please help x
From this line we get we multiply by sinӨ(1 + cosӨ) to get:
(1 - cosӨ)(1 + cosӨ) = sin²Ө
=> 1 - cos²Ө = sin²Ө
=> sin²Ө + cos²Ө = 1......which is true for all Ө and a standard trig identity.
:)>
First we define sinÓ¨ = 0 and cosÓ¨ = -1 as being outside the domain of this equation
(1-cosÓ¨)/sinÓ¨
= sinÓ¨(1-cosÓ¨)/sin^2Ó¨ .............. (multliply top and bottom by sinÓ¨. Allowed since sinÓ¨ is not 0)
= sinÓ¨(1-cosÓ¨)/(1-cos^2Ó¨) ....... (sin^2Ó¨ = 1-cos^2Ó¨. That well known identity)
= sinÓ¨(1-cosÓ¨)/[(1-cosÓ¨)(1+cosÓ¨)].... (difference of two squares on the bottom)
= sinÓ¨/(1+cosÓ¨)............................. (cancel 1-cosÓ¨. Allowed because cosÓ¨ is not -1)
QED
Be careful to use brackets. I understand your problem but others might not..
sin^2(Ó¨) + cos^2(Ó¨) = 1
sin^2(Ó¨) = 1 - cos^2(Ó¨)
sin^2(Ó¨) = (1 + cos(Ó¨) ) ( 1- cos(Ó¨) )
sin(Ó¨) sin(Ó¨) = (1 + cos(Ó¨) ) ( 1- cos(Ó¨) )
sin(Ó¨) / (1 + cos(Ó¨) ) = ( 1- cos(Ó¨) ) / sin(Ó¨)
Copyright © 2024 1QUIZZ.COM - All rights reserved.
Answers & Comments
Verified answer
From this line we get we multiply by sinӨ(1 + cosӨ) to get:
(1 - cosӨ)(1 + cosӨ) = sin²Ө
=> 1 - cos²Ө = sin²Ө
=> sin²Ө + cos²Ө = 1......which is true for all Ө and a standard trig identity.
:)>
First we define sinÓ¨ = 0 and cosÓ¨ = -1 as being outside the domain of this equation
(1-cosÓ¨)/sinÓ¨
= sinÓ¨(1-cosÓ¨)/sin^2Ó¨ .............. (multliply top and bottom by sinÓ¨. Allowed since sinÓ¨ is not 0)
= sinÓ¨(1-cosÓ¨)/(1-cos^2Ó¨) ....... (sin^2Ó¨ = 1-cos^2Ó¨. That well known identity)
= sinÓ¨(1-cosÓ¨)/[(1-cosÓ¨)(1+cosÓ¨)].... (difference of two squares on the bottom)
= sinÓ¨/(1+cosÓ¨)............................. (cancel 1-cosÓ¨. Allowed because cosÓ¨ is not -1)
QED
Be careful to use brackets. I understand your problem but others might not..
sin^2(Ó¨) + cos^2(Ó¨) = 1
sin^2(Ó¨) = 1 - cos^2(Ó¨)
sin^2(Ó¨) = (1 + cos(Ó¨) ) ( 1- cos(Ó¨) )
sin(Ó¨) sin(Ó¨) = (1 + cos(Ó¨) ) ( 1- cos(Ó¨) )
sin(Ó¨) / (1 + cos(Ó¨) ) = ( 1- cos(Ó¨) ) / sin(Ó¨)