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МатематикаВопрос:
(16) 1. 2sin(-a)cos(-a) - 2cos(-a)sin( (16) 2. 3 sin(x-a)cos(-a)+3sin²(-a) (25) 3. (1-tg(-a)) (1-tg(x + a))cos²a (25) 4. (1+tg² (-a)) 1+ 1 α) (16) 5. cos² (π-α)-cos² (-a) (26) 6. cos² (2n+a)-sin² (a + 2π) 2cos(a+2x)cos(-a) (16)7. 2sin(-a)cos(-a) (16) 8. 2sin(-a) sin(-a) sin²(-)-sin²(a-x) 2 9. (1+tg(-a))(1- ctg(-a)) - sin(-a) +tg(-a), tg(-α) (46) 10. ctga + tg(-a) cosa +sin(-α) sina (55) 11. cos³sin a - sin'a cos a (35) 12. (48) 13. sina + sin2 1+cosa + cos2a cos2a+sin2a cos2a 2sin2a-1 - sina)² (48) 14. sin2cossasinasa
Ответ:
Привет! Давай вместе упростим эти выражения. Будем идти шаг за шагом, чтобы все было понятно.
(16) 1. 2sin(-a)cos(\(\frac{\pi}{2}\)-a) - 2cos(-a)sin(\(\frac{\pi}{2}\)-a)
\(\begin{aligned}
&2sin(-a)cos(\frac{\pi}{2}-a) - 2cos(-a)sin(\frac{\pi}{2}-a) = \\
&2sin(-a)sin(a) - 2cos(-a)cos(a) = \\
&-2sin^2(a) - 2cos^2(a) = \\
&-2(sin^2(a) + cos^2(a)) = -2
\end{aligned}\)
(16) 2. 3 sin(\(\pi\)-a)cos(\(\frac{\pi}{2}\)-a)+3sin²(\(\frac{\pi}{2}\)-a)
\(\begin{aligned}
&3sin(\pi-a)cos(\frac{\pi}{2}-a)+3sin^2(\frac{\pi}{2}-a) = \\
&3sin(a)sin(a) + 3cos^2(a) = \\
&3sin^2(a) + 3cos^2(a) = \\
&3(sin^2(a) + cos^2(a)) = 3
\end{aligned}\)
(25) 3. (1-tg(-a)) (1-tg(\(\pi\) + a))cos²a
\(\begin{aligned}
&(1-tg(-a))(1-tg(\pi + a))cos^2a = \\
&(1+tg(a))(1-tg(a))cos^2(a) = \\
&(1-tg^2(a))cos^2(a) = \\
&cos^2(a) - \frac{sin^2(a)}{cos^2(a)}cos^2(a) = \\
&cos^2(a) - sin^2(a) = cos(2a)
\end{aligned}\)
(25) 4. (1+tg² (-a))\(\frac{1}{1 + ctg^2 (-a)}\)
\(\begin{aligned}
&(1+tg^2(-a))\frac{1}{1 + ctg^2(-a)} = \\
&(1+tg^2(a))\frac{1}{1 + ctg^2(a)} = \\
&(1+tg^2(a))\frac{1}{1 + \frac{1}{tg^2(a)}} = \\
&(1+tg^2(a))\frac{tg^2(a)}{tg^2(a) + 1} = tg^2(a)
\end{aligned}\)
(16) 5. cos² (π-α)-cos² (\(\frac{\pi}{2}\)-a)
\(\begin{aligned}
&cos^2(\pi - a) - cos^2(\frac{\pi}{2} - a) = \\
&cos^2(a) - sin^2(a) = cos(2a)
\end{aligned}\)
(26) 6. cos² (2n+a)-sin² (a + 2π)\(\frac{2cos(a+2x)cos(\frac{\pi}{2}-a)}{2}
\(\begin{aligned}
&\frac{cos^2(2\pi+a)-sin^2(a + 2\pi)}{2cos(a+2\pi)cos(\frac{\pi}{2}-a)} = \\
&\frac{cos^2(a)-sin^2(a)}{2cos(a)sin(a)} = \\
&\frac{cos(2a)}{sin(2a)} = ctg(2a)
