Math, asked by sabikakhatun007, 7 months ago

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

Answers

Answered by shagun1639
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.

Answered by mathdude500
1

\huge\pink{\boxed{\blue{\boxed{ \purple{ \boxed{{\pink{Answer}}}}}}}} \\ \large\pink{\boxed{\blue{\boxed{ \purple{ \boxed{{\pink{Your~answer↓}}}}}}}}

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♛}

\red{\textsf{(✪MARK BRAINLIEST✿)}} \\ \huge\fcolorbox{aqua}{aqua}{\red{FolloW ME}}

Similar questions