one full cycle.

• For $y = \sin x$ and $y = \cos x$, the period is $2\pi$.

• For $y = \tan x$ and $y = \cot x$, the period is $\pi$.

• For a function $y = \sin(kx)$, the period is $\frac{2\pi}{|k|}$.

10. The amplitude of the function $y = 3\sin x$ is:

1. 1

2. 3

3. $2\pi$

4. $\frac{1}{3}$

Show me the answer

Answer: 2. 3

Explanation:

• The amplitude of a trigonometric function is half the distance between its maximum and minimum values.

• For $y = A \sin x$ or $y = A \cos x$, the amplitude is $|A|$.

• Here, $A = 3$, so the amplitude is 3.

• The graph oscillates between -3 and 3.

Inverse Trigonometric Functions

11. The principal value range of $\sin^{-1} x$ is:

1. $[0, \pi]$

2. $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

3. $[0, \pi]$ (excluding $\frac{\pi}{2}$)

4. $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ excluding 0

Show me the answer

Answer: 2. $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

Explanation:

• To make trigonometric functions invertible, we restrict their domains:

  • $\sin^{-1} x$ has domain $[-1, 1]$ and range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

  • $\cos^{-1} x$ has domain $[-1, 1]$ and range $[0, \pi]$.

  • $\tan^{-1} x$ has domain $(-\infty, \infty)$ and range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

12. $\sin^{-1}\left(\sin\left(\frac{2\pi}{3}\right)\right)$ equals:

1. $\frac{2\pi}{3}$

2. $\frac{\pi}{3}$

3. $-\frac{\pi}{3}$

4. $\pi$

Show me the answer

Answer: 2. $\frac{\pi}{3}$

Explanation:

• Since $\frac{2\pi}{3}$ is not in the principal range of $\sin^{-1}$ ($\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$), we find an equivalent angle in that range.

• $\sin\left(\frac{2\pi}{3}\right) = \sin\left(\pi - \frac{2\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right)$

• $\frac{\pi}{3}$ is within the principal range.

• Therefore, $\sin^{-1}\left(\sin\left(\frac{2\pi}{3}\right)\right) = \sin^{-1}\left(\sin\left(\frac{\pi}{3}\right)\right) = \frac{\pi}{3}$.

Trigonometric Equations

13. The general solution of $\sin x = 0$ is:

1. $x = n\pi$, where $n \in \mathbb{Z}$

2. $x = 2n\pi$, where $n \in \mathbb{Z}$

3. $x = \frac{\pi}{2} + n\pi$, where $n \in \mathbb{Z}$

4. $x = n\pi + (-1)^n \frac{\pi}{2}$, where $n \in \mathbb{Z}$

Show me the answer

Answer: 1. $x = n\pi$, where $n \in \mathbb{Z}$

Explanation:

• $\sin x = 0$ when $x$ is an integer multiple of $\pi$.

• The solutions occur at $x = 0, \pm\pi, \pm 2\pi, \ldots$, which can be written as $x = n\pi$ for any integer $n$.

• Compare with:

  • $\cos x = 0 \Rightarrow x = \frac{\pi}{2} + n\pi$

  • $\sin x = 1 \Rightarrow x = \frac{\pi}{2} + 2n\pi$

  • $\cos x = 1 \Rightarrow x = 2n\pi$

14. The general solution of $\cos x = \frac{1}{2}$ is:

1. $x = \frac{\pi}{3} + 2n\pi$

2. $x = \pm \frac{\pi}{3} + 2n\pi$, where $n \in \mathbb{Z}$

3. $x = \frac{\pi}{6} + 2n\pi$

4. $x = \pm \frac{\pi}{6} + 2n\pi$, where $n \in \mathbb{Z}$

Show me the answer

Answer: 2. $x = \pm \frac{\pi}{3} + 2n\pi$, where $n \in \mathbb{Z}$

Explanation:

• The principal solutions of $\cos x = \frac{1}{2}$ are $x = \frac{\pi}{3}$ and $x = -\frac{\pi}{3}$ (or $\frac{5\pi}{3}$).

