Question

If sinA+sin2A=x and cosA+COS2A=y, then
(x2 + y²). (x^2+ y2 - 3)​

Answer 1

sinA + sin2A = x and cosA + cos2A = y

Then we have to find the value of (x² + y²)(x² + y² - 3).

solution : given sinA + sin2A = x and cosA + cos2A = y

x² + y² = (sinA + sin2A)² + (cosA + cos2A)²

= sin²A + sin²2A + 2sinAsin2A + cos²A + cos²2A + 2cosAcos2A

= (sin²A + cos²A) + (sin²2A + cos²2A) + 2sinAsin2A + 2cosAcos2A

= 1 + 1 + 2[sinAsin2A + cosAcos2A]

= 2 + 2[ sinA(2sinAcosA) + cosA(1 - 2sin²A)]

= 2 + 2[2sin²A cosA + cosA - 2sin²AcosA ]

= 2 + 2cosA

= 2(1 + cosA) .........(1)

now x² + y² - 3 = 2(1 + cosA) - 3 .......(2)

(x² + y²)(x² + y² - 3)

from equations (1) and (2) we get,

= [2(1 + cosA)] [2(1 + cosA) - 3]

= 4(1 + cosA)² - 6(1 + cosA)

= 4(1 + cos²A + 2cosA) - 6 - 6cosA

= 4 + 4cos²A + 8cosA - 6 - 6cosA

= 4cos²A + 2cosA - 2

= 2(2cos²A -1) + 2cosA

= 2cos2A + 2cosA

= 2[cosA + cos2A]

but it is given that cosA + cos2A = y

so 2[cosA + cos2A] = 2y

Therefore the value of (x² + y²)(x² + y² - 3) = 2y

Answer 2

Answer:

cos a =

$1 \div 2( {x}^{2} + {y}^{2} - 2)$