prove [sin(x) - sin(y)] / [cos(x) + cos(y)] = tan(x-y/2)
B baselramjet New member Joined Mar 22, 2007 Messages 34 Mar 31, 2007 #1 prove [sin(x) - sin(y)] / [cos(x) + cos(y)] = tan(x-y/2)
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Mar 31, 2007 #2 Hello, baselramjet! You'll need some sum-to-product identities: . . \(\displaystyle \sin(A)\,-\,\sin(B)\:=\:2\cdot\cos\left(\frac{A+B}{2}\right)\cdot\sin\left(\frac{A-B}{2}\right)\) . . \(\displaystyle \cos(A)\,+\,\cos(B)\:=\:2\cdot\cos\left(\frac{A+B}{2}\right)\cdot\cos\left(\frac{A-B}{2}\right)\) Prove: \(\displaystyle \L\:\frac{\sin(x)\,-\,\sin(y)}{\cos(x)\,+\,\cos(y)}\:=\:\tan\left(\frac{x-y}{2}\right)\) Click to expand... \(\displaystyle \L\frac{\sin(x)\,-\,\sin(y)}{\cos(x)\,+\,\cos(yu)}\;=\;\frac{2\cdot\cos\left(\frac{x+y}{2}\right)\cdot\sin\left(\frac{x-y}{2}\right)}{2\cdot\cos\left(\frac{x+y}{2}\right)\cdot\cos\left(\frac{x-y}{2}\right)} \;=\;\frac{\sin\left(\frac{x-y}{2}\right)}{\cos\left(\frac{x-y}{2}\right)} \;=\;\tan\left(\frac{x-y}{2}\right)\)
Hello, baselramjet! You'll need some sum-to-product identities: . . \(\displaystyle \sin(A)\,-\,\sin(B)\:=\:2\cdot\cos\left(\frac{A+B}{2}\right)\cdot\sin\left(\frac{A-B}{2}\right)\) . . \(\displaystyle \cos(A)\,+\,\cos(B)\:=\:2\cdot\cos\left(\frac{A+B}{2}\right)\cdot\cos\left(\frac{A-B}{2}\right)\) Prove: \(\displaystyle \L\:\frac{\sin(x)\,-\,\sin(y)}{\cos(x)\,+\,\cos(y)}\:=\:\tan\left(\frac{x-y}{2}\right)\) Click to expand... \(\displaystyle \L\frac{\sin(x)\,-\,\sin(y)}{\cos(x)\,+\,\cos(yu)}\;=\;\frac{2\cdot\cos\left(\frac{x+y}{2}\right)\cdot\sin\left(\frac{x-y}{2}\right)}{2\cdot\cos\left(\frac{x+y}{2}\right)\cdot\cos\left(\frac{x-y}{2}\right)} \;=\;\frac{\sin\left(\frac{x-y}{2}\right)}{\cos\left(\frac{x-y}{2}\right)} \;=\;\tan\left(\frac{x-y}{2}\right)\)
