How do you prove \[\sin \left( {{180}^{\circ }}-a \right)=\sin a\]?
Answer
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Hint: In this type of question we have to use the basic simplification of trigonometric functions. Here, we have to use a simple trigonometric formula \[\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B\]. Also we have to use the values of the standard angles that are, \[\sin {{180}^{\circ }}=0\] and \[\cos {{180}^{\circ }}=\left( -1 \right)\]. In this question we first consider the LHS and then by using above mentioned formula and values we can obtain the required result.
Complete step-by-step solution:
Now, we have to prove that \[\sin \left( {{180}^{\circ }}-a \right)=\sin a\]
For this, let us take the LHS of the above equation
\[\Rightarrow LHS=\sin \left( {{180}^{\circ }}-a \right)\]
Now, to simplify this trigonometric expression we have to use the compound angle formula of sine.
Also we know that, \[\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B\] by applying it to LHS that is to \[\sin \left( {{180}^{\circ }}-a \right)\], we can write the LHS as
\[\Rightarrow LHS=\sin {{180}^{\circ }}\cos a-\cos {{180}^{\circ }}\sin a\]
Now, we know the values of \[\sin {{180}^{\circ }}\] and \[\cos {{180}^{\circ }}\] which are \[\sin {{180}^{\circ }}=0\] and \[\cos {{180}^{\circ }}=\left( -1 \right)\]. Hence, by substituting these values in the above expression we can write our LHS as
\[\Rightarrow LHS=\left( 0 \right)\cos a-\left( -1 \right)\sin a\]
By simplifying the trigonometric expression further we get,
\[\Rightarrow LHS=\sin a\]
Also by given \[RHS=\sin a\]
\[\Rightarrow LHS=RHS\]
Hence we have proved, \[\sin \left( {{180}^{\circ }}-a \right)=\sin a\]
Note: In this type of question students should remember the formulas of trigonometry. Students have to remember the basic algebraic rules and trigonometric identities which will help them in simplification of such trigonometric expressions. Students should have a good grip over the basic trigonometric identities and formulae. Also students have to note that the identity which we have proved in the given question is further used in many questions as a direct result and has a wide range of applications.
Complete step-by-step solution:
Now, we have to prove that \[\sin \left( {{180}^{\circ }}-a \right)=\sin a\]
For this, let us take the LHS of the above equation
\[\Rightarrow LHS=\sin \left( {{180}^{\circ }}-a \right)\]
Now, to simplify this trigonometric expression we have to use the compound angle formula of sine.
Also we know that, \[\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B\] by applying it to LHS that is to \[\sin \left( {{180}^{\circ }}-a \right)\], we can write the LHS as
\[\Rightarrow LHS=\sin {{180}^{\circ }}\cos a-\cos {{180}^{\circ }}\sin a\]
Now, we know the values of \[\sin {{180}^{\circ }}\] and \[\cos {{180}^{\circ }}\] which are \[\sin {{180}^{\circ }}=0\] and \[\cos {{180}^{\circ }}=\left( -1 \right)\]. Hence, by substituting these values in the above expression we can write our LHS as
\[\Rightarrow LHS=\left( 0 \right)\cos a-\left( -1 \right)\sin a\]
By simplifying the trigonometric expression further we get,
\[\Rightarrow LHS=\sin a\]
Also by given \[RHS=\sin a\]
\[\Rightarrow LHS=RHS\]
Hence we have proved, \[\sin \left( {{180}^{\circ }}-a \right)=\sin a\]
Note: In this type of question students should remember the formulas of trigonometry. Students have to remember the basic algebraic rules and trigonometric identities which will help them in simplification of such trigonometric expressions. Students should have a good grip over the basic trigonometric identities and formulae. Also students have to note that the identity which we have proved in the given question is further used in many questions as a direct result and has a wide range of applications.
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