Question

cos [π/4 -x]cos[π/4-x] - sin[π/4-x]sin[π/4-y]=sin(x+y)

prove that​

Answer 1

cosxcosy-sinxsiny=cos(X+y) ,RHS cos (π/4-x+π/4-y)=cos (π/2-(x+y))=sin(x+y). [cos(90-x)=sinx

Answer 2

Answer:

LHS : cos [x/4 -x]cos[π/4-x] - sin[π/4-x]sin[π/4-y]

Dividing and multiplying by 2.

⇒ ½ × 2 {cos [x/4-x]cos[π/4-x] - sin[π/4-x]sin[π/4-y]}

⇒ ½ {2cos(π/4-x)cos(π/4-x)} + ½{-2sin(π/4-x)sin(π/4-y)}

Using the identities :

• 2cos x cos y = cos(x+y) + cos(x-y)
• -2sin x sin y = cos(x+y) - cos(x-y)

⇒ ½ [cos{π/4-x+π/4-y} + cos{π/4-x-(π/4-y)}] + ½ [cos{π/4-x+π/4-y} - cos{π/4-x-(π/4-y)}]

(The second term and the fourth term cancels out because of opposite signs)

⇒ ½ × 2 [cos{π/4-x+π/4-y}]

⇒ cos{π/4-x+π/4-y}

⇒ cos{π/4-(x+y)}

⇒ sin(x+y) : RHS

SOME IMPORTANT IDENTITIES –

• 2cos x cos y = cos(x+y) + cos(x-y)
• -2sin x sin y = cos(x+y) - cos(x-y)
• 2sin x cos y = sin(x+y) + sin(x-y)
• 2cos x sin y = sin(x+y) - sin(x-y)
• cos(A+B) = cosAcosB - sinAsinB
• cos(A-B) = cosAcosB + sinAsinB