• Since cosine has period $2\pi$, we add $2n\pi$ to each principal solution.

• The general solution is: $x = 2n\pi \pm \frac{\pi}{3}$, for any integer $n$.

Height and Distance Applications

15. The angle of elevation of the top of a tower from a point 20 meters away from its base is 45°. The height of the tower is:

1. 10 meters

2. 20 meters

3. 40 meters

4. $20\sqrt{3}$ meters

Show me the answer

Answer: 2. 20 meters

Explanation:

• Let height be $h$.

• Distance from base = 20 m.

• Angle of elevation = 45°.

• $\tan 45^\circ = \frac{h}{20}$.

• Since $\tan 45^\circ = 1$, we have $h = 20$ meters.

16. A ladder leaning against a wall makes a 60° angle with the ground. If the foot of the ladder is 5 meters from the wall, the length of the ladder is:

1. 5 meters

2. 10 meters

3. $5\sqrt{3}$ meters

4. $\frac{10}{\sqrt{3}}$ meters

Show me the answer

Answer: 2. 10 meters

Explanation:

• Let ladder length be $L$.

• Distance from wall = 5 m (adjacent side to 60° angle).

• $\cos 60^\circ = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{5}{L}$.

• $\cos 60^\circ = \frac{1}{2}$, so $\frac{1}{2} = \frac{5}{L} \Rightarrow L = 10$ meters.

Trigonometric Ratios of Sum and Difference

17. $\sin(A + B)$ equals:

1. $\sin A \cos B + \cos A \sin B$

2. $\sin A \cos B - \cos A \sin B$

3. $\cos A \cos B - \sin A \sin B$

4. $\cos A \cos B + \sin A \sin B$

Show me the answer

Answer: 1. $\sin A \cos B + \cos A \sin B$

Explanation:

• Sum and difference formulas:

  • $\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$

  • $\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$

  • $\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$

18. The value of $\cos 75^\circ$ using sum/difference formulas is:

1. $\frac{\sqrt{6} + \sqrt{2}}{4}$

2. $\frac{\sqrt{6} - \sqrt{2}}{4}$

3. $\frac{\sqrt{3} + 1}{2\sqrt{2}}$

4. $\frac{\sqrt{3} - 1}{2\sqrt{2}}$

Show me the answer

Answer: 2. $\frac{\sqrt{6} - \sqrt{2}}{4}$

Explanation:

• $\cos 75^\circ = \cos(45^\circ + 30^\circ)$

• Using formula: $\cos(45^\circ + 30^\circ) = \cos 45^\circ \cos 30^\circ - \sin 45^\circ \sin 30^\circ$

• $= \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right)$

• $= \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} - 1}{2\sqrt{2}} = \frac{\sqrt{6} - \sqrt{2}}{4}$ (after rationalizing).

Transformation Formulas

19. The product $2 \sin A \cos B$ can be expressed as:

1. $\sin(A + B) + \sin(A - B)$

2. $\sin(A + B) - \sin(A - B)$

3. $\cos(A + B) + \cos(A - B)$

4. $\cos(A + B) - \cos(A - B)$

Show me the answer

Answer: 1. $\sin(A + B) + \sin(A - B)$

Explanation:

• Product-to-sum formulas:

  • $2 \sin A \cos B = \sin(A + B) + \sin(A - B)$

  • $2 \cos A \sin B = \sin(A + B) - \sin(A - B)$

  • $2 \cos A \cos B = \cos(A + B) + \cos(A - B)$

  • $2 \sin A \sin B = \cos(A - B) - \cos(A + B)$

20. The sum $\sin C + \sin D$ can be expressed as:

1. $2 \sin\left(\frac{C + D}{2}\right) \cos\left(\frac{C - D}{2}\right)$

2. $2 \cos\left(\frac{C + D}{2}\right) \sin\left(\frac{C - D}{2}\right)$

3. $2 \cos\left(\frac{C + D}{2}\right) \cos\left(\frac{C - D}{2}\right)$

4. $2 \sin\left(\frac{C + D}{2}\right) \sin\left(\frac{C - D}{2}\right)$

Show me the answer

Answer: 1. $2 \sin\left(\frac{C + D}{2}\right) \cos\left(\frac{C - D}{2}\right)$

Explanation:

