Question

Show that $\sin 600^\circ \cos 330^\circ + \cos 120^\circ \sin 150^\circ = - 1$

Answer 1

Hint: In this question we have to prove that the trigonometric expression given on both sides is equal. For that we are going to solve using trigonometric identities in angle and ratio. And also we are going to multiply and add the trigonometric identities in complete step-by-step solutions.
Trigonometric is a function that deals with the relationship between the sides and angles of triangles.

Formula used: In, trigonometric angle formulas, \[ = - (\sin 90^\circ )\]
There are six functions of an angle commonly used in trigonometry, they are sine, cosine, tangent, cosecant, secant, cotangent. In this sum we are going to see only the sine and cosine angle and ratio formula. The formulas are
 $\sin (n.360^\circ - \theta ) = - \sin \theta $
$\cos (n.360^\circ - \theta ) = \cos \theta $
$\sin (180^\circ - \theta ) = \sin \theta $
$\cos (180^\circ - \theta ) = - \cos \theta $
\[\sin 60^\circ = \dfrac{{\sqrt 3 }}{2}\]
\[\cos 30^\circ = \dfrac{{\sqrt 3 }}{2}\]
\[\sin 30^\circ = \dfrac{1}{2}\]
\[\cos 60^\circ = \dfrac{1}{2}\]

Complete step-by-step answer:
Let us consider the equation $\sin 600^\circ \cos 330^\circ + \cos 120^\circ \sin 150^\circ $
Here, we applying trigonometry angle formulas on the expression, the sine and cosine angle commonly known as $\sin $ and $\cos $,
$ \Rightarrow \sin 600^\circ = \sin (2.360^\circ - 60^\circ ) = - \sin 60^\circ - - - - - - - - - - \left( 1 \right)$
$ \Rightarrow \cos 330^\circ = \cos (360^\circ - 30^\circ ) = \cos 30^\circ - - - - - - - - - - \left( 2 \right)$
$ \Rightarrow \cos 120^\circ = \cos (180^\circ - 60^\circ ) = - \cos 60^\circ - - - - - - - - - - \left( 3 \right)$
$ \Rightarrow \sin 150^\circ = \sin (180^\circ - 30^\circ ) = \sin 30^\circ - - - - - - - - - - \left( 4 \right)$
Substitute the equation \[1,{\text{ }}2,{\text{ }}3\] \[\&\] \[4\] in the expression we get,
$ \Rightarrow ( - \sin 60^\circ \cos 30^\circ ) + ( - \cos 60^\circ \sin 30^\circ )$
$ \Rightarrow - \sin 60^\circ \cos 30^\circ - \cos 60^\circ \sin 30^\circ $
We know that, trigonometric ratios mentioned on the formula used
Substitute the values in the expression, we get
\[ \Rightarrow \left( { - \dfrac{{\sqrt 3 }}{2}} \right) \times \dfrac{{\sqrt 3 }}{2} \times \left( { - \dfrac{1}{2}} \right) \times \dfrac{1}{2}\]
By Multiply two same root numbers we obtain, natural number
$ \Rightarrow \sqrt 3 \times \sqrt 3 = 3$
Product of two negative number is positive number,
Applying fraction addition, we get
\[ \Rightarrow - \dfrac{3}{4} - \dfrac{1}{4}\]
Denominators are same in both terms of fractions,
\[ \Rightarrow \dfrac{{ - 3 - 1}}{4}\]
Subtracting the terms we get,
\[ \Rightarrow \dfrac{{ - 4}}{4}\]
\[ \Rightarrow - 1\]
$\therefore $Thus the value of $\sin 600^\circ \cos 330^\circ + \cos 120^\circ \sin 150^\circ = - 1$
Hence we have proved the given relation.

Note: Here it is another method to proving this problem,
Formulas used:
$\sin (x + y) = \sin x\cos y + \cos x\sin y$
$\sin 90^\circ = 1$
Other formulas are same as above method
Now, we solve the formulas in the trigonometric expression
Like above method similarly using identities in the expression
$ \Rightarrow - \sin 60^\circ \cos 30^\circ - \cos 60^\circ \sin 30^\circ $
Taking \[\left( - \right)\] commonly we get,
\[ \Rightarrow - (\sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ )\]
Applying, formulas used we get,
\[ \Rightarrow - (\sin 60^\circ + \sin 30^\circ )\]
By adding sin angles we only and angle values, we get
\[ \Rightarrow - (\sin 90^\circ )\]
Substitute values mentioned in formula used,
\[ \Rightarrow - \left( 1 \right)\]
\[ \Rightarrow - 1\]
$\therefore $LHS = RHS,
Hence we proved the relation $\sin 600^\circ \cos 330^\circ + \cos 120^\circ \sin 150^\circ = - 1$.