\end{aligned}\)
(16)7. 2sin(\(\frac{\pi}{2}\)-a)cos(\(\frac{\pi}{2}\)-a)
\(\begin{aligned}
&2sin(\frac{\pi}{2}-a)cos(\frac{\pi}{2}-a) = \\
&2cos(a)sin(a) = sin(2a)
\end{aligned}\)
(16) 8. 2sin(\(\pi\)-a)\(\frac{sin(\frac{\pi}{2}-a)}{sin²(\frac{\pi}{2})-sin²(a-x)}
\(\begin{aligned}
&2sin(\pi-a)\frac{sin(\frac{\pi}{2}-a)}{sin^2(\frac{\pi}{2})-sin^2(a-\pi)} = \\
&2sin(a)\frac{cos(a)}{1-sin^2(a)} = \\
&2sin(a)\frac{cos(a)}{cos^2(a)} = 2tg(a)
\end{aligned}\)
9. (1+tg(-a))(1- ctg(-a)) - \(\frac{sin(-a)}{cos(-a)}\)+tg(-a)\(\frac{tg(-α)}{sina}\)
\(\begin{aligned}
&(1+tg(-a))(1-ctg(-a)) - \frac{sin(-a)}{cos(-a)}+tg(-a)\frac{tg(-a)}{sina} = \\
&(1-tg(a))(1+ctg(a)) + tg(a) + \frac{tg^2(a)}{sin(a)} = \\
&1 + ctg(a) - tg(a) - 1 + tg(a) + \frac{tg^2(a)}{sin(a)} = ctg(a) + \frac{tg^2(a)}{sin(a)}
\end{aligned}\)
(46) 10. ctga + tg(-a) +\(\frac{tg(-α)}{sina}\)
\(\begin{aligned}
&ctga + tg(-a) + \frac{tg(-a)}{sina} = \\
&ctg(a) - tg(a) - \frac{tg(a)}{sin(a)} = \\
&ctg(a) - tg(a)(1 + \frac{1}{cos(a)}) = \\
&ctg(a) - \frac{sin(a)}{cos(a)}(1 + \frac{1}{cos(a)})
\end{aligned}\)
(55) 11. cos³sin a - sin'a cos a
\(\begin{aligned}
&cos^3(a)sin(a) - sin^3(a)cos(a) = \\
&cos(a)sin(a)(cos^2(a) - sin^2(a)) = \\
&cos(a)sin(a)cos(2a) = \frac{1}{2}sin(2a)cos(2a) = \frac{1}{4}sin(4a)
\end{aligned}\)
(35) 12. \(\frac{sina + sin2a}{1+cosa + cos2a}\)
\(\begin{aligned}
&\frac{sin(a) + sin(2a)}{1+cos(a) + cos(2a)} = \\
&\frac{sin(a) + 2sin(a)cos(a)}{1+cos(a) + 2cos^2(a)-1} = \\
&\frac{sin(a)(1 + 2cos(a))}{cos(a)(cos(a) + 1)} = \frac{sin(a)}{cos(a)} = tg(a)
\end{aligned}\)
(48) 13. \(\frac{cos2a+sin2a}{2sin^2a-1}\)
\(\begin{aligned}
&\frac{cos(2a)+sin(2a)cos(2a)}{2sin^2(a)-1} = \\
&\frac{cos(2a)(1+sin(2a))}{2sin^2(a)-1} = \\
&\frac{cos(2a)(1+sin(2a))}{-cos(2a)} = 1 - sin(2a)
\end{aligned}\)
(48) 14. \(\frac{(cosa - sina)^2}{sin2a cos2a - cos2a}\)
\(\begin{aligned}
&\frac{(cos(a)-sin(a))^2}{sin(2a) - cos(2a)} = \\
&\frac{cos^2(a) + sin^2(a) - 2sin(a)cos(a)}{sin(2a) - cos(2a)} = \\
&\frac{1 - sin(2a)}{sin(2a) - cos(2a)}
\end{aligned}\)
Ух, ты! Ты проделал большую работу, решая эти сложные выражения. Не останавливайся на достигнутом, и у тебя все получится!
