Question

Prove the following:
$\sin \left( {n + 1} \right)x\sin \left( {n + 2} \right)x + \cos \left( {n + 1} \right)x\cos \left( {n + 2} \right)x = \cos x$

Answer 1

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

Complete step-by-step answer:
We need to prove $\sin \left( {n + 1} \right)x\sin \left( {n + 2} \right)x + \cos \left( {n + 1} \right)x\cos \left( {n + 2} \right)x = \cos x$
Let us look at the LHS.
$LHS = \sin \left( {n + 1} \right)x\sin \left( {n + 2} \right)x + \cos \left( {n + 1} \right)x\cos \left( {n + 2} \right)x$
It is of the form $\sin A\sin B + \cos A\cos B$ where $A = \left( {n + 1} \right)x$ and $B = \left( {n + 2} \right)x$
We know that $\cos \left( {A - B} \right) = \cos \left( A \right)\cos \left( B \right) + \sin \left( A \right)\sin \left( B \right)$
We can substitute the values,
 $ \Rightarrow \sin \left( {n + 1} \right)x\sin \left( {n + 2} \right)x + \cos \left( {n + 1} \right)x\cos \left( {n + 2} \right)x = \cos \left( {\left( {n + 1} \right)x - \left( {n + 2} \right)x} \right)$
Then the LHS becomes,
\[ \Rightarrow LHS = \cos \left( {\left( {n + 1} \right)x - \left( {n + 2} \right)x} \right)\]
We can simplify of the terms inside the cos function
On doing the multiplication, we get,
\[LHS = \cos \left( {\left( {nx + x} \right) - \left( {nx + 2x} \right)} \right)\]
Opening the brackets, we get,
\[LHS = \cos \left( {nx + x - nx - 2x} \right)\]
After simplification, we get,
\[LHS = \cos \left( { - x} \right)\]
We know that \[\cos \left( { - x} \right) = \cos \left( x \right)\]
$ \Rightarrow LHS = \cos \left( x \right)$.
RHS is also equal to\[\cos x\]. 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.${\sin ^2}x + {\cos ^2}x = 1$
\[\cos \left( { - x} \right) = \cos \left( x \right)\]
We must know the values of trigonometric functions at common angles. Adding $\pi $ or multiples of $\pi $ with the angle retains the ratio and adding $\dfrac{\pi }{2}$ or odd multiples of $\dfrac{\pi }{2}$ will change the ratio.