Value of $\cos 15\pi$ is (a) 1 (b) 0 (c) -1 (d) 2

> > [!TLDR] > $\cos(n\pi) = (-1)^n$ for integer $n$; with $n=15$ (odd), $\cos(15\pi) = -1$. > …

Value of cos 15π is (a) 1 (b) 0 (c) -1 (d) 2

Question of 150

\cos 15\pi

(a) 1

(b) 0

(d) 2

Jharkhand JacJAC Intermediate Board (1st Year) 2025MCQ· 1mImportance★★★★★

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Concept understanding — Trigonometric Functions in Quadrants

Trigonometric Functions in Quadrants

Imagine standing at the centre of a circle, facing east. If you turn by some angle, you end up pointing in a certain direction. That direction has both a horizontal component (east-west) and a vertical component (north-south). Trigonometric functions are just a way to describe those components — and whether they are positive or negative depends entirely on which quadrant you're facing.

The Four Quadrants

The coordinate plane is split into four quadrants, numbered anticlockwise starting from the top-right:

Quadrant I (0° to 90°): x > 0, y > 0
Quadrant II (90° to 180°): x < 0, y > 0
Quadrant III (180° to 270°): x < 0, y < 0
Quadrant IV (270° to 360°): x > 0, y < 0

Now, recall the definitions on the unit circle (radius = 1):

$\cos \theta$ = x-coordinate of the point on the circle
$\sin \theta$ = y-coordinate of that point
$\tan \theta = \frac{\sin \theta}{\cos \theta}$

So the sign of $\cos \theta$ follows the sign of x, and the sign of $\sin \theta$ follows the sign of y. That's all there is to it.

The Sign Pattern

Quadrant	$\sin \theta$	$\cos \theta$	$\tan \theta$
I (0–90)	+	+	+
II (90–180)	+	–	–
III (180–270)	–	–	+
IV (270–360)	–	+	–

Tip

The mnemonic "All Students Take Coffee" helps you remember which functions are positive in each quadrant, starting from QI and going anticlockwise: All (all positive), Sine (sin positive), Tan (tan positive), Cos (cos positive).

Why This Matters

Suppose you're solving $\sin \theta = \frac{1}{2}$ . The calculator gives you $\theta = 30^\circ$ , but that's only one solution. Because sine is positive in both QI and QII, there's a second angle: $180^\circ - 30^\circ = 150^\circ$ . If you forget the quadrant rule, you lose half the answers.

Similarly, if $\cos \theta = -\frac{\sqrt{3}}{2}$ , cosine is negative in QII and QIII. So the solutions are $150^\circ$ and $210^\circ$ (plus full rotations). …

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$\cos(n\pi) = (-1)^n$ for integer $n$ ; with $n=15$ (odd), $\cos(15\pi) = -1$ .

…

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 54 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$\sin \dfrac{31\pi}{3}$
(a) $\dfrac{\sqrt{3}}{2}$
(b) $-\dfrac{\sqrt{3}}{2}$
(c) $\dfrac{1}{2}$
(d) $\dfrac{1}{\sqrt{2}}$
›Reveal solutionSolution

Subtract multiples of $2\pi$ (period of sine) until the angle lies in $[0, 2\pi)$ ; $31\pi/3$ reduces to $\pi/3$ .

Since sine has period $2\pi = \dfrac{6\pi}{3}$ , subtract multiples of $\dfrac{6\pi}{3}$ from $\dfrac{31\pi}{3}$ : …

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CBSE 2026Set ANNUAL1 markMCQ
$\tan x = \dfrac{3}{4}$
(a) $-\dfrac{1}{\sqrt{10}}$
(b) $\dfrac{1}{\sqrt{10}}$
(c) $\dfrac{3}{\sqrt{10}}$
(d) None of these
›Reveal solutionSolution

Find $\cos x$ from $\tan x$ (both sin, cos negative in Q3), then use the half-angle formula, choosing the sign from the quadrant of $x/2$ .

Given $\tan x = \dfrac{3}{4}$ with $\pi < x < \dfrac{3\pi}{2}$ ( $x$ in the third quadrant, where both $\sin x$ and $\cos x$ are negative). Using the 3-4-5 triangle: $\sin x = -\dfrac{3}{5}$ , $\cos x = -\dfrac{4}{5}$ .

Since $\pi < x < \dfrac{3\pi}{2}$ , dividing by 2: $\dfrac{\pi}{2} < \dfrac{x}{2} < \dfrac{3\pi}{4}$ — so $x/2$ lies in the second quadrant, where sine is positive.

…

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CBSE 2026Set ANNUAL1 markMCQ
$\tan \frac{19\pi}{3}$
(a) $\sqrt{3}$
(b) $-\sqrt{3}$
(c) $\frac{1}{\sqrt{3}}$
(d) $-\frac{1}{\sqrt{3}}$
›Reveal solutionSolution

$\tan\frac{19\pi}{3} = \sqrt{3}$ , option (a).

Since $\tan\theta$ has period $\pi$ , we can subtract integer multiples of $\pi$ from the angle without changing its value.

$\frac{19\pi}{3} - 6\pi = \frac{19\pi}{3} - \frac{18\pi}{3} = \frac{\pi}{3}$ …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. If tan x = -5/12 and x lies in 2nd quadrant, then the value of sin x is:$
(a) 5/13
(b) 12/13
(c) -5/13
(d) -12/13
›Reveal solutionSolution

In the 2nd quadrant sine is positive; using the 5-12-13 right triangle from tan x = -5/12 gives sin x = 5/13.

