докажите тождество - страница 37
Докажите тождество: \(sin^2(\frac{15\pi}{8}-2\alpha) - cos^2(\frac{17\pi}{8}-2\alpha)= -\frac{cos4\alpha}{\sqrt2} \)
Решение: $$ sin(\frac{15\pi }{8}-2 \alpha )=sin(\frac{16\pi -\pi }{8}-2 \alpha )=sin(2\pi -\frac{\pi}{8}-2 \alpha )=\\\\=sin(-\frac{\pi}{8}-2 \alpha )=-sin(\frac{\pi}{8}+2 \alpha )\\\\cos(\frac{17\pi }{8}-2 \alpha )=cos(\frac{16\pi +\pi }{8}-2 \alpha )=cos(2\pi +\frac{\pi}{8}-2 \alpha )=cos(\frac{\pi}{8}-2 \alpha ) \\ sin^2(\frac{15\pi}{8}-2 \alpha )-cos^2(\frac{17\pi}{8}-2 \alpha )=sin^2(\frac{\pi}{8}+2 \alpha )-cos^2(\frac{\pi}{8}-2 \alpha )=\\\\=(sin (\frac{\pi}{8}+2 \alpha )-cos(\frac{\pi}{8}-2 \alpha ))\cdot (sin(\frac{\pi}{8}+ \alpha )+cos(\frac{\pi}{8}-2 \alpha ))=\\\\=[\, \frac{\pi}{8}-2 \alpha =\frac{\pi}{2}-(\frac{3\pi}{8}+2 \alpha )\, ]=\\\\=(sin(\frac{\pi}{8}+2 \alpha )-sin(\frac{3\pi}{8}+2 \alpha ))\cdot (sin(\frac{\pi}{8}+2 \alpha )+sin(\frac{3\pi}{8}+2 \alpha ))= \\ =2sin(-\frac{\pi}{8})\cdot cos(\frac{\pi}{4}+2 \alpha )\cdot 2sin(\frac{\pi}{4}+2 \alpha )cos(\frac{-\pi}{8})=\\\\=(-2sin\frac{\pi}{8}\cdot cos\frac{\pi}{8})\cdot (2sin(\frac{\pi}{4}+2 \alpha )\cdot cos(\frac{\pi}{4}+2 \alpha ))=\\\\=-sin\frac{\pi}{4}\cdot sin(\frac{\pi}{2}+4 \alpha )=-\frac{\sqrt2}{2}\cdot (-cos4 \alpha )= \frac{cos4 \alpha }{\sqrt2} $$Докажите тождество:
sin15° + tg30° * cos15° = √6/3
Решение: Для того, чтобы доказать тождество, для начала решим его левую часть.
sin15° + tg30° * cos15°=
$$ \frac{ \sqrt{3}-1 }{ 2\sqrt{2} } + \frac{1}{ \sqrt{3} } * \frac{ \sqrt{3}+1 }{2 \sqrt{2} } =\frac{ \sqrt{3}-1 }{ 2\sqrt{2} } + \frac{ \sqrt{3} +1}{2 \sqrt{6} } =\\= \frac{3- \sqrt{3}+ \sqrt{3} +1 }{2 \sqrt{6} } = \frac{4}{2 \sqrt{6} } = \frac{2}{ \sqrt{6} } = \frac{2 \sqrt{6} }{6} = \frac{ \sqrt{6} }{3} \\ \frac{ \sqrt{6} }{3} = \frac{ \sqrt{6} }{3} $$
Тождество доказано.Докажите тождество sinx+cosx tgx/cosx+sinx tgx=2tgx
Решение: Будем делать по частям.