• Sum-to-product formulas:

  • $\sin C + \sin D = 2 \sin\left(\frac{C + D}{2}\right) \cos\left(\frac{C - D}{2}\right)$

  • $\sin C - \sin D = 2 \cos\left(\frac{C + D}{2}\right) \sin\left(\frac{C - D}{2}\right)$

  • $\cos C + \cos D = 2 \cos\left(\frac{C + D}{2}\right) \cos\left(\frac{C - D}{2}\right)$

  • $\cos C - \cos D = -2 \sin\left(\frac{C + D}{2}\right) \sin\left(\frac{C - D}{2}\right)$

Periodicity and Symmetry

21. Which of the following is true for all $x$?

1. $\sin(-x) = \sin x$

2. $\cos(-x) = -\cos x$

3. $\tan(-x) = -\tan x$

4. $\csc(-x) = \csc x$

Show me the answer

Answer: 3. $\tan(-x) = -\tan x$

Explanation:

• Trigonometric functions have specific parity (even/odd properties):

  • Odd functions: $\sin(-x) = -\sin x$, $\tan(-x) = -\tan x$, $\csc(-x) = -\csc x$, $\cot(-x) = -\cot x$

  • Even functions: $\cos(-x) = \cos x$, $\sec(-x) = \sec x$

22. $\sin(\pi + x)$ equals:

1. $\sin x$

2. $-\sin x$

3. $\cos x$

4. $-\cos x$

Show me the answer

Answer: 2. $-\sin x$

Explanation:

• Trigonometric functions of $\pi \pm x$:

  • $\sin(\pi \pm x) = \mp \sin x$

  • $\cos(\pi \pm x) = -\cos x$

• Specifically:

  • $\sin(\pi + x) = -\sin x$

  • $\sin(\pi - x) = \sin x$

  • $\cos(\pi + x) = -\cos x$

  • $\cos(\pi - x) = -\cos x$

Solving Triangles

23. In triangle ABC, the Law of Sines states that:

1. $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

2. $\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$

3. $a^2 = b^2 + c^2 - 2bc \cos A$

4. Both 1 and 2 are equivalent forms

Show me the answer

Answer: 4. Both 1 and 2 are equivalent forms

Explanation:

• The Law of Sines: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$, where R is the circumradius.

• Equivalently: $\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$.

• It is used when we know:

  • Two angles and one side (AAS or ASA), or

  • Two sides and a non-included angle (SSA - ambiguous case).

24. In triangle ABC, if $a = 7$, $b = 5$, and $\angle C = 60^\circ$, then side $c$ is:

1. $\sqrt{39}$

2. $\sqrt{29}$

3. $\sqrt{19}$

4. $\sqrt{49}$

Show me the answer

Answer: 1. $\sqrt{39}$

Explanation:

• Use the Law of Cosines: $c^2 = a^2 + b^2 - 2ab \cos C$

• Substitute: $c^2 = 7^2 + 5^2 - 2(7)(5)\cos 60^\circ$

• $c^2 = 49 + 25 - 70 \times \frac{1}{2}$

• $c^2 = 74 - 35 = 39$

• $c = \sqrt{39}$

Trigonometric Limits

25. $\lim_{x \to 0} \frac{\sin x}{x}$ equals:

1. 0

2. 1

3. $\infty$

4. Does not exist

Show me the answer

Answer: 2. 1

Explanation:

• This is a fundamental trigonometric limit: $\lim_{x \to 0} \frac{\sin x}{x} = 1$.

• Similarly: $\lim_{x \to 0} \frac{\tan x}{x} = 1$.

• Also: $\lim_{x \to 0} \frac{1 - \cos x}{x} = 0$, but $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$.

Trigonometric Inequalities

26. The solution set of $\sin x > 0$ in $[0, 2\pi]$ is:

1. $(0, \pi)$

2. $\left(0, \frac{\pi}{2}\right) \cup \left(\frac{3\pi}{2}, 2\pi\right)$

3. $(0, \pi) \setminus \{\pi\}$

4. $\left(0, \frac{\pi}{2}\right)$

Show me the answer

Answer: 1. $(0, \pi)$

Explanation:

• Sine is positive in Quadrants I and II.

