# 6.1 MCQs-Trigonometry *** ## 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°): **A**ll ratios are positive. * Quadrant II (90° to 180°): **S**ine and cosecant are positive. * Quadrant III (180° to 270°): **T**angent and cotangent are positive. * Quadrant IV (270° to 360°): **C**osine 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$$.