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Question 1

Question 2

Решите задачу:

$4) \frac{cos^{-2}60^ \circ}{2sin135^\circ+1} = \frac{( \frac{1}{2} )^{-2} }{2* \frac{ \sqrt{2} }{2} +1} = \frac{2^{2} }{ \sqrt{2} +1} =\frac{4 }{ \sqrt{2} +1} \\ \\ 5) \frac{6sin2^ {\circ}sin88^ \circ}{sin176^ \circ} =\frac{6sin(90^ {\circ}-88^ {\circ})sin88^ \circ}{2cos88^ {\circ}sin88^ \circ} =\frac{6cos88^ {\circ}sin88^ \circ}{2cos88^ {\circ}sin88^ \circ} =3 \\ \\ 6)cos^2a-sin^2a=cos^22a$

$7)8 \sqrt{3} sin \frac{ \pi }{24}*cos \frac{ \pi }{24}*cos \frac{ \pi }{12}=4 \sqrt{3} sin \frac{2 \pi }{24}*cos \frac{ \pi }{12}= \\ =4 \sqrt{3} sin \frac{ \pi }{12}*cos \frac{ \pi }{12}=2 \sqrt{3} sin \frac{ 2\pi }{12}=2\sqrt{3} sin \frac{ \pi }{6}=2 \sqrt{3} * \frac{1}{2} =\sqrt{3$

$8) \frac{1 -2sin10^\circ*cos80^\circ}{ cos20^\circ} = \frac{1 -2sin10^\circ*cos(90^\circ-10^\circ)}{ cos20^\circ} = \frac{1 -2sin10^\circ*sin10^\circ}{ cos20^\circ} = \\ =\frac{1 -2sin^210^\circ}{ cos20^\circ}=\frac{cos20^\circ}{ cos20^\circ}=1 \\ \\ 9)3sin \frac{ \pi }{8} cos \frac{ \pi }{8} = \frac{3*2sin \frac{ \pi }{8} cos \frac{ \pi }{8} }{2} = \frac{3*sin \frac{2 \pi }{8} }{2} = \frac{3*sin \frac{ \pi }{4} }{2} = \frac{3* \frac{ \sqrt{2} }{2} }{2} = \frac{3 \sqrt{2} }{4}$

$10) cos43^\circ+cos47^\circ=2*cos \frac{43^\circ+47^\circ}{2} *cos \frac{43^\circ-47^\circ}{2} = \\ =2*cos 45^\circ *cos (-2^\circ) =2* \frac{ \sqrt{2} }{2} *cos 2^\circ= \sqrt{2}cos 2^\circ$