• In $[0, 2\pi]$:

  • $\sin x = 0$ at $x = 0, \pi, 2\pi$

  • $\sin x > 0$ for $0 < x < \pi$

• Therefore, the solution is $(0, \pi)$.

Maximum and Minimum Values

27. The maximum value of $3\sin x + 4\cos x$ is:

1. 3

2. 4

3. 5

4. 7

Show me the answer

Answer: 3. 5

Explanation:

• Expressions of the form $a\sin x + b\cos x$ can be written as $R\sin(x + \alpha)$ where $R = \sqrt{a^2 + b^2}$.

• Here, $a = 3$, $b = 4$, so $R = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.

• The maximum value is $R = 5$, and the minimum is $-R = -5$.

Area of Triangle

28. The area of triangle ABC with sides $a$, $b$ and included angle $C$ is:

1. $\frac{1}{2} ab \sin C$

2. $\frac{1}{2} bc \sin A$

3. $\frac{1}{2} ac \sin B$

4. All of the above

Show me the answer

Answer: 4. All of the above

Explanation:

• The area of a triangle can be calculated using different combinations:

  • $\text{Area} = \frac{1}{2} ab \sin C$

  • $\text{Area} = \frac{1}{2} bc \sin A$

  • $\text{Area} = \frac{1}{2} ac \sin B$

• This formula is derived from the fact that the height to side $a$ is $b \sin C$.

Trigonometric Series

29. $\cos \alpha + \cos(\alpha + \beta) + \cos(\alpha + 2\beta) + \ldots + \cos(\alpha + (n-1)\beta)$ equals:

1. $\frac{\sin\left(\frac{n\beta}{2}\right)}{\sin\left(\frac{\beta}{2}\right)} \cos\left(\alpha + \frac{(n-1)\beta}{2}\right)$

2. $\frac{\sin\left(\frac{n\beta}{2}\right)}{\sin\left(\frac{\beta}{2}\right)} \sin\left(\alpha + \frac{(n-1)\beta}{2}\right)$

3. $\frac{\cos\left(\frac{n\beta}{2}\right)}{\cos\left(\frac{\beta}{2}\right)} \cos\left(\alpha + \frac{(n-1)\beta}{2}\right)$

4. $\frac{\cos\left(\frac{n\beta}{2}\right)}{\cos\left(\frac{\beta}{2}\right)} \sin\left(\alpha + \frac{(n-1)\beta}{2}\right)$

Show me the answer

Answer: 1. $\frac{\sin\left(\frac{n\beta}{2}\right)}{\sin\left(\frac{\beta}{2}\right)} \cos\left(\alpha + \frac{(n-1)\beta}{2}\right)$

Explanation:

• This is the formula for the sum of a cosine series in arithmetic progression.

• The corresponding sine series sum is: $\sin \alpha + \sin(\alpha + \beta) + \ldots + \sin(\alpha + (n-1)\beta) = \frac{\sin\left(\frac{n\beta}{2}\right)}{\sin\left(\frac{\beta}{2}\right)} \sin\left(\alpha + \frac{(n-1)\beta}{2}\right)$

Miscellaneous

30. If $\sin x + \csc x = 2$, then $\sin^n x + \csc^n x$ equals:

1. 2

2. $2^n$

3. 1

4. $n$

Show me the answer

Answer: 1. 2

Explanation:

• Given: $\sin x + \frac{1}{\sin x} = 2$

• Let $t = \sin x$, then $t + \frac{1}{t} = 2$

• Multiply by $t$: $t^2 + 1 = 2t \Rightarrow t^2 - 2t + 1 = 0 \Rightarrow (t-1)^2 = 0$

• So $t = 1$, meaning $\sin x = 1$, and $\csc x = 1$

• Therefore, $\sin^n x + \csc^n x = 1^n + 1^n = 1 + 1 = 2$ for any positive integer $n$.