сначала числитель:
Sin x + Cos x tg x = Sin x +Cos x · Sin x/Cos x= Sin x + Sin x = 2 Sin x
Теперь знаменатель:
Cos x + Sin x tg x = Cos x + Sin x· Sin x /Cos x= (Cos ² x +Sin² x)/ Сos x=
=1/Cos
Теперь сама дробь: 2Sin x : 1/Cos x = 2Sin x Cos x= Sin 2x
$$ \frac{sinx+coxtgx}{cosx+sinxtgx} = \frac{sinx+cosx \frac{sinx}{cosx} }{cosx+sinx \frac{sinx}{cosx} } = \frac{sinx+sinx}{cosx+ \frac{sin ^{2}x }{cosx} } = \frac{2sinx}{ \frac{cos ^{2} x+sin ^{2} x}{cosx} } = \\ = \frac{2sinx}{ \frac{1}{cosx} } =2sinxcosx=sin2x $$
Докажите тождество: tg(в квадрате) a-sin(в квадрате) a= tg (в квадрате)a * sin (в квадрате) а
Решение: Приведём левую часть к виду правой$$ {tg^{2}\alpha - sin^{2}\alpha} = \\ {sin^{2}\alpha \over cos^{2}\alpha} - sin^{2}\alpha = \\ {sin^{2}\alpha \over cos^{2}\alpha} - {sin^{2}\alpha \cdot cos^{2}\alpha\over cos^{2}\alpha} =\\ $$
$$ {sin^{2}\alpha - sin^{2}\alpha \cdot cos^{2}\alpha \over cos^{2}\alpha} = \\ {sin^{2}\alpha(1 - cos^{2}\alpha) \over cos^{2}\alpha}=\\ {sin^{2}\alpha \cdot sin^{2}\alpha \over cos^{2}\alpha} = \\ $$
$$ {({sin^{2}\alpha \over cos^{2}\alpha}) \cdot sin^{2}\alpha}=\\ {tg^{2}\alpha \cdot sin^{2}\alpha.} $$
Что и требовалось доказать.
1) известно что:\( sin(\frac{\pi}{2} +\alpha)=-\frac{1}{2}; \pi < \alpha < \frac{3\pi}{2} \) найти sin(30 + a)
2) упростите выражения \( a) 2cos^2\alpha-(tg\alpha cos\alpha)^2 - (ctg\alpha sin\alpha)^2;\\ b) \frac{sin\alpha + sin3\alpha}{cos\alpha + cos3\alpha}(1+cos4\alpha) \)
3) докажите тождество \( ctg2\alpha -\frac{2tg\alpha}{1+tg^2\alpha}=cos2\alpha \)
Решение: $$ sin(\frac{\pi}{2}+a)=-\frac{1}{2}\\ sin(30+a)\\ \\ sin(\frac{\pi}{2}+a)=cosa\\ cosa=-\frac{1}{2}\\ a=-\frac{2\pi}{3}+2\pi*n=\frac{4pi}{3}\\\\ sin(30+240)=sin(270)=-1\\\\ 2cos^2a-(tga*cosa)^2-(ctga*sina)^2=\\ 2cos^2a-(\frac{sina}{cosa}*cosa)^2-(\frac{cosa}{sina}*sina)^2=\\ 2cos^2a-sin^2a-cos^2a=\\ cos^2a-sin^2a=cos2a\\ \\ \\ \frac{sina+sin3a}{cosa+cos3a}*(1+cos4a)=\\ \frac{2sin2a*cosa}{2cos2a*cosa}*(1+cos4a)=\\ \frac{sin2a}{cos2a}*(1+cos(2*2a))=\\ \frac{sin2a}{cos2a}*(2cos^2(2a))=\\ 2sin2a*cos2a=sin4a\\ \\ ctg2a*\frac{2tga}{1+tg^2a}=\\ ctg2a*sin2a=\\ \frac{cos2a}{sin2a}*sin2a=cos2a $$
1)
