Q.$\cos\dfrac{2\pi}{15}\,\cos\dfrac{4\pi}{15}\,\cos\dfrac{8\pi}{15}\,\cos\dfrac{16\pi}{15} = \dfrac{1}{16}$.

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

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). …