Find-1-sin30-2-sin30-4-sin30-6-

Question Number 202356 by hardmath last updated on 25/Dec/23

$$\mathrm{Find}: \\ $$$$\mathrm{1}−\left(\mathrm{sin30}°\right)^{\mathrm{2}} \:+\:\left(\mathrm{sin30}°\right)^{\mathrm{4}} \:−\:\left(\mathrm{sin30}°\right)^{\mathrm{6}} \:+\:… \\ $$

Answered by Rasheed.Sindhi last updated on 25/Dec/23

1−(sin30°)^2 + (sin30°)^4 − (sin30°)^6 + ... (1+(sin30°)^4 +(sin30°)^8 +...)−{(sin30°)^2 +(sin30°)^6 +(sin30°)^(10) ...} ={1+((1/2))^4 +((1/2))^8 +...} −{((1/2))^2 +((1/2))^6 +((1/2))^(10) +...} In 1st series: a=1,r=((1/2))^4 In 2nd series: a=((1/2))^2 ,r=((1/2))^4 =(1/(1−(1/(16))))−((1/4)/(1−(1/(16))))=(((1−(1/4)))/((1−(1/4))(1+(1/4)))) =(1/(5/4))=(4/5) ✓ Formula used: determinant (((a+ar+ar^2 +...=(a/(1−r)) ; ∣r∣<1)))

$$\mathrm{1}−\left(\mathrm{sin30}°\right)^{\mathrm{2}} \:+\:\left(\mathrm{sin30}°\right)^{\mathrm{4}} \:−\:\left(\mathrm{sin30}°\right)^{\mathrm{6}} \:+\:… \\ $$$$\left(\mathrm{1}+\left(\mathrm{sin30}°\right)^{\mathrm{4}} +\left(\mathrm{sin30}°\right)^{\mathrm{8}} +…\right)−\left\{\left(\mathrm{sin30}°\right)^{\mathrm{2}} +\left(\mathrm{sin30}°\right)^{\mathrm{6}} +\left(\mathrm{sin30}°\right)^{\mathrm{10}} …\right\} \\ $$$$=\left\{\mathrm{1}+\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{4}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{8}} +…\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:−\left\{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{6}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{10}} +…\right\} \\ $$$${In}\:\mathrm{1st}\:\mathrm{series}:\:{a}=\mathrm{1},{r}=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{4}} \\ $$$${In}\:\mathrm{2nd}\:\:\mathrm{series}:\:{a}=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} ,{r}=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{4}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{16}}}−\frac{\frac{\mathrm{1}}{\mathrm{4}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{16}}}=\frac{\cancel{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)}}{\cancel{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}/\mathrm{4}}=\frac{\mathrm{4}}{\mathrm{5}}\:\checkmark \\ $$$$\mathcal{F}{ormula}\:{used}: \\ $$$$\begin{array}{|c|}{{a}+{ar}+{ar}^{\mathrm{2}} +…=\frac{{a}}{\mathrm{1}−{r}}\:;\:\mid{r}\mid<\mathrm{1}}\\\hline\end{array}\: \\ $$

Commented by hardmath last updated on 25/Dec/23

$$ \\ $$Thank you dear professor, what are the values of a and r in the exercise…

Commented by Rasheed.Sindhi last updated on 25/Dec/23

$${Have}\:{added}\:{two}\:{red}\:{lines}\:{to}\:{explain}. \\ $$

Answered by Sutrisno last updated on 25/Dec/23

s=(1/(1−(−(sin30°)^2 ))) s=(1/(1−(−(1/4)))) s=(4/5)

$${s}=\frac{\mathrm{1}}{\mathrm{1}−\left(−\left({sin}\mathrm{30}°\right)^{\mathrm{2}} \right)} \\ $$$${s}=\frac{\mathrm{1}}{\mathrm{1}−\left(−\frac{\mathrm{1}}{\mathrm{4}}\right)} \\ $$$${s}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$ \\ $$

Answered by Rajpurohith last updated on 26/Dec/23

Call x=sin(30°) then the expression =1−x^2 +x^4 −x^6 +...=((1/(1+x^2 )))=(1/(1+sin^2 (30°))) =(1/(1+(1/4)))=(4/5)

$${Call}\:{x}={sin}\left(\mathrm{30}°\right) \\ $$$${then}\:{the}\:{expression} \\ $$$$=\mathrm{1}−{x}^{\mathrm{2}} +{x}^{\mathrm{4}} −{x}^{\mathrm{6}} +…=\left(\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)=\frac{\mathrm{1}}{\mathrm{1}+{sin}^{\mathrm{2}} \left(\mathrm{30}°\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$ \\ $$

Answered by MathematicalUser2357 last updated on 29/Dec/23

=Σ_(x=0) ^∞ ((−1)^n sin^(2n) (π/6))=(4/5)

$$=\underset{{x}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\left(−\mathrm{1}\right)^{{n}} \mathrm{sin}^{\mathrm{2}{n}} \frac{\pi}{\mathrm{6}}\right)=\frac{\mathrm{4}}{\mathrm{5}} \\ $$

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