Найдите значения: sina/2, cosa/2, tga/2, если sina 14/50 и П/2 <а<П. ​

[tex] \sin(a) = \frac{14}{50} [/tex]

[tex] \frac{\pi}{2} < a < \pi [/tex]

a принадлежит второй четверти, в которой косинус отрицателен,

тогда [tex] \cos(a) < 0 [/tex]

[tex] \cos^2(a) + \sin^2(a) \equiv 1 [/tex]

[tex] \cos^2(a) \equiv 1 - \sin^2(a) [/tex]

[tex] \cos(a) = \pm\sqrt{1 - \sin^2(a)} [/tex]

но т.к. [tex] \cos(a) < 0 [/tex], то

[tex] \cos(a) = -\sqrt{1-\sin^2(a)} [/tex]

[tex] \cos(a) = -\sqrt{1 - \left( \frac{14}{50}\right)^2} = [/tex]

[tex] = -\sqrt{1 - \frac{14^2}{50^2}} = -\sqrt{1 - \frac{196}{2500}} = [/tex]

[tex] = -\sqrt{\frac{2500 - 196}{2500}} = -\sqrt{\frac{2304}{2500}} = [/tex]

[tex] = -\frac{\sqrt{2304}}{\sqrt{2500}} = -\frac{48}{50} [/tex]

итак [tex] \cos(a) = -\frac{48}{50} [/tex]

теперь найдем [tex] \sin(\frac{a}{2}) [/tex] и [tex] \cos(\frac{a}{2}) [/tex].

[tex] \cos(a) = \cos^2(\frac{a}{2}) - \sin^2(\frac{a}{2}) = [/tex]

[tex] = 1 - \sin^2(\frac{a}{2}) - \sin^2(\frac{a}{2}) = [/tex]

[tex] = 1 - 2\sin^2(\frac{a}{2}) [/tex]

[tex] 2\sin^2(\frac{a}{2}) = 1 - \cos(a) [/tex]

[tex] \sin^2(\frac{a}{2}) = \frac{1 - \cos(a)}{2} [/tex]

[tex] \sin(\frac{a}{2} = \pm\sqrt{\frac{1-\cos(a)}{2}} [/tex]

[tex] \frac{\pi}{2} < a < \pi [/tex] ⇔ [tex] \frac{\pi}{4} < \frac{a}{2} < \frac{\pi}{2} [/tex]

а это значит [tex] \frac{a}{2} [/tex] принадлежит первой четверти в которой и синус и косинус положительны, поэтому

[tex] \sin(\frac{a}{2}) = \sqrt{\frac{1 - \cos(a)}{2}} = [/tex]

[tex] = \sqrt{\frac{1 - (-\frac{48}{50})}{2}} = [/tex]

[tex] = \sqrt{\frac{1 + \frac{48}{50}}{2}} = [/tex]

[tex] = \sqrt{\frac{50 + 48}{2\cdot 50}} = [/tex]

[tex] = \sqrt{\frac{98}{100}} = \sqrt{\frac{49}{50}} = \frac{\sqrt{49}}{\sqrt{50}} = [/tex]

[tex] = \frac{7}{5\cdot\sqrt{2}} = \frac{7\sqrt{2}}{10} [/tex]

[tex] \cos(a) = \cos^2(\frac{a}{2}) - \sin^2(\frac{a}{2}) = [/tex]

[tex] = \cos^2(\frac{a}{2}) - (1 - \cos^2(\frac{a}{2})) = [/tex]

[tex] = 2\cos^2(\frac{a}{2}) - 1 [/tex]

[tex] 2\cos^2(\frac{a}{2}) = \cos(a) + 1 [/tex]

[tex] \cos^2(\frac{a}{2}) = \frac{\cos(a) + 1}{2}  = \frac{-\frac{48}{50} + 1}{2} = [/tex]

[tex] = \frac{-48 + 50}{2\cdot 50} = \frac{2}{2\cdot 50} = \frac{1}{50} [/tex]

[tex] \cos(\frac{a}{2}) = \sqrt{\frac{1}{50}} = \frac{1}{\sqrt{50}} = \frac{1}{5\sqrt{2}} = [/tex]

[tex] = \frac{\sqrt{2}}{10} [/tex]

[tex] \mathrm{tg}(\frac{a}{2}) = \frac{\sin(\frac{a}{2})}{\cos(\frac{a}{2})} = [/tex]

[tex] = \frac{\frac{7\sqrt{2}}{10}}{\frac{\sqrt{2}}{10}} = 7 [/tex]

Ответ:

<a<π

a принадлежит второй четверти, в которой косинус отрицателен,

тогда \cos(a) < 0cos(a)<0

\cos^2(a) + \sin^2(a) \equiv 1cos2(a)+sin2(a)≡1

\cos^2(a) \equiv 1 - \sin^2(a)cos2(a)≡1−sin2(a)

\cos(a) = \pm\sqrt{1 - \sin^2(a)}cos(a)=±1−sin2(a)

но т.к. \cos(a) < 0cos(a)<0 , то

\cos(a) = -\sqrt{1-\sin^2(a)}cos(a)=−1−sin2(a)

\cos(a) = -\sqrt{1 - \left( \frac{14}{50}\right)^2} =cos(a)=−1−(5014)2=

= -\sqrt{1 - \frac{14^2}{50^2}} = -\sqrt{1 - \frac{196}{2500}} ==−1−502142=−1−2500196=

= -\sqrt{\frac{2500 - 196}{2500}} = -\sqrt{\frac{2304}{2500}} ==−25002500−196=−25002304=

= -\frac{\sqrt{2304}}{\sqrt{2500}} = -\frac{48}{50}=−25002304=−5048

итак \cos(a) = -\frac{48}{50}cos(a)=−5048

теперь найдем \sin(\frac{a}{2})sin(2a) и \cos(\frac{a}{2})cos(2a) .

\cos(a) = \cos^2(\frac{a}{2}) - \sin^2(\frac{a}{2}) =cos(a)=cos2(2a)−sin2(2a)=

= 1 - \sin^2(\frac{a}{2}) - \sin^2(\frac{a}{2}) ==1−sin2(2a)−sin2(2a)=

= 1 - 2\sin^2(\frac{a}{2})=1−2sin2(2a)

2\sin^2(\frac{a}{2}) = 1 - \cos(a)2sin2(2a)=1−cos(a)

\sin^2(\frac{a}{2}) = \frac{1 - \cos(a)}{2}sin2(2a)=21−cos(a)

\sin(\frac{a}{2} = \pm\sqrt{\frac{1-\cos(a)}{2}}sin(2a=±21−cos(a)

\frac{\pi}{2} < a < \pi2π<a<π ⇔ \frac{\pi}{4} < \frac{a}{2} < \frac{\pi}{2}4π<2a<

не за что