Question

If \[x=30{}^\circ \] verify that \[\sin x=\sqrt{\dfrac{1-\cos 2x}{2}}\].

Answer 1

Hint: The left-hand side and right-hand side are calculated by putting the value of x. The right-hand side can be obtained by using the value of x.
Formulas used:
The value of \[\sin 30{}^\circ =\dfrac{1}{2}\] and the value of \[\cos 60{}^\circ =\dfrac{1}{2}\].


Complete step-by-step answer:
First step will be considering \[\sin x\] as the left-hand side and put the value of x.
The right-hand side be \[\sqrt{\dfrac{1-\cos 2x}{2}}\] and put the value of x, and obtain the result.
The verification can be obtained as,
\[\begin{align}
  & \sin x=\sqrt{\dfrac{1-\cos 2x}{2}} \\
 & L.HS.=\sin x \\
 & L.HS.=\sin 30{}^\circ =\dfrac{1}{2} \\
 & R.H.S=\sqrt{\dfrac{1-\cos 2x}{2}} \\
 & R.H.S=\sqrt{\dfrac{1-\cos \left( 2\cdot 30 \right){}^\circ }{2}} \\
 & R.H.S=\sqrt{\dfrac{1-\cos 60{}^\circ }{2}} \\
 & R.H.S=\sqrt{\dfrac{1-\dfrac{1}{2}}{2}} \\
 & R.H.S=\sqrt{\dfrac{\dfrac{1}{2}}{2}} \\
 & R.H.S=\sqrt{\dfrac{1}{4}} \\
 & R.H.S=\dfrac{1}{2} \\
\end{align}\]
     \[Therefore,\text{ }L.H.S=R.H.S\text{ }\left[ verified \right]\].
Thus, the above expression is verified.

Note: The standard identities of trigonometric ratio have some special value for standard angle. The value of the trigonometric ratios is different for different angles.