Question

Question−​

$\sf \: Solve : \: {sin}^{ - 1} \bigg( cos\dfrac{6\pi}{5} \bigg)$​

Answer 1

Answer

Thermodynamic equilibrium is an axiomatic concept of thermodynamics. ... In thermodynamic equilibrium there are no net macroscopic flows of matter or of energy, either within a system or between systems. In a system that is in its own state of internal thermodynamic equilibrium, no macroscopic change occurs.

Answer 2

$\large\underline{\bold{Given \:Question - }}$

$\sf \: Solve : \: {sin}^{ - 1} \bigg( cos\dfrac{6\pi}{5} \bigg)$

$\large\underline{\bf{Solution-}}$

We know, that

$\boxed{ \bf \: cos(2\pi \: - x) = cosx}$

$\boxed{ \bf \: {sin}^{ - 1} (sinx) = x \: \: if \: x \in \: \bigg[-\dfrac{\pi}{2}\:, \dfrac{\pi}{2}\bigg]}$

$\boxed{ \bf \: \: {cos} \bigg( \dfrac{\pi}{2} - x \bigg) = sinx}$

$\boxed{ \bf \: {sin}^{ - 1} ( - x) = - {sin}^{ - 1} x}$

Let's solve the problem now!!

Consider,

$\rm :\longmapsto\:{sin}^{ - 1} \bigg( cos\dfrac{6\pi}{5} \bigg)$

$\: \: \sf \: = \: \: {sin}^{ - 1} \bigg( cos \bigg(2\pi \: - \: \dfrac{4\pi}{5} \bigg) \bigg)$

$\: \: \sf \: = \: \: {sin}^{ - 1} \bigg( cos\dfrac{4\pi}{5} \bigg)$

$\: \: \sf \: = \: \: {sin}^{ - 1} \bigg( sin \bigg(\dfrac{\pi}{2} - \dfrac{4\pi}{5} \bigg) \bigg)$

$\: \: \sf \: = \: \: {sin}^{ - 1} \bigg( sin \bigg(\dfrac{5\pi \: - \: 8\pi}{10} \bigg) \bigg)$

$\: \: \sf \: = \: \: {sin}^{ - 1} \bigg( sin \bigg(\dfrac{ - \: 3\pi}{10} \bigg) \bigg)$

$\: \: \sf \: = \: \: {sin}^{ - 1} \bigg( - sin \bigg(\dfrac{\: 3\pi}{10} \bigg) \bigg)$

$\: \: \sf \: = \: - \: {sin}^{ - 1} \bigg(sin \bigg(\dfrac{\: 3\pi}{10} \bigg) \bigg)$

$\: \: \sf \: = \: \: - \: \dfrac{3\pi}{10}$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf function & \bf Range \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf {sin}^{ - 1} x &\bigg[-\dfrac{\pi}{2}\:, \dfrac{\pi}{2}\bigg] \sf \\ \\ \sf {cos}^{ - 1}x & \sf [0 \: , \pi] \\ \\ \sf {tan}^{ - 1}x & \sf \bigg( - \dfrac{\pi}{2} , \dfrac{\pi}{2} \bigg) \\ \\ \sf {cot}^{ - 1}x & \sf (0, \: \pi)\\ \\ \sf {sec}^{ - 1}x & \sf [0 \: , \: \pi] - \{\dfrac{\pi}{2} \} \\ \\ \sf {cosec}^{ - 1}x & \sf \bigg[-\dfrac{\pi}{2}\:, \dfrac{\pi}{2}\bigg] - \{0 \} \end{array}} \\ \end{gathered}\end{gathered}$