Question

sin^2(45°- theta) + sin^2(45°+ theta) = 1​

Answer 1

Answer:

To Prove :

Sin²(45-θ)+ Sin²(45+θ) = 1

Let A= 45 - θ, B = 45+θ and θ > 0

=> For θ>0, (45 - θ) will be an acute angle.

For an angle in the 1st Quadrant i.e., for A <90° :

Sin A = Cos (90-A)

=> Sin (45 -θ) =Cos (90-(45-θ))

=> Sin (45 -θ) = Cos (45 +θ)

=> Sin²(45-θ) = Cos²(45+θ) -----(1)

Sin²(45-θ) +Sin²(45+θ) = Cos²(45+θ) + Sin²(45+θ)

= Cos²B + Sin²B

= 1

=> Sin²(45-θ) + Sin²(45+θ) =1. Hence Proved.

Answer 2

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To prove :-

$\large\bold\green{♧ \: sin²(45 - θ) + sin²(45 + θ) = 1}$

Identity used :-

$\large\bold\green{♧ \: {sin}^{2}x + {cos}^{2} x = 1 }$

$\large\bold\green{♧ \: sin(90 - x) \: = cosx}$

$\large\bold\blue{♧ \: Solution \: ♧}$

$sin²(45° - θ) + sin²(45° + θ) \\ </p><p> = sin²(45° - θ) + sin²[90° - (45° - θ)] \\ = </p><p>sin²(45 - θ) + cos²(45 - θ)$

$\small\bold\red{ = 1}$

$\large\bold\green{♧ \: Hence, proved. \: ♧}$

$\huge \fcolorbox{black}{cyan}{♛Hope it helps U♛}$

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