Q.Find the value of $\tan\frac{13\pi}{12}$.

$\cos\dfrac{5\pi}{3} = \dfrac12$.

Write $\dfrac{5\pi}{3} = 2\pi - \dfrac{\pi}{3}$. This angle lies in the fourth quadrant, where cosine is positive.

Using $\cos(2\pi-\theta) = \cos\theta$:

$\cos\dfrac{5\pi}{3} = \cos\left(2\pi-\dfrac{\pi}{3}\right) = \cos\dfrac{\pi}{3} = \dfrac12$

$\text{cosec}(-1410^\circ) = 2$.

Since cosec is an odd function, $\text{cosec}(-1410^\circ) = -\text{cosec}(1410^\circ)$.

Reduce $1410^\circ$ modulo $360^\circ$: $1410^\circ - 3\times360^\circ = 1410^\circ - 1080^\circ = 330^\circ$. So $\text{cosec}(1410^\circ) = \text{cosec}(330^\circ)$.

$\sin(330^\circ) = \sin(360^\circ - 30^\circ) = -\sin(30^\circ) = -\dfrac{1}{2}$, so $\text{cosec}(330^\circ) = \dfrac{1}{\sin(330^\circ)} = -2$.

Therefore $\text{cosec}(-1410^\circ) = -(-2) = 2$.

$\sin\dfrac{19\pi}{4} = \dfrac{1}{\sqrt2}$.

Since sine has period $2\pi = \dfrac{8\pi}{4}$, subtract full periods: $\dfrac{19\pi}{4} - 2\times2\pi = \dfrac{19\pi}{4} - \dfrac{16\pi}{4} = \dfrac{3\pi}{4}$.

So $\sin\dfrac{19\pi}{4} = \sin\dfrac{3\pi}{4} = \sin\left(\pi-\dfrac{\pi}{4}\right) = \sin\dfrac{\pi}{4} = \dfrac{1}{\sqrt2}$.

The value is $-\dfrac12$.

$\sin\dfrac{\pi}{6}=\dfrac12 \Rightarrow \sin^2\dfrac{\pi}{6}=\dfrac14$.

$\cos\dfrac{\pi}{3}=\dfrac12 \Rightarrow \cos^2\dfrac{\pi}{3}=\dfrac14$.

$\tan\dfrac{\pi}{4}=1 \Rightarrow \tan^2\dfrac{\pi}{4}=1$.

Sum: $\dfrac14+\dfrac14-1 = \dfrac12-1=-\dfrac12$.

Reducing the angle using the $2\pi$ periodicity of sine, $\sin\left(-\dfrac{11\pi}{3}\right)=\sin\dfrac{\pi}{3}=\dfrac{\sqrt3}{2}$.

Step 1: Reduce the angle.

$-\dfrac{11\pi}{3} + 4\pi = -\dfrac{11\pi}{3}+\dfrac{12\pi}{3} = \dfrac{\pi}{3}$

Since $4\pi$ is exactly two full periods of $2\pi$, and sine is $2\pi$-periodic,

$\sin\left(-\dfrac{11\pi}{3}\right) = \sin\left(-\dfrac{11\pi}{3}+4\pi\right) = \sin\dfrac{\pi}{3}$

Step 2: Evaluate.

$\sin\dfrac{\pi}{3} = \dfrac{\sqrt3}{2}$

Adding $4\pi$ (two full rotations) to $-\dfrac{11\pi}{3}$ gives the coterminal angle $\dfrac{\pi}{3}$, so the value equals $\sin\dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}$.

Since sine has period $2\pi$, we can add any integer multiple of $2\pi$ to the angle without changing its value.

$-\dfrac{11\pi}{3} + 4\pi = -\dfrac{11\pi}{3} + \dfrac{12\pi}{3} = \dfrac{\pi}{3}$

So $\sin\left(-\dfrac{11\pi}{3}\right) = \sin\dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}$.

(Check by the odd-function route: $\sin(-\theta) = -\sin\theta$, and $\dfrac{11\pi}{3} = 2\pi + \dfrac{5\pi}{3}$ lies in the fourth quadrant where sine is negative, so $\sin\dfrac{11\pi}{3} = -\sin\dfrac{\pi}{3} = -\dfrac{\sqrt3}{2}$, giving $\sin\left(-\dfrac{11\pi}{3}\right) = -\left(-\dfrac{\sqrt3}{2}\right) = \dfrac{\sqrt3}{2}$ — consistent.)