$$ \sin(\frac{\pi}{2}+\alpha)=-\frac{1}{2}; \\ \pi<\alpha<\frac{3\pi}{2};\ \alpha\in(\pi;\frac{2\pi}{3});\\ \sin(30^0+\alpha)-;\\ \frac{\pi}{2}+\alpha=(-1)^n\arcsin(-\frac{1}{2})+\pi n;\ n\in Z\\ \alpha=-(-1)^n\arcsin(\frac{1}{2})-\frac{\pi}{2}+\pi n;\\ \alpha=(-1)^{n+1}\cdot\frac{\pi}{6}-\frac{\pi}{2}+\pi n;\\ n=2: \ \alpha=-\frac{\pi}{6}-\frac{\pi}{2}+2\pi=\frac{-\pi-3\pi+12\pi}{6}=\frac{8\pi}{6}=\frac{4\pi}{3}\in(\pi;\frac{2\pi}{3});\\ \\ \sin(30^0+\alpha)=\sin(\frac{\pi}{6}+\frac{4\pi}{3})=\sin(\frac{\pi+8\pi}{6})=\\ =\sin\frac{9\pi}{6}=\sin{\frac{3\pi}{2}}-1;\\ or: \sin\frac{\pi}{6}\cos\frac{4\pi}{3}+\sin{\frac{4\pi}{3}}\cos\frac{\pi}{6}=\\ =\frac{1}{2}\cdot(-\frac{1}{2})+(-\frac{\sqrt{3}}{2})\cdot\frac{\sqrt{3}}{2}=-\frac{1}{4}-\frac{3}{4}=-\frac{4}{4}=-1 \\ \sin(30^0+\alpha)=-1;\\ $$
2)
a)
$$ 2\cos^2\alpha-(tg\alpha\cdot\cos\alpha)^2-(ctg\alpha\cdot\sin\alpha)^2=\\ =2\cos^2\alpha-\left(\left(\frac{\sin\alpha}{\cos\alpha}\cdot\cos\alpha\right)^2+\left(\frac{\cos\alpha}{\sin\alpha}\cdot sin\alpha\right)^2\right)=\\ =2\cos^2\alpha-\left(\sin^2\alpha+\cos^2\alpha\right)=2\cos^2\alpha-1=\cos2\alpha;\\ $$
б)
$$ \frac{\sin\alpha+\sin3\alpha}{\cos\alpha+\cos3\alpha}\cdot(1+\cos4\alpha)=\\ =\frac{\sin\alpha+\sin(\alpha+2\alpha)}{\cos\alpha+\cos(\alpha+2\alpha)}\cdot(1+2\cos^22\alpha-1)=\\ =\frac{\sin\alpha+\sin\alpha\cos2\alpha+\cos\alpha\sin2\alpha}{\cos\alpha+\cos\alpha\cos2\alpha-\sin\alpha\sin2\alpha}\cdot2\cos^22\alpha=\\ ==\frac{\sin\alpha+\sin\alpha\cos2\alpha+2\cos^2\alpha\sin\alpha}{\cos\alpha+\cos\alpha\cos2\alpha-2\sin^2\alpha\cos\alpha}\cdot2\cos^22\alpha=\\ \\ =\frac{\sin\alpha(1+\cos2\alpha+2\cos^2\alpha)}{\cos\alpha(1+\cos2\alpha-2\sin^2\alpha)}\cdot2\cos^22\alpha=\\ \frac{\sin\alpha(1+2\cos^2\alpha-1+2\cos^2\alpha)}{\cos\alpha(\cos2\alpha+1-2\sin^2\alpha)}\cdot2\cos^22\alpha=\\ =\frac{\sin\alpha\cdot4\cos^2\alpha}{\cos\alpha\cdot2\cos2\alpha}\cdot2\cos^22\alpha=\\ =4\sin\alpha\cos\alpha\cos2\alpha=2\sin2\alpha2\cos2\alpha=\sin4\alpha;\\ $$
3)
$$ ctg2\alpha\frac{2tg\alpha}{1+tg^2\alpha}=\cos2\alpha;\\ \frac{\cos2\alpha}{\sin2\alpha}\cdot\frac{2\frac{sin\alpha}{\cos\alpha}}{1+(\frac{\sin\alpha}{\cos\alpha})^2}=\frac{\cos2\alpha}{\sin2\alpha}\cdot\frac{2\frac{\sin\alpha}{\cos\alpha}}{\frac{\cos^2\alpha+\sin^2\alpha}{\cos^2\alpha}}=2\cdot\frac{\cos2\alpha}{\sin2\alpha}\cdot\frac{\frac{\sin\alpha}{\cos\alpha}}{\frac{1}{\cos^2\alpha}}=\\ =2\cdot\frac{\cos2\alpha}{\sin2\alpha}\cdot\frac{\sin\alpha\cdot\cos^2\alpha}{\cos\alpha}= \\ =2\cdot\frac{\cos2\alpha}{\sin2\alpha}\cdot\sin\alpha\cdot\cos\alpha=\frac{\cos2\alpha}{\sin2\alpha}\cdot2\sin\alpha\cos\alpha=\\=\frac{\cos2\alpha}{\sin2\alpha}\cdot\sin2\alpha=\cos2\alpha; $$
