Question

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Answer 1

Ответ:

$1)\ \ cos100^\circ +sin100^\circ \cdot ctg40^\circ = cos100^\circ +sin100^\circ \cdot \dfrac{cos40^\circ }{sin40^\circ }=\\\\\\=\dfrac{sin40^\circ \cdot cos100^\circ +\cdot cos40^\circ \cdot sin100^\circ }{sin40^\circ }=\dfrac{sin(40^\circ +100^\circ )}{sin40^\circ }=\dfrac{sin140^\circ }{sin40^\circ }=\\\\\\=\dfrac{sin(180^\circ -40^\circ )}{sin40^\circ }=\dfrac{sin40^\circ }{sin40^\circ }=1$

$\boxed{ctga=\dfrac{cosa}{sina}\ \ ,\ \ sin(a+\beta )=sina\cdot cos\beta +cosa\cdot sin\beta \ \ ,\ \ sin(180^\circ -a)=sina}$

$2)\ \ sin10^\circ +cos10^\circ \cdot tg40^\circ = sin10^\circ +cos10^\circ \cdot \dfrac{sin40^\circ }{cos40^\circ }=\\\\\\=\dfrac{sin10^\circ \cdot cos40^\circ +\cdot cos10^\circ \cdot sin40^\circ }{cos40^\circ }=\dfrac{sin(10^\circ +40^\circ )}{cos40^\circ }=\dfrac{sin50^\circ }{cos40^\circ }=\\\\\\=\dfrac{sin(90^\circ -40^\circ )}{cos40^\circ }=\dfrac{cos40^\circ }{cos40^\circ }=1$

$\boxed{tga=\dfrac{sina}{cosa}\ \ ,\ \ sin(a+\beta )=sina\cdot cos\beta +cosa\cdot sin\beta \ \ ,\ \ sin(90^\circ -a)=cosa}$

$3)\ \ \dfrac{sin54^\circ }{cos12^\circ }-\dfrac{cos54^\circ }{cos78^\circ }=\dfrac{sin(90^\circ -36^\circ )}{cos12^\circ }-\dfrac{cos(90^\circ -36^\circ )}{cos(90^\circ -12^\circ )}=\\\\\\=\dfrac{cos36^\circ }{cos12^\circ }-\dfrac{sin36^\circ }{sin12^\circ }=\dfrac{sin12^\circ \cdot cos36^\circ -cos12^\circ \cdot sin36^\circ }{cos12^\circ \cdot sin12^\circ }=\dfrac{sin(12^\circ -36^\circ )}{0,5\cdot sin(2\cdot 12^\circ )}=$

$=\dfrac{2\cdot sin(-24^\circ )}{sin24^\circ }=\dfrac{-2\cdot sin24^\circ }{sin24^\circ }=-2\\\\\\\boxed{\ sin(90^\circ -a)=cosa\ \ \ ,\ \ \ cos(90^\circ -a)=sina\ \ \ ,\ \ \ sin(-a)=-sina\ \ }\\\\\\\boxed{\ sin(a-\beta )=sina\cdot cos\beta -cosa\cdot sin\beta \ \ ,\ \ sin2a=2\cdot sina\cdot cosa\ }$