Answer:
The correct prove will be:
[tex]sin(\frac{\pi }{14})sin(\frac{3\pi }{14})sin(\frac{5\pi }{14})[/tex] [tex]= \frac{1}{8}[/tex]
Step-by-step explanation:
[tex]sin(\frac{\pi }{14})sin(\frac{3\pi }{14})sin(\frac{5\pi }{14})[/tex]
Multiply and Divide by [tex]2cos(\frac{\pi }{14})[/tex]
= [tex]\frac{2sin(\frac{\pi }{14})cos(\frac{\pi }{14})sin(\frac{3\pi }{14})sin(\frac{5\pi }{14})}{2cos(\frac{\pi }{14})}[/tex]
⇒ Let [tex]\frac{\pi }{14}[/tex] = Ф and 7Ф = [tex]\frac{\pi }{2}[/tex]
= sin2Ф sin3Ф sin5Ф ÷ 2cosФ
= 1/2(2sin2Фsin5Ф)sin3Ф ÷ 2cosФ
= (cos3Ф - cos7Ф) sin3Ф ÷ 4cosФ
= 1/2(2cos3Ф sin3Ф) ÷ 4cosФ ∵cos7Ф = 0
= sin6Ф ÷ 8cosФ
= sin(7Ф - Ф) ÷ 8cosФ
= (sin7ФcosФ - cos7ФsinФ) ÷ 8cosФ ∵sin7Ф = 1
= cosФ ÷ 8cosФ
= 1/8
Hence, The correct prove will be:
[tex]sin(\frac{\pi }{14})sin(\frac{3\pi }{14})sin(\frac{5\pi }{14})[/tex] [tex]= \frac{1}{8}[/tex]
Keywords: prove, sinФ, cosФ
Learn more about trigonometric functions from brainly.com/question/7331447
#learnwithBrainly