Given $\tan x = -\dfrac{5}{12}$ with $x$ in the 2nd quadrant.

In the 2nd quadrant: $\sin x > 0$ , $\cos x < 0$ , and $\tan x = \dfrac{\sin x}{\cos x} < 0$ — consistent with the given negative value.

Treat $5$ and $12$ as the magnitudes of the opposite and adjacent sides; the hypotenuse is $\sqrt{5^2+12^2} = \sqrt{25+144} = \sqrt{169} = 13$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$\sin x$
(a) Both Assertion (A) and Reason (R) are correct and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are correct, but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is correct, but Reason (R) is incorrect.
(d) Assertion (A) is incorrect, but Reason (R) is correct.
›Reveal solutionSolution

Assertion (A) is true, but Reason (R) is false — $\sin x$ increases, not decreases, on $[0,\pi/2]$ — so (R) cannot even be a valid explanation of (A).

Checking Assertion (A): In the third quadrant ( $\pi < x < \tfrac{3\pi}{2}$ ) and fourth quadrant ( $\tfrac{3\pi}{2} < x < 2\pi$ ), the value of $\sin x$ is indeed negative (sine is positive only in the first and second quadrants). So (A) is TRUE.

…

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CBSE 2026Set ANNUAL1 markMCQ
$\sin(-x)$
(a) $\dfrac{1-\tan^2 x}{1+\tan^2 x}$
(b) $\dfrac{2\tan x}{1-\tan^2 x}$
(c) $\dfrac{2\tan x}{1+\tan^2 x}$
(d) $-\sin x$
(e) $\sin x$
(f) $-\cos x$
(g) $\cos x$
›Reveal solutionSolution

Sine is an odd function: $\sin(-x)=-\sin x$ for all $x$ .

This follows directly from the standard trigonometric identity for negative angles, provable from the unit-circle definition: reflecting the angle across the x-axis flips the sign of the y-coordi …

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CBSE 2026Set ANNUAL1 markMCQ
$\cos\left(\dfrac{\pi}{2} - x\right)$
(a) $\dfrac{1-\tan^2 x}{1+\tan^2 x}$
(b) $\dfrac{2\tan x}{1-\tan^2 x}$
(c) $\dfrac{2\tan x}{1+\tan^2 x}$
(d) $-\sin x$
(e) $\sin x$
(f) $-\cos x$
(g) $\cos x$
›Reveal solutionSolution

By the co-function identity, $\cos\left(\dfrac{\pi}{2}-x\right)=\sin x$ .

This is a standard complementary-angle identity: the cosine of $90^\circ$ (i.e. $\tfrac{\pi}{2}$ ) minus an angle equals the sine of that angle, following from the right-trian …

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CBSE 2026Set ANNUAL1 markMCQ
$\cos(\pi - x)$
(a) $\dfrac{1-\tan^2 x}{1+\tan^2 x}$
(b) $\dfrac{2\tan x}{1-\tan^2 x}$
(c) $\dfrac{2\tan x}{1+\tan^2 x}$
(d) $-\sin x$
(e) $\sin x$
(f) $-\cos x$
(g) $\cos x$
›Reveal solutionSolution

By the supplementary-angle identity, $\cos(\pi-x)=-\cos x$ .

This follows since $\pi-x$ lies in the second quadrant when $x$ is a first-quadrant angle, where cosine is negative, and the reference angle is $x$ itself, …

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CBSE 2026Set 1A1 mark
$\sin\left(-\frac{11\pi}{3}\right)$
›Reveal solutionSolution

$-\tfrac{11\pi}{3}+4\pi=\tfrac{\pi}{3}$ , so the value is $\sin\tfrac{\pi}{3}=\tfrac{\sqrt3}{2}$ .

Using periodicity $\sin\theta=\sin(\theta+2\pi k)$ :

$-\frac{11\pi}{3}+4\pi=-\frac{11\pi}{3}+\frac{12\pi}{3}=\frac{\pi}{3}.$ …

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CBSE 2025Set ANNUAL1 markMCQ
$\sin \dfrac{3\pi}{2} =$
(a) 0
(b) -1
(c) 1
(d) 1/2
›Reveal solutionSolution

$\sin\dfrac{3\pi}{2} = -1$ .

$\dfrac{3\pi}{2}$ radians $= 270°$ . On the unit circle, $270°$ corresponds to the point $(0, -1)$ on the negative $y$ -axis, where $\sin\theta$ equal …

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CBSE 2025Set ANNUAL1 markMCQ
$\sin(\pi + x) =$
(a) $\cos x$
(b) $-\cos x$
(c) $\sin x$
(d) $-\sin x$
›Reveal solutionSolution

$\sin(\pi + x) = -\sin x$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$\cos \dfrac{5\pi}{2} =$
(a) 0
(b) 1
(c) -1
(d) 1/2
›Reveal solutionSolution

$\cos\dfrac{5\pi}{2} = 0$ .

Since cosine has period $2\pi$ : $\dfrac{5\pi}{2} - 2\pi = \dfrac{5\pi}{2} - \dfrac{4\pi}{2} = \dfrac{\pi}{2}$ .

…

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Q.Value of cos⁡15π\cos 15\picos15π is

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