displaystylesinleft(arccosleft(1 ight) ight)=0 Explanation: displaystyley=arccosleft(x ight)is defined as the value ofdisplaystyleyinleft<0,pi ight) ...

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displaystyleleft(360npm40 ight)^o,n=0,pm1,pm2,pm3,ldots Explanation:Had it been arc cos (cos a), the answer is that a.Here,displaystylesin50^o=cosleft(360npm40 ight)^o,n=0,pm1,pm2,pm3,ldots ...
Have a look:Explanation:Consider that: displaystylecosleft(2 heta ight)=cos^2left( heta ight)-sin^2left( heta ight) So: displaystylecos^2left( heta ight)-sin^2left( heta ight)=1-2sin^2left( heta ight) ...
How bởi vì you express the complex number in trigonometric size displaystyle2left(cos0°+isin0° ight) ?
https://socratic.org/questions/how-do-you-express-the-complex-number-in-trigonometric-form-2-cos-0-i-sin-0
Gió May 8, năm ngoái I am not sure if I got it right. Your number is in trigonometric size so maybe you want to change it into standard formdisplaystylea+bi ...?If is so, you know ...
displaystyle3sinleft(A ight)=1+cosleft(2A ight)quad:quadA=fracpi6+2pin,:A=frac5pi6+2pin Explanation: displaystyle3sinleft(A ight)=1+cosleft(2A ight) ...

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Basically you can use infinite series to calculate approximation of inverse trigonometric functions. arcsin z = z+ left( frac 12 ight) z^3 over 3 + left( 1 cdot 3 over 2 cdot 4 ight)z^5 over 5 + left( 1 cdot 3 cdot 5 over 2 cdot 4 cdot 6 ight)z^7 over 7 + ... ; = sum_n=0^infty inom2nnz^2n+1 over 4^n(2n+1); ;;; |z| le 1 ...
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left< eginarray l l 2 và 3 \ 5 & 4 endarray ight> left< eginarray l l l 2 & 0 và 3 \ -1 & 1 & 5 endarray ight>

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