undefinedundefinedundefinedneed help=====) (sin2x+sin4x)/(cos2x+cos4x)=tan3x
Aug 4, 2005 #1 Dainy 2 0 undefinedundefinedundefinedneed help=====) (sin2x+sin4x)/(cos2x+cos4x)=tan3x
Related Introductory Physics Homework Help News on Phys.org A steaming cauldron follows the dinosaurs' demise Scientists develop method to help epidemiologists map spread of COVID-19 Mergers between galaxies trigger activity in their core
Related Introductory Physics Homework Help News on Phys.org A steaming cauldron follows the dinosaurs' demise Scientists develop method to help epidemiologists map spread of COVID-19 Mergers between galaxies trigger activity in their core
Aug 4, 2005 #2 VietDao29 Homework Helper 1,423 2 You can use these to prove that trig identity: [tex]\cos \alpha + \cos \beta = 2 \cos \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2}[/tex] [tex]\cos \alpha - \cos \beta = - 2 \sin \frac{\alpha + \beta}{2} \sin \frac{\alpha - \beta}{2}[/tex] [tex]\sin \alpha + \sin \beta = 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2}[/tex] [tex]\sin \alpha - \sin \beta = 2 \cos \frac{\alpha + \beta}{2} \sin \frac{\alpha - \beta}{2}[/tex] Choose 2 of the above equations, and use them. Can you go from there? Viet Dao,
You can use these to prove that trig identity: [tex]\cos \alpha + \cos \beta = 2 \cos \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2}[/tex] [tex]\cos \alpha - \cos \beta = - 2 \sin \frac{\alpha + \beta}{2} \sin \frac{\alpha - \beta}{2}[/tex] [tex]\sin \alpha + \sin \beta = 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2}[/tex] [tex]\sin \alpha - \sin \beta = 2 \cos \frac{\alpha + \beta}{2} \sin \frac{\alpha - \beta}{2}[/tex] Choose 2 of the above equations, and use them. Can you go from there? Viet Dao,