Question

Prove that \[\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}} = \dfrac{{\cos 2x}}{{1 + \sin 2x}}\]

Answer 1

Hint:The given question deals with proving trigonometric equality using the basic and simple trigonometric formulae and identities such as $\cos 2x = {\cos ^2}x - {\sin ^2}x$. So, we will first multiply the numerator and denominator of the left side of the result with $\cos x + \sin x$. Then, we will evaluate the whole square of $\cos x + \sin x$ in the denominator and will use the double angle formula for cosine in the numerator to get the required result.

Complete step by step answer:
Given, we have to prove \[\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}} = \dfrac{{\cos 2x}}{{1 + \sin 2x}}\]
We know, the formulas that,
\[\sin 2x = 2\sin x\cos x\]
\[\Rightarrow \cos 2x = {\cos ^2}x - {\sin ^2}x\]
\[\Rightarrow \tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}\],
Using these basic identities and some other methods we can prove the given question.

At first multiplying \[\cos x + \sin x\]to both numerator and denominator of left hand side, we get
\[\dfrac{{\left( {\cos x - \sin x} \right)\left( {\cos x + \sin x} \right)}}{{\left( {\cos x + \sin x} \right)\left( {\cos x + \sin x} \right)}}\]
Now, using the formula \[\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}\]and \[\left( {a + b} \right)\left( {a + b} \right) = {\left( {a + b} \right)^2}\], we get
\[\dfrac{{\left( {\cos x - \sin x} \right)\left( {\cos x + \sin x} \right)}}{{\left( {\cos x + \sin x} \right)\left( {\cos x + \sin x} \right)}}\]\[ = \dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\left( {\cos x + \sin x} \right)}^2}}}\]

Then using the formulas of double angle identities,
\[\cos 2x = {\cos ^2}x - {\sin ^2}x\], we get
\[\Rightarrow \dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\left( {\cos x + \sin x} \right)}^2}}} = \dfrac{{\cos 2x}}{{{{\left( {\cos x + \sin x} \right)}^2}}} \]
Now expanding the denominator in the formula of \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\], we get
\[\dfrac{{\cos 2x}}{{{{\cos }^2}x + 2\sin x\cos x + {{\sin }^2}x}}\]
We know that \[{\cos ^2}x\]\[ + {\sin ^2}x\]\[ = 1\], so using this formula, we get,
\[\dfrac{{\cos 2x}}{{1 + 2\sin x\cos x}}\]
Now from the formula of double angle identities, \[\sin 2x = 2\sin x\cos x\], we get
\[\dfrac{{\cos 2x}}{{1 + \sin 2x}}\]

Thus \[\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}} = \dfrac{{\cos 2x}}{{1 + \sin 2x}}\](proved).

Note:There are many forms of cosine double angle formulae such as $\cos 2x = {\cos ^2}x - {\sin ^2}x$, $\cos 2x = 2{\cos ^2}x - 1$, $\cos 2x = 1 - 2{\sin ^2}x$, $\cos 2x = \left( {\dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}}} \right)$. Hence, it is very crucial to understand which formula should be applied in the given problem. Since, we get, ${\cos ^2}x - {\sin ^2}x$ in the numerator after multiplying and dividing the left side by $\cos x + \sin x$, hence we apply $\cos 2x = {\cos ^2}x - {\sin ^2}x$ to simplify the expression. We should also remember the double angle formula for sine as $\sin 2x = 2\sin x\cos x$ to solve the given problem.