Question

Prove that: $\sin 600{}^\circ \cos 390{}^\circ +\cos 480{}^\circ \sin 150{}^\circ =-1$.

Answer 1

Hint: For solving this problem, first converting the angle given in the problem statement in the range 0 to 90 degree by using suitable trigonometric properties. Now, we use the trigonometric table of values to prove the equivalence of both the sides.

Complete step-by-step answer:
Some of the important trigonometric formulas used in solving this problem are:
\[\begin{align}
  & \sin \left( 360-\theta \right)=-\sin \theta \\
 & \cos \left( 360+\theta \right)=\cos \theta \\
 & \cos \left( 180-\theta \right)=-\cos \theta \\
 & \sin \left( 180-\theta \right)=\sin \theta \\
\end{align}\]
The specific value of functions of sin and cos can be illustrated as:
$\begin{align}
  & \sin 30{}^\circ =\dfrac{1}{2} \\
 & \sin 60{}^\circ =\dfrac{\sqrt{3}}{2} \\
 & \cos 30{}^\circ =\dfrac{\sqrt{3}}{2} \\
 & \cos 60{}^\circ =\dfrac{1}{2} \\
\end{align}$
According to the problem statement, we consider the left-hand side of the equation for proving equivalence of both sides. First, we convert the $\sin 600{}^\circ ,\cos 390{}^\circ \text{ and }\cos 480{}^\circ $ by using the formula \[\sin \left( 360-\theta \right)=-\sin \theta \ \text{ and }\cos \left( 360+\theta \right)=\cos \theta \].
$\begin{align}
  & \sin 600{}^\circ =\sin \left( 2\times 360-120 \right){}^\circ \\
 & \therefore \sin 600{}^\circ =-\sin \left( 120 \right){}^\circ \\
 & \cos 390{}^\circ =\cos \left( 360+30 \right){}^\circ \\
 & \therefore \cos 390{}^\circ =\cos \left( 30 \right){}^\circ \\
 & \cos 480{}^\circ =\cos \left( 360+120 \right){}^\circ \\
 & \therefore \cos 480{}^\circ =\cos \left( 120 \right){}^\circ \\
\end{align}$
On replacing the above obtained values in the left-hand side, the required expression reduces to: $\Rightarrow \left( -\sin 120{}^\circ \right)\left( \cos 30{}^\circ \right)+\left( \cos 120{}^\circ \right)\left( \sin 150{}^\circ \right)$
Now, we try to convert the above functions into the functions given in the form of tables having respective angles between the range of 0 to 90 in degrees. For doing so, we use the formulas \[\cos \left( 180-\theta \right)=-\cos \theta \text{ and }\sin \left( 180-\theta \right)=\sin \theta \]. Now, we get
$\begin{align}
  & \sin 120{}^\circ =\sin \left( 180-60 \right){}^\circ \\
 & \therefore \sin 120{}^\circ =\sin \left( 60 \right){}^\circ \\
 & \cos 120{}^\circ =\cos \left( 180-60 \right){}^\circ \\
 & \therefore \cos 120{}^\circ =-\cos \left( 60 \right){}^\circ \\
 & \sin 150{}^\circ =\sin \left( 180-30 \right){}^\circ \\
 & \therefore \sin 150{}^\circ =\sin \left( 30 \right){}^\circ \\
\end{align}$
On replacing the above obtained values in the left-hand side, the required expression reduces to: $\Rightarrow -\sin 60{}^\circ \cos 30{}^\circ -\cos 60{}^\circ \sin 30{}^\circ $
Now, putting the values from the table mentioned above, we get
$\begin{align}
  & \Rightarrow -\dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}-\dfrac{1}{2}\times \dfrac{1}{2} \\
 & \Rightarrow -\dfrac{3}{4}-\dfrac{1}{4} \\
 & \Rightarrow \dfrac{-4}{4}=-1 \\
\end{align}$
Hence, we proved the equivalence of both sides by considering the expression of the left side.

Note: Students must remember the trigonometric table and the trigonometric formulas associated with different functions. The conversion of the respective function should be done carefully, and the magnitude of the required quantity must be copied correctly in the final expression for avoiding calculation error.