Establish the identity: 1. Cos(α-β)cos(α+β) = cos2α-sin2β
R rolita225 New member Joined Mar 2, 2006 Messages 1 Mar 2, 2006 #1 Establish the identity: 1. Cos(α-β)cos(α+β) = cos2α-sin2β
R royhaas Full Member Joined Dec 14, 2005 Messages 832 Mar 2, 2006 #2 Expand the left hand side using the formulas for the cosine of the sum and difference of two angles. Then just work through the algebra and simplify.
Expand the left hand side using the formulas for the cosine of the sum and difference of two angles. Then just work through the algebra and simplify.
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Mar 2, 2006 #3 Hello, Rolita! If no one is responding, it's because it is not an identity. I think this is what you meant: cos(A − B)⋅cos(A + B) = cos2A − sin2B\displaystyle \;\cos(A\,-\,B)\cdot\cos(A\,+\,B)\;=\;\cos^2A\,-\,\sin^2Bcos(A−B)⋅cos(A+B)=cos2A−sin2B The left side is: [cosA⋅cosB + sinA⋅sinB][cosA⋅cosB − sinA⋅sinB]\displaystyle \:[\cos A\cdot\cos B\,+\,\sin A\cdot\sin B][\cos A\cdot\cos B\,-\,\sin A\cdot\sin B][cosA⋅cosB+sinA⋅sinB][cosA⋅cosB−sinA⋅sinB] = cos2A⋅cos2B⏟ − sin2A⏟⋅sin2B\displaystyle \;\;\;=\;\;\;\cos^2A\cdot\underbrace{\cos^2B}\;\;\;-\;\;\;\underbrace{\sin^2A}\cdot\sin^2B=cos2A⋅cos2B−sin2A⋅sin2B = cos2A(1 − sin2B⏞) − (1 −cos2A⏞)sin2B\displaystyle \;\;\;=\;\cos^2A(\overbrace{1\,-\,\sin^2B})\,-\,(\overbrace{1\,-\cos^2A})\sin^2B=cos2A(1−sin2B)−(1−cos2A)sin2B = cos2A − cos2A⋅sin2B − sin2B + cos2A⋅sin2B\displaystyle \;\;\;=\;\cos^2A\,-\,\cos^2A\cdot\sin^2B\,-\,\sin^2B \,+\,\cos^2A\cdot\sin^2B=cos2A−cos2A⋅sin2B−sin2B+cos2A⋅sin2B = cos2A − sin2B\displaystyle \;\;\;=\;\cos^2A\,-\,\sin^2B=cos2A−sin2B
Hello, Rolita! If no one is responding, it's because it is not an identity. I think this is what you meant: cos(A − B)⋅cos(A + B) = cos2A − sin2B\displaystyle \;\cos(A\,-\,B)\cdot\cos(A\,+\,B)\;=\;\cos^2A\,-\,\sin^2Bcos(A−B)⋅cos(A+B)=cos2A−sin2B The left side is: [cosA⋅cosB + sinA⋅sinB][cosA⋅cosB − sinA⋅sinB]\displaystyle \:[\cos A\cdot\cos B\,+\,\sin A\cdot\sin B][\cos A\cdot\cos B\,-\,\sin A\cdot\sin B][cosA⋅cosB+sinA⋅sinB][cosA⋅cosB−sinA⋅sinB] = cos2A⋅cos2B⏟ − sin2A⏟⋅sin2B\displaystyle \;\;\;=\;\;\;\cos^2A\cdot\underbrace{\cos^2B}\;\;\;-\;\;\;\underbrace{\sin^2A}\cdot\sin^2B=cos2A⋅cos2B−sin2A⋅sin2B = cos2A(1 − sin2B⏞) − (1 −cos2A⏞)sin2B\displaystyle \;\;\;=\;\cos^2A(\overbrace{1\,-\,\sin^2B})\,-\,(\overbrace{1\,-\cos^2A})\sin^2B=cos2A(1−sin2B)−(1−cos2A)sin2B = cos2A − cos2A⋅sin2B − sin2B + cos2A⋅sin2B\displaystyle \;\;\;=\;\cos^2A\,-\,\cos^2A\cdot\sin^2B\,-\,\sin^2B \,+\,\cos^2A\cdot\sin^2B=cos2A−cos2A⋅sin2B−sin2B+cos2A⋅sin2B = cos2A − sin2B\displaystyle \;\;\;=\;\cos^2A\,-\,\sin^2B=cos2A−sin2B
