3.57 trigonometric identities

[Eigenvalue]

See attachment:

[Guest]

[tex]\frac{(sin5x+sinx) +sin3x}{(cos5x+cosx) +cos3x}[/tex]=

=[tex]\frac{2sin \frac{5x+x}{2} .cos \frac{5x-x}{2} +sin3x}{2cos \frac{5x+x}{2} .cos \frac{5x-x}{2} +cos3x }[/tex]=

=[tex]\frac{2sin3x.cos2x+sin3x}{2cos3x.cos2x+cos3x}[/tex]=

=[tex]\frac{sin3x( 2cos2x+1)}{cos3x( 2cos2x+1)}[/tex] (B) =

=[tex]\frac{sin3x}{cos3x}[/tex] =tan3x

In (B) we consider the permissible values .
Must cos3x[tex]\ne[/tex]0 and cos2x[tex]\ne[/tex] - [tex]\frac{1}{2}[/tex] [tex]\Rightarrow[/tex] 3x[tex]\ne \frac{ \pi }{2}+k \pi[/tex] and 2x[tex]\ne \pm \frac{2 \pi }{3} +2k \pi[/tex]

x[tex]\ne \frac{ \pi }{6} + \frac{k \pi }{3}[/tex] and x[tex]\ne \pm \frac{ \pi }{3} +k \pi[/tex] ,k[tex]\in[/tex] Z

[nycmath]

[tex]\frac{(sin5x+sinx) +sin3x}{(cos5x+cosx) +cos3x}[/tex]=

=[tex]\frac{2sin \frac{5x+x}{2} .cos \frac{5x-x}{2} +sin3x}{2cos \frac{5x+x}{2} .cos \frac{5x-x}{2} +cos3x }[/tex]=

=[tex]\frac{2sin3x.cos2x+sin3x}{2cos3x.cos2x+cos3x}[/tex]=

=[tex]\frac{sin3x( 2cos2x+1)}{cos3x( 2cos2x+1)}[/tex] (B) =

=[tex]\frac{sin3x}{cos3x}[/tex] =tan3x

In (B) we consider the permissible values .
Must cos3x[tex]\ne[/tex]0 and cos2x[tex]\ne[/tex] - [tex]\frac{1}{2}[/tex] [tex]\Rightarrow[/tex] 3x[tex]\ne \frac{ \pi }{2}+k \pi[/tex] and 2x[tex]\ne \pm \frac{2 \pi }{3} +2k \pi[/tex]

x[tex]\ne \frac{ \pi }{6} + \frac{k \pi }{3}[/tex] and x[tex]\ne \pm \frac{ \pi }{3} +k \pi[/tex] ,k[tex]\in[/tex] Z

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[Guest]

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[nycmath]

I like to use my time to help students with math .
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Count on help from Bulgaria .

I understand.