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Julli [10]
3 years ago
13

Find the multiplicative order of 5 and 7 in z16

Mathematics
1 answer:
marusya05 [52]3 years ago
3 0
5\equiv5\mod{16}
5^2=25\equiv9\mod{16}
5^3\equiv5\cdot9\equiv45\equiv48-3\equiv-3\equiv13\mod{16}
5^4\equiv9^2\equiv81\equiv1\mod{16}
5^5\equiv5\cdot1\equiv5\mod{16}

7\equiv7\mod{16}
7^2\equiv49\equiv48+1\equiv1\mod{16}
7^3\equiv7\cdot1\equiv7\mod{16}
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Prove the following
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Answer:

Step-by-step explanation:

\large\underline{\sf{Solution-}}

<h2 /><h2><u>Consider</u></h2>

\rm \: \cos \bigg( \dfrac{3\pi}{2} + x \bigg) \cos \: (2\pi + x) \bigg \{ \cot \bigg( \dfrac{3\pi}{2} - x \bigg) + cot(2\pi + x) \bigg \}cos(23π+x)cos(2π+x)

<h2><u>W</u><u>e</u><u> </u><u>K</u><u>n</u><u>o</u><u>w</u><u>,</u></h2>

\rm \: \cos \bigg( \dfrac{3\pi}{2} + x \bigg) = sinx

\rm \: {cos \: (2\pi + x) }

\rm \: \cot \bigg( \dfrac{3\pi}{2} - x \bigg) \: = \: tanx

\rm \: cot(2\pi + x) \: = \: cotx

So, on substituting all these values, we get

\rm \: = \: sinx \: cosx \: (tanx \: + \: cotx)

\rm \: = \: sinx \: cosx \: \bigg(\dfrac{sinx}{cosx} + \dfrac{cosx}{sinx}

\rm \: = \: sinx \: cosx \: \bigg(\dfrac{ {sin}^{2}x + {cos}^{2}x}{cosx \: sinx}

\rm \: = \: 1=1

<h2>Hence,</h2>

\boxed{\tt{ \cos \bigg( \frac{3\pi}{2} + x \bigg) \cos \: (2\pi + x) \bigg \{ \cot \bigg( \frac{3\pi}{2} - x \bigg) + cot(2\pi + x) \bigg \} = 1}}

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

<h2>ADDITIONAL INFORMATION :-</h2>

Sign of Trigonometric ratios in Quadrants

  • sin (90°-θ)  =  cos θ
  • cos (90°-θ)  =  sin θ
  • tan (90°-θ)  =  cot θ
  • csc (90°-θ)  =  sec θ
  • sec (90°-θ)  =  csc θ
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  • tan (90°+θ)  =  -cot θ
  • csc (90°+θ)  =  sec θ
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  • sec (180°+θ)  =  -sec θ
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