Question

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

Answer 1

Hint: We will first write the formulas we will require in this question related to trigonometric ratios. Now, we will rewrite the given expression, so that we can simplify it into such values whose trigonometric values we know.

Complete step-by-step answer:
Let us first write some formulas, which we will require:-
$\sin ({360^ \circ } + \theta ) = \sin \theta $ ……………(1)
$\sin ({180^ \circ } - \theta ) = \sin \theta $ ……………(2)
$\cos ({360^ \circ } - \theta ) = \cos \theta $ …………….(3)
$\cos ({180^ \circ } - \theta ) = - \cos \theta $ ………….(4)
$\sin ({180^ \circ } + \theta ) = - \sin \theta $ …………(5)
Now, coming back to the question.
Consider the LHS which is:- $\sin {600^ \circ }\cos {330^ \circ } + \cos {120^ \circ }\sin {150^ \circ }$ …………..(6)
Consider $\sin {600^ \circ }$, we can write it as: $\sin {600^ \circ } = \sin ({360^ \circ } + {240^ \circ })$
So, using (1), we get:- $\sin {600^ \circ } = \sin {240^ \circ }$
We can rewrite it as:- $\sin {600^ \circ } = \sin {240^ \circ } = \sin ({180^ \circ } + {60^ \circ })$
By using (5), we will get:- $\sin {600^ \circ } = \sin ({180^ \circ } + {60^ \circ }) = - \sin {60^ \circ }$
We know that $\sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2}$.
Hence, $\sin {600^ \circ } = - \sin {60^ \circ } = - \dfrac{{\sqrt 3 }}{2}$ ………..(7)
Now consider $\cos {330^ \circ }$, we can write it as: $\cos {330^ \circ } = \cos ({360^ \circ } - {30^ \circ })$
So, using (3), we get:- $\cos {330^ \circ } = \cos {30^ \circ }$
We know that $\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$.
Hence, $\cos {330^ \circ } = \cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$ ………..(8)
Now consider $\cos {120^ \circ }$, we can write it as: $\cos {120^ \circ } = \cos ({180^ \circ } - {60^ \circ })$
So, using (4), we get:- $\cos {120^ \circ } = - \cos {60^ \circ }$
We know that $\cos {60^ \circ } = \dfrac{1}{2}$.
Hence, $\cos {120^ \circ } = - \cos {60^ \circ } = - \dfrac{1}{2}$ ………..(9)
Now consider $\sin {150^ \circ }$, we can write it as: $\sin {150^ \circ } = \sin ({180^ \circ } - {30^ \circ })$
So, using (2), we get:- $\sin {150^ \circ } = \sin {30^ \circ }$
We know that $\sin {30^ \circ } = \dfrac{1}{2}$.
Hence, $\sin {150^ \circ } = \sin {30^ \circ } = \dfrac{1}{2}$ ………..(10)
Now, putting in (7), (8), (9), and (10) in (6), we will get:-
\[\sin {600^ \circ }\cos {330^ \circ } + \cos {120^ \circ }\sin {150^ \circ } = \left( { - \dfrac{{\sqrt 3 }}{2}} \right)\left( {\dfrac{{\sqrt 3 }}{2}} \right) + \left( { - \dfrac{1}{2}} \right)\left( {\dfrac{1}{2}} \right) = - \dfrac{3}{4} - \dfrac{1}{4} = - 1\]
Hence, proved.

Note: The students must not try to skip steps in between because that may create confusion and mistakes. If you see questions like this, always try to bring the functions in simplest form, whose value we already know and then just put in the values.
The word “Trigonometry” originated from the greek word “trigono” which means triangle and the word “metry” which means measure. It is a field of mathematics which develops a relationship between the sides of a triangle and its angles.
Trigonometry has numerous applications in the real life you would have/ will study in the chapter “Applications of Trigonometry”. It is used by architects to get a better result. It is used by air forces to understand the angles toward the land.