Question

cos x + sin x = root2 cos (x-x/4)=root2 sin (x+x/4)

Answer 1

Appropriate Question :-

$\rm \: cosx + sinx = \sqrt{2}cos\bigg(x - \dfrac{\pi}{4} \bigg) = \sqrt{2}sin\bigg(x + \dfrac{\pi}{4} \bigg)$

$\large\underline{\sf{Solution-}}$

Consider,

$\rm \: cosx + sinx$

can be rewritten as

$\rm \: = \: \sqrt{2}\bigg(\dfrac{1}{ \sqrt{2} } cosx + \frac{1}{ \sqrt{2} } sinx\bigg)$

can be further rewritten as,

$\rm \: = \: \sqrt{2}\bigg(cos\dfrac{\pi}{4} \: cosx + sinx \: sin\dfrac{\pi}{4}\bigg)$

$\rm \: = \: \sqrt{2}\bigg(cosx \: cos\dfrac{\pi}{4} + sinx \: sin\dfrac{\pi}{4}\bigg)$

We know,

$\boxed{ \rm{ \:cosx \: cosy \: + \: sinx \: siny \: = \: cos(x - y) \: }}$

So, using this result, we get

$\color{green}\rm \: = \: \sqrt{2}cos\bigg(x - \dfrac{\pi}{4} \bigg)$

We know,

$\boxed{ \rm{ \:cosx = sin\bigg(\dfrac{\pi}{2} + x\bigg) \: }}$

So, using this result, we get

$\rm \: = \: \sqrt{2}sin\bigg(\dfrac{\pi}{2} + x - \dfrac{\pi}{4} \bigg)$

$\color{blue}\rm \: = \: \sqrt{2}sin\bigg(\dfrac{\pi}{4} + x \bigg)$

Hence,

$\rm\implies \: cosx + sinx = \sqrt{2}cos\bigg(x - \dfrac{\pi}{4} \bigg) = \sqrt{2}sin\bigg(x + \dfrac{\pi}{4} \bigg)$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sin(x - y) = sinx \: cosy \: - \: siny \: cosx}\\ \\ \bigstar \: \bf{sin(x + y) = sinx \: cosy \: + \: siny \: cosx}\\ \\ \bigstar \: \bf{cos(x + y) = cosx \: cosy \: - \: sinx \: siny}\\ \\ \bigstar \: \bf{cos(x - y) = cosx \: cosy \:+\: siny \: sinx}\\ \\ \bigstar \: \bf{tan(x + y) = \dfrac{tanx + tany}{1 - tanx \: tany} }\\ \\ \bigstar \: \bf{tan(x - y) = \dfrac{tanx - tany}{1 + tanx \: tany} }\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$