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\(\begin{array}{l}\cos 75^\circ = \frac{{\sqrt 6 - \sqrt 2 }}{4}\\\tan \left( { - \frac{{19\pi }}{6}} \right) = - \frac{{\sqrt 3 }}{3}\end{array}\)
a) \(\cos \frac{{3\pi }}{7} = 0,22252\); \(\tan ( - {37^ \circ }25') = 0,765018\)
b) \(179^o23'30"\approx3,130975234\left(rad\right)\)
c) \(\frac{{7\pi }}{9} = {140^ \circ }\)
\(\begin{array}{l}A = \sin \left( {a - 17^\circ } \right)\cos \left( {a + 13^\circ } \right) - \sin \left( {a + 13^\circ } \right)\cos \left( {a - 17^\circ } \right)\\A = \sin \left( {a - 17^\circ - a - 13^\circ } \right) = \sin \left( { - 30^\circ } \right) = - \frac{1}{2}\end{array}\)
\(\begin{array}{l}B = \cos \left( {b + \frac{\pi }{3}} \right)\cos \left( {\frac{\pi }{6} - b} \right) - \sin \left( {b + \frac{\pi }{3}} \right)\sin \left( {\frac{\pi }{6} - b} \right)\\B = \cos \left( {b + \frac{\pi }{3} + \frac{\pi }{6} - b} \right) = \cos \frac{\pi }{2} = 0\end{array}\)
\(a,cos\left(\dfrac{5\pi}{12}\right)=cos\left(\dfrac{\pi}{4}+\dfrac{\pi}{6}\right)=cos\left(\dfrac{\pi}{4}\right)cos\left(\dfrac{\pi}{6}\right)-sin\left(\dfrac{\pi}{4}\right)sin\left(\dfrac{\pi}{6}\right)=\dfrac{\sqrt{2}}{2}\cdot\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{2}}{2}\cdot\dfrac{1}{2}=\dfrac{\sqrt{6}-\sqrt{2}}{4}\\ sin\left(\dfrac{5\pi}{12}\right)=sin\left(\dfrac{\pi}{4}+\dfrac{\pi}{6}\right)=sin\left(\dfrac{\pi}{4}\right)cos\left(\dfrac{\pi}{6}\right)+cos\left(\dfrac{\pi}{4}\right)sin\left(\dfrac{\pi}{6}\right)=\dfrac{\sqrt{2}}{2}\cdot\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{2}\cdot\dfrac{1}{2}=\dfrac{\sqrt{6}+\sqrt{2}}{4}\\ tan\left(\dfrac{5\pi}{12}\right)=\dfrac{sin\left(\dfrac{5\pi}{12}\right)}{cos\left(\dfrac{5\pi}{12}\right)} =2-\sqrt{3}\\ cot\left(\dfrac{5\pi}{12}\right)=\dfrac{1}{tan\left(\dfrac{5\pi}{12}\right)}=\dfrac{1}{2-\sqrt{3}}\)
\(b,cos\left(-555^o\right)=cos\left(3\pi+\dfrac{\pi}{12}\right)=-cos\left(\dfrac{\pi}{12}\right)=-cos\left(\dfrac{\pi}{3}-\dfrac{\pi}{4}\right)=-\left[cos\left(\dfrac{\pi}{3}\right)cos\left(\dfrac{\pi}{4}\right)+sin\left(\dfrac{\pi}{3}\right)sin\left(\dfrac{\pi}{4}\right)\right]=-\dfrac{\sqrt{6}+\sqrt{2}}{4}\\ sin\left(-555^o\right)=sin\left(3\pi+\dfrac{\pi}{12}\right)=sin\left(\dfrac{\pi}{12}\right)=sin\left(\dfrac{\pi}{3}-\dfrac{\pi}{4}\right)=sin\left(\dfrac{\pi}{3}\right)cos\left(\dfrac{\pi}{4}\right)-cos\left(\dfrac{\pi}{3}\right)sin\left(\dfrac{\pi}{4}\right)=\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}\cdot\dfrac{\sqrt{2}}{2}=\dfrac{\sqrt{6}-\sqrt{2}}{4}\\ tan\left(-555^o\right)=\dfrac{sin\left(-555^o\right)}{cos\left(-555^o\right)}=-2+\sqrt{3}\\ cot\left(-555^o\right)=\dfrac{1}{tan\left(-555^o\right)}=\dfrac{1}{-2+\sqrt{3}}=-2-\sqrt{3}\)
c.
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\(\Leftrightarrow cot\left(2x-\frac{3\pi}{4}\right)=cot\left(\frac{2\pi}{3}-x\right)\)
\(\Leftrightarrow2x-\frac{3\pi}{4}=\frac{2\pi}{3}-x+k\pi\)
\(\Leftrightarrow x=\frac{17\pi}{36}+\frac{k\pi}{3}\)
d.
\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)=cos\left(\frac{3\pi}{4}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{3}=\frac{3\pi}{4}-x+k2\pi\\2x+\frac{\pi}{3}=x-\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{36}+\frac{k2\pi}{3}\\x=-\frac{13\pi}{12}+k2\pi\end{matrix}\right.\)
a.
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\(\Leftrightarrow tan\left(3x-\frac{\pi}{3}\right)=tan\left(-x\right)\)
\(\Leftrightarrow3x-\frac{\pi}{3}=-x+k\pi\)
\(\Leftrightarrow x=\frac{\pi}{12}+\frac{k\pi}{4}\)
b.
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\(\Leftrightarrow cot\left(x-\frac{\pi}{4}\right)=cot\left(-x\right)\)
\(\Leftrightarrow x-\frac{\pi}{4}=-x+k\pi\)
\(\Leftrightarrow x=\frac{\pi}{8}+\frac{k\pi}{2}\)
\(\begin{array}{l}\sin \left( { - \frac{{2\pi }}{3}} \right) = - \frac{{\sqrt 3 }}{2}\\\tan 495^\circ = - 1\end{array}\)
a) \(\tan ( - {75^ \circ }) = - 2 - \sqrt 3 \)
b) \(\cot \left( { - \frac{\pi }{5}} \right) \approx - 1,376\)