# Find the value of $x$ if $\sin 2x = \sin {60^ \circ }\cos {30^ \circ } - \sin {30^ \circ }\cos {60^ \circ }$

A.${20^ \circ }$

B.${15^ \circ }$

C.${30^ \circ }$

D.${45^ \circ }$

Last updated date: 23rd Mar 2023

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Answer

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Hint : Use the trigonometric identity $\sin a\cos b - \cos a\sin b = \sin (a - b)$.

Given,

$\sin 2x = \sin {60^ \circ }\cos {30^ \circ } - \sin {30^ \circ }\cos {60^ \circ }$ …..(i)

As we know

$\sin a\cos b - \cos a\sin b = \sin (a - b)$

When we put $a = 60,b = 30$ we get the above equation after assigning value as,

$\sin {60^ \circ }\cos {30^ \circ } - \sin {30^ \circ }\cos {60^ \circ } = \sin ({60^ \circ } - {30^ \circ }) = \sin {30^ \circ }$ …(ii)

From (i) & (ii) We get,

$\sin {30^ \circ } = \sin 2x$

$

2x = {30^ \circ } \\

x = {15^ \circ } \\

$

Hence the correct option is B.

Note :- In these types of questions we have to apply the basic identities of trigonometry and solve the asked question. We can also assign values of the angles and find the asked value.

Given,

$\sin 2x = \sin {60^ \circ }\cos {30^ \circ } - \sin {30^ \circ }\cos {60^ \circ }$ …..(i)

As we know

$\sin a\cos b - \cos a\sin b = \sin (a - b)$

When we put $a = 60,b = 30$ we get the above equation after assigning value as,

$\sin {60^ \circ }\cos {30^ \circ } - \sin {30^ \circ }\cos {60^ \circ } = \sin ({60^ \circ } - {30^ \circ }) = \sin {30^ \circ }$ …(ii)

From (i) & (ii) We get,

$\sin {30^ \circ } = \sin 2x$

$

2x = {30^ \circ } \\

x = {15^ \circ } \\

$

Hence the correct option is B.

Note :- In these types of questions we have to apply the basic identities of trigonometry and solve the asked question. We can also assign values of the angles and find the asked value.

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