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dimulka [17.4K]
2 years ago
10

What are the exact values of sin(2π3radians) and cos(2π3radians)

Mathematics
1 answer:
deff fn [24]2 years ago
8 0
We know that
2π/3 radians-------> convert to degrees-----> 2*180/3---> 120°
120°=90°+30°

Part a) Find <span>sin(2π/3)
</span>sin(2π/3)=sin (90°+30°)
we know that
sin (A+B)=sin A*cos B+cos A*sin B
so
sin (90°+30°)=sin 90*cos 30+cos 90*sin 30
sin 90=1
cos 30=√3/2
cos 90=0
sin 30=1/2
sin (90°+30°)=1*√3/2+0*1/2-----> √3/2

the answer part a) is
sin(2π/3)=√3/2

Part b) Find cos (2π/3)
cos (2π/3)=cos (90°+30°)
we know that
cos (A+B)=cos A*cos B-sin A*sin B
so
cos (90°+30°)=cos 90*cos 30-sin 90*sin 30
sin 90=1
cos 30=√3/2
cos 90=0
sin 30=1/2
cos (90°+30°)=0*√3/2-1*1/2----> -1/2

the answer part b) is
cos (2π/3)=-1/2
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Step-by-step explanation:

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A geometric sequence has a constant ratio and is defined by

a_n=a_1\cdot r^{n-1}

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\frac{6}{8}=\frac{3}{4},\:\quad \frac{4.5}{6}=\frac{3}{4}

\mathrm{The\:ratio\:of\:all\:the\:adjacent\:terms\:is\:the\:same\:and\:equal\:to}

r=\frac{3}{4}

\mathrm{The\:first\:element\:of\:the\:sequence\:is}

a_1=8

\mathrm{Therefore,\:the\:}n\mathrm{th\:term\:is\:computed\:by}\:

a_n=8\left(\frac{3}{4}\right)^{n-1}

\mathrm{Geometric\:sequence\:sum\:formula:}

a_1\frac{1-r^n}{1-r}

\mathrm{Plug\:in\:the\:values:}

n=25,\:\spacea_1=8,\:\spacer=\frac{3}{4}

=8\cdot \frac{1-\left(\frac{3}{4}\right)^{25}}{1-\frac{3}{4}}

\mathrm{Multiply\:fractions}:\quad \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c}

=\frac{\left(1-\left(\frac{3}{4}\right)^{25}\right)\cdot \:8}{1-\frac{3}{4}}

=\frac{8\left(-\left(\frac{3}{4}\right)^{25}+1\right)}{\frac{1}{4}}

\mathrm{Apply\:exponent\:rule}:\quad \left(\frac{a}{b}\right)^c=\frac{a^c}{b^c}

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=\frac{32\left(-\frac{3^{25}}{4^{25}}+1\right)}{1}

=\frac{32\cdot \frac{4^{25}-3^{25}}{4^{25}}}{1}               ∵ \mathrm{Join}\:1-\frac{3^{25}}{4^{25}}:\quad \frac{4^{25}-3^{25}}{4^{25}}

=32\cdot \frac{4^{25}-3^{25}}{4^{25}}

=\frac{\left(4^{25}-3^{25}\right)\cdot \:32}{4^{25}}

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so

=\frac{4^{25}-3^{25}}{2^{45}}        ∵ \mathrm{Cancel\:}\frac{\left(4^{25}-3^{25}\right)\cdot \:2^5}{2^{50}}:\quad \frac{4^{25}-3^{25}}{2^{45}}

\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a\pm \:b}{c}=\frac{a}{c}\pm \frac{b}{c}

=\frac{4^{25}}{2^{45}}-\frac{3^{25}}{2^{45}}      

=32-\frac{3^{25}}{2^{45}}            ∵  \frac{4^{25}}{2^{45}}=32

=32-0.024        ∵  \frac{3^{25}}{2^{45}}=0.024

=31.98            

Therefore, the sum of the first 25 terms in this geometric series: 31.98

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