Question

Prove that :-

$\sf {cos \ \bigg(\dfrac{\pi}{4}\ -\ x \bigg)\ cos\ \bigg(\dfrac{\pi}{4}\ -\ y \bigg)\ -\ sin\ \bigg(\dfrac{\pi}{4}\ -\ x \bigg)\ sin\ \bigg(\dfrac{\pi}{4}\ -\ y \bigg)\ =\ sin(x\ +\ y)}$

Answer 1

Step-by-step explanation:

$\underline{\underline{\maltese\: \: \textbf{\textsf{To\ Prove}}}}$

$\sf \circ\ {cos \ \bigg(\dfrac{\pi}{4}\ -\ x \bigg)\ cos\ \bigg(\dfrac{\pi}{4}\ -\ y \bigg)\ -\ sin\ \bigg(\dfrac{\pi}{4}\ -\ x \bigg)\ sin\ \bigg(\dfrac{\pi}{4}\ -\ y \bigg)\ =\ sin(x\ +\ y)}$

$\underline{\underline{\maltese\: \: \textbf{\textsf{Solution}}}}$

By taking the L.H.S

➤ We know that :-

• $\sf {cos\ (A\ +\ B)\ =\ cos\ A\ cos\ B\ -\ sin\ A\ sin\ B}$

➠ The equation given in question is of this form, where,

• $\sf {A\ =\ \bigg(\dfrac{\pi}{4}\ -\ x \bigg)\ B\ =\ \bigg(\dfrac{\pi}{4}\ -\ y \bigg)}$

➤ Therefore,

$\sf \rightarrow {cos \ \bigg(\dfrac{\pi}{4}\ -\ x \bigg)\ cos\ \bigg(\dfrac{\pi}{4}\ -\ y \bigg)\ -\ sin\ \bigg(\dfrac{\pi}{4}\ -\ x \bigg)\ sin\ \bigg(\dfrac{\pi}{4}\ -\ y \bigg)}$

$\sf \rightarrow {cos\ \bigg[\bigg(\dfrac{\pi}{4}\ -\ x \bigg)\ +\ \bigg(\dfrac{\pi}{4}\ -\ y \bigg) \bigg]}$

$\sf \rightarrow {cos\ \bigg[\dfrac{\pi}{4}\ -\ x\ +\ \dfrac{\pi}{4}\ -\ y \bigg]}$

$\sf \rightarrow {cos\ \bigg[\dfrac{\pi}{4}\ +\ \dfrac{\pi}{4}\ -\ x\ -\ y \bigg]}$

$\sf \rightarrow {cos\ \bigg[\dfrac{\pi}{2}\ -\ (x\ +\ y) \bigg]}$

• $\sf {Now,\ putting\ {\pi}\ =\ 180^{\circ}}$

$\sf \rightarrow {cos\ \bigg[\dfrac{180^{\circ}}{2}\ -\ (x\ +\ y) \bigg]}$

$\sf \rightarrow {cos\ [90^{\circ}\ -\ (x\ +\ y)]}$

• $\sf {cos\ (90^{\circ}\ -\ \theta)\ =\ sin\ \theta}$

$\sf \rightarrow {sin\ (x\ +\ y)}$

$\sf \rightarrow {R.H.S}$

$\large {\therefore{\underline{\boxed{\boxed{\sf{cos \ \bigg(\dfrac{\pi}{4}\ -\ x \bigg)\ cos\ \bigg(\dfrac{\pi}{4}\ -\ y \bigg)\ -\ sin\ \bigg(\dfrac{\pi}{4}\ -\ x \bigg)\ sin\ \bigg(\dfrac{\pi}{4}\ -\ y \bigg)\ =\ sin(x\ +\ y)}}}}}}$

$\underline{\textbf{\textsf{Hence\ Proved...!!!}}}$

Answer 2

⠀⠀${\rm{\pink{\underline{\underline{★Step-by-step -Explaination★}}}}}$

$\underline{\underline{\maltese\: \: \textbf{\textsf{To\ Prove}}}}$

$\sf \circ\ {cos \ \bigg(\dfrac{\pi}{4}\ -\ x \bigg)\ cos\ \bigg(\dfrac{\pi}{4}\ -\ y \bigg)\ -\ sin\ \bigg(\dfrac{\pi}{4}\ -\ x \bigg)\ sin\ \bigg(\dfrac{\pi}{4}\ -\ y \bigg)\ =\ sin(x\ +\ y)}$

$\underline{\underline{\maltese\: \: \textbf{\textsf{Solution}}}}$

${\rm{\red{\underline{\underline{From\: L.H.S : }}}}}$

➤ We know that :-

$\sf {cos\ (A\ +\ B)\ =\ cos\ A\ cos\ B\ -\ sin\ A\ sin\ B}$

➠ The equation given in question is of this form, where,

$\sf {A\ =\ \bigg(\dfrac{\pi}{4}\ -\ x \bigg)\ B\ =\ \bigg(\dfrac{\pi}{4}\ -\ y \bigg)}$

➤ Therefore,

$\sf \rightarrow {cos \ \bigg(\dfrac{\pi}{4}\ -\ x \bigg)\ cos\ \bigg(\dfrac{\pi}{4}\ -\ y \bigg)\ -\ sin\ \bigg(\dfrac{\pi}{4}\ -\ x \bigg)\ sin\ \bigg(\dfrac{\pi}{4}\ -\ y \bigg)}$

$\sf \rightarrow {cos\ \bigg[\bigg(\dfrac{\pi}{4}\ -\ x \bigg)\ +\ \bigg(\dfrac{\pi}{4}\ -\ y \bigg) \bigg]}$

$\sf \rightarrow {cos\ \bigg[\dfrac{\pi}{4}\ -\ x\ +\ \dfrac{\pi}{4}\ -\ y \bigg]}$

$\sf \rightarrow {cos\ \bigg[\dfrac{\pi}{4}\ +\ \dfrac{\pi}{4}\ -\ x\ -\ y \bigg]}$

$\sf \rightarrow {cos\ \bigg[\dfrac{\pi}{2}\ -\ (x\ +\ y) \bigg]}$

$\sf {Now,\ putting\ {\pi}\ =\ 180^{\circ}}$

$\sf \rightarrow {cos\ \bigg[\dfrac{180^{\circ}}{2}\ -\ (x\ +\ y) \bigg]}$

$\sf \rightarrow {cos\ [90^{\circ}\ -\ (x\ +\ y)]}$

$\sf {cos\ (90^{\circ}\ -\ \theta)\ =\ sin\ \theta}$

$\sf \rightarrow {sin\ (x\ +\ y)}$

$\sf \rightarrow {R.H.S}$

$\large {\therefore{\underline{\boxed{\boxed{\sf{cos \ \bigg(\dfrac{\pi}{4}\ -\ x \bigg)\ cos\ \bigg(\dfrac{\pi}{4}\ -\ y \bigg)\ -\ sin\ \bigg(\dfrac{\pi}{4}\ -\ x \bigg)\ sin\ \bigg(\dfrac{\pi}{4}\ -\ y \bigg)\ =\ sin(x\ +\ y)}}}}}}$

$\underline{\textbf{\textsf{Hence\ Proved...!!!}}}$