
How do you prove: \[\sin \left( {{{180}^ \circ } + a} \right) = - \sin a\] ?
Answer
531k+ views
Hint: The given question deals with basic simplification of trigonometric functions by using some of the simple trigonometric formulae such as $ \sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B $ . Basic algebraic rules and trigonometric identities are to be kept in mind while doing simplification in the given problem and proving the result given to us.
Complete step-by-step answer:
In the given problem, we have to prove a trigonometric equality that can be further used in many questions and problems as a direct result and has wide ranging applications. For proving the desired result, we need to have a good grip over the basic trigonometric formulae and identities.
Now, we need to make the left and right sides of the equation equal.
L.H.S. $ = \sin \left( {{{180}^ \circ } + a} \right) $
Now, we have to apply the compound angle formula of sine so as to simplify the trigonometric expression.
Now, we know that $ \sin \left( {A + B} \right) $ can be expanded as $ \left( {\sin A\cos B + \cos A\sin B} \right) $ .
So, we get the left side of the equation as
$ = \sin \left( {{{180}^ \circ } + a} \right) $
$ = \sin {180^ \circ }\cos a + \cos {180^ \circ }\sin a $
So, we know that the value of $ \sin {180^ \circ } $ is zero and $ \cos {180^ \circ } $ is $ \left( { - 1} \right) $ .
Now, we substitute the values of $ \sin {180^ \circ } $ and $ \cos {180^ \circ } $ , we get,
$ = \left( 0 \right)\cos a + \left( { - 1} \right)\sin a $
Simplifying the trigonometric expression further, we get,
$ = - \sin a $
Also, R.H.S $ = - \sin a $
As the left side of the equation is equal to the right side of the equation, we have,
\[\sin \left( {{{180}^ \circ } + a} \right) = - \sin a\]
Hence, Proved.
Note: Given problem deals with trigonometric functions. For solving such problems, trigonometric formulae should be remembered by heart. Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such types of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations.
Complete step-by-step answer:
In the given problem, we have to prove a trigonometric equality that can be further used in many questions and problems as a direct result and has wide ranging applications. For proving the desired result, we need to have a good grip over the basic trigonometric formulae and identities.
Now, we need to make the left and right sides of the equation equal.
L.H.S. $ = \sin \left( {{{180}^ \circ } + a} \right) $
Now, we have to apply the compound angle formula of sine so as to simplify the trigonometric expression.
Now, we know that $ \sin \left( {A + B} \right) $ can be expanded as $ \left( {\sin A\cos B + \cos A\sin B} \right) $ .
So, we get the left side of the equation as
$ = \sin \left( {{{180}^ \circ } + a} \right) $
$ = \sin {180^ \circ }\cos a + \cos {180^ \circ }\sin a $
So, we know that the value of $ \sin {180^ \circ } $ is zero and $ \cos {180^ \circ } $ is $ \left( { - 1} \right) $ .
Now, we substitute the values of $ \sin {180^ \circ } $ and $ \cos {180^ \circ } $ , we get,
$ = \left( 0 \right)\cos a + \left( { - 1} \right)\sin a $
Simplifying the trigonometric expression further, we get,
$ = - \sin a $
Also, R.H.S $ = - \sin a $
As the left side of the equation is equal to the right side of the equation, we have,
\[\sin \left( {{{180}^ \circ } + a} \right) = - \sin a\]
Hence, Proved.
Note: Given problem deals with trigonometric functions. For solving such problems, trigonometric formulae should be remembered by heart. Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such types of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations.
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