Question

Ex 7 și 8
Dau coroana

Answer 1

 

$\displaystyle\bf\\7)\\a)\\cos\,65^o+cos\,66^o+cos\,114^o+cos\,115^o\\\\\textbf{Unghiurile de }114^o~si~de~115^o~\textbf{sunt in cadranul 2.}\\\textbf{In cadranul 2 cosinusul este negativ.}\\cos\,114^o=cos(180^o-66^o)=-cos\,66^o\\cos\,115^o=cos(180^o-65^o)=-cos\,65^o\\\\cos\,65^o+cos\,66^o+cos\,114^o+cos\,115^o=\\=cos\,65^o+cos\,66^o-cos\,66^o-cos\,65^o=\boxed{\bf0}$

.

$\displaystyle\bf\\b)\\\Big(sin\,18^o+sin\,31^o+3\Big)\Big(cos\,50^o+cos\,130^o\Big)=\\\\=\Big(sin\,18^o+sin\,31^o+3\Big)\Big(cos\,50^o+cos(180^o-50^o)\Big)=\\\\=\Big(sin\,18^o+sin\,31^o+3\Big)\Big(cos\,50^o-cos\,50^o\Big)=\\\\=\Big(sin\,18^o+sin\,31^o+3\Big)\times0=\boxed{\bf0}$

.

$\displaystyle\bf\\c)\\sin\70^o+cos\,70^o-sin\,110^o+cos\,110^o\\Unghiul~de~110^o~este~in~cadranul~2.\\In~cadranul~2~sinusul~este~pozitiv~si~cosinusul~este~negativ.\\sin\,70^o+cos\,70^o-sin\,110^o+cos\,110^o=\\=sin\,70^o+cos\,70^o-sin\,(180^o-70^o)+cos\,(180^o-70^o)=\\=sin\,70^o+cos\,70^o-sin\,70^o-cos\,70^o=\boxed{\bf0}$

.

$\displaystyle\bf\\8)\\a)\\\\a\in\Big(\frac{\pi}{2},~\pi\Big)\in C2;~~~b\in\Big(\frac{0;~\pi}{2}\Big)\in C1\\\\sin\,a=\frac{12}{13}\implies cos\,a=-\sqrt{1-\Big(\frac{12}{13}\Big)^2}=-\sqrt\frac{13^2-12^2}{13^2}}=-\frac{5}{13}\\\\sin\,b=\frac{4}{5}\implies cos\,b=\sqrt{1-\Big(\frac{4}{5}\Big)^2}=\sqrt\frac{5^2-4^2}{5^2}}=\frac{3}{5}$

.

$\displaystyle\bf\\sin(a+b)=sin\,a\,cos\,b+sin\,b\,cos\,a=\frac{12}{13}\times\frac{3}{5}+\frac{4}{5}\times\Big(-\frac{5}{13}\Big)=\\\\=\frac{12}{13}\times\frac{3}{5}-\frac{4}{5}\times\frac{5}{13}=\\\\=\frac{12\times3-4\times5}{5\times13}=\frac{36-20}{65}=\boxed{\bf\frac{16}{65}}$

.

$\displaystyle\bf\\sin(a-b)=sin\,a\,cos\,b-sin\,b\,cos\,a=\frac{12}{13}\times\frac{3}{5}-\frac{4}{5}\times\Big(-\frac{5}{13}\Big)=\\\\=\frac{12}{13}\times\frac{3}{5}+\frac{4}{5}\times\frac{5}{13}=\\\\=\frac{12\times3+4\times5}{5\times13}=\frac{36+20}{65}=\boxed{\bf\frac{56}{65}}$

.

$\displaystyle\bf\\cos(a-b)=cos\,a\,cos\,b+sin\,a\,sin\,b=-\frac{5}{13}\times\frac{3}{5}+\frac{12}{13}\times\frac{4}{5}=\\\\=\frac{-5\times3+12\times4}{5\times13}=\frac{-15+48}{65}=\boxed{\bf-\frac{33}{65}}\\\\\\cos(a+b)=cos\,a\,cos\,b-sin\,a\,sin\,b=-\frac{5}{13}\times\frac{3}{5}-\frac{12}{13}\times\frac{4}{5}=\\\\=\frac{-5\times3-12\times4}{5\times13}=\frac{-15-48}{65}=\boxed{\bf-\frac{63}{65}}$

.

$\displaystyle\bf\\8)\\b)\\\\a\in\Big(\pi,~\frac{3\pi}{2}\Big)\in C3;~~~b\in\Big(\frac{0;~\pi}{2}\Big)\in C1\\\\sin\,a=-\frac{3}{5}\implies cos\,a=-\sqrt{1-\Big(-\frac{3}{5}\Big)^2}=-\sqrt\frac{5^2-(-3)^2}{5^2}}=-\frac{4}{5}\\\\cos\,b=\frac{5}{13}\implies sin\,b=\sqrt{1-\Big(\frac{5}{13}\Big)^2}=\sqrt\frac{13^2-5^2}{13^2}}=\frac{12}{13}$

.

$\displaystyle\bf\\sin(a+b)=sin\,a\,cos\,b+sin\,b\,cos\,a=-\frac{3}{5}\times\frac{5}{13}+\frac{12}{13}\times\Big(-\frac{4}{5}\Big)=\\\\=-\frac{3}{5}\times\frac{5}{13}-\frac{12}{13}\times\frac{4}{5}=\\\\=\frac{-3\times5-12\times4}{5\times13}=\frac{-15-48}{65}=\boxed{\bf-\frac{63}{65}}$

.

$\displaystyle\bf\\sin(a-b)=sin\,a\,cos\,b-sin\,b\,cos\,a=-\frac{3}{5}\times\frac{5}{13}-\frac{12}{13}\times\Big(-\frac{4}{5}\Big)=\\\\=-\frac{3}{5}\times\frac{5}{13}+\frac{12}{13}\times\frac{4}{5}=\\\\=\frac{-3\times5+12\times4}{5\times13}=\frac{-15+48}{65}=\boxed{\bf\frac{33}{65}}$

.

$\displaystyle\bf\\cos(a-b)=cos\,a\,cos\,b+sin\,a\,sin\,b=-\frac{4}{5}\times\frac{5}{13}+\frac{-3}{5}\times\frac{12}{13}=\\\\=\frac{-4\times5-3\times12}{5\times13}=\frac{-20-36}{65}=\boxed{\bf-\frac{56}{65}}\\\\\\cos(a+b)=cos\,a\,cos\,b-sin\,a\,sin\,b=-\frac{4}{5}\times\frac{5}{13}-\frac{-3}{5}\times\frac{12}{13}=\\\\=\frac{-4\times5+3\times12}{5\times13}=\frac{-20+36}{65}=\boxed{\bf\frac{16}{65}}$