Question

Prove the following:
$\cos \left( {\dfrac{{3\pi }}{4} + x} \right) - \cos \left( {\dfrac{{3\pi }}{4} - x} \right) = - \sqrt 2 \sin x$

Answer 1

Hint: We can expand the LHS of the given equation using the trigonometric identities $\cos \left( {A - B} \right) = \cos \left( A \right)\cos \left( B \right) + \sin \left( A \right)\sin \left( B \right)$and $\cos \left( {A + B} \right) = \cos \left( A \right)\cos \left( B \right) - \sin \left( A \right)\sin \left( B \right)$. On simplification and further calculations, we will obtain the RHS of the equation. We can say the equation is true when $LHS = RHS$

Complete step-by-step answer:
We need to prove that $\cos \left( {\dfrac{{3\pi }}{4} + x} \right) - \cos \left( {\dfrac{{3\pi }}{4} - x} \right) = - \sqrt 2 \sin x$
Let us look at the LHS.
$LHS = \cos \left( {\dfrac{{3\pi }}{4} + x} \right) - \cos \left( {\dfrac{{3\pi }}{4} - x} \right)$
It is of the form $\cos \left( {A + B} \right) - \cos \left( {A - B} \right)$ .
We know that $\cos \left( {A - B} \right) = \cos \left( A \right)\cos \left( B \right) + \sin \left( A \right)\sin \left( B \right)$and $\cos \left( {A + B} \right) = \cos \left( A \right)\cos \left( B \right) - \sin \left( A \right)\sin \left( B \right)$
 $ \Rightarrow \cos \left( {A + B} \right) - \cos \left( {A - B} \right) = \cos \left( A \right)\cos \left( B \right) - \sin \left( A \right)\sin \left( B \right) - \left( {\cos \left( A \right)\cos \left( B \right) + \sin \left( A \right)\sin \left( B \right)} \right)$
On further simplification, we get,
$ \Rightarrow \cos \left( {A + B} \right) - \cos \left( {A - B} \right) = - 2\sin \left( A \right)\sin \left( B \right)$
We can substitute the values,
\[ \Rightarrow \cos \left( {\dfrac{{3\pi }}{4} + x} \right) - \cos \left( {\dfrac{{3\pi }}{4} - x} \right) = - 2\sin \left( {\dfrac{{3\pi }}{4}} \right)\sin x\]
Then the LHS will become,
\[ \Rightarrow LHS = - 2\sin \left( {\dfrac{{3\pi }}{4}} \right)\sin x\]
We know that $\sin \dfrac{{3\pi }}{4} = \dfrac{1}{{\sqrt 2 }}$. On substituting, we get,
\[ \Rightarrow LHS = - 2\left( {\dfrac{1}{{\sqrt 2 }}} \right)\sin x\]
On simplification,
\[ \Rightarrow LHS = - \sqrt 2 \sin x\]
RHS is also equal to\[ - \sqrt 2 \sin x\]. So, we can write,
$LHS = RHS$.
Hence the equation is proved.

Note: We must be familiar with the following trigonometric identities used in this problem.
1. $\cos \left( {A \pm B} \right) = \cos \left( A \right)\cos \left( B \right) \mp \sin \left( A \right)\sin \left( B \right)$
2.$\cot \left( {A \pm B} \right) = \dfrac{{\cot A\cot B \mp 1}}{{\cot B \pm \cot A}}$
3.$\cos \left( {A + B} \right) - \cos \left( {A - B} \right) = - 2\sin \left( A \right)\sin \left( B \right)$
4.\[\cos \left( { - x} \right) = \cos \left( x \right)\]