Question

Fórmulas de arco duplo

Answer 1

$\large\boxed{\begin{array}{l}\underline{\sf seno,cosseno\,e\,tangente}\\\underline{\sf da\,soma\,de\,arcos }\\\sf sendo\,\bf{a}~\rm e\bf{\,b}\,\sf dois\,arcos\,quaisquer\\\sf temos:\\\sf sen(a+b)=sen(a)\cdot cos(b)+sen(b)\cdot cos(a)\\\sf cos(a+b)=cos(a)\cdot cos(b)-sen(a)\cdot sen(b)\\\sf tg(a+b)=\dfrac{ tg(a)+tg(b)}{1-tg(a)\cdot tg(b)}\end{array}}$

$\Large\boxed{\begin{array}{l}\underline{\sf F\acute ormulas\,do\,arco\,duplo}\\\sf fazendo\,b=a~,vamos\,substituir\\\sf nas\,f\acute ormulas\,anteriores:\\\sf sen(a+a)=sen(a)\cdot cos(a)+sen (a)\cdot cos(a)\\\boxed{\boxed{\boxed{\sf sen(2a)=2\cdot sen(a)\cdot cos(a)}}}\\\sf cos(a+a)=cos(a)\cdot cos(a)-sen(a)\cdot sen(a)\\\boxed{\boxed{\boxed{\sf cos(2a)=cos^2(a)-sen^2(a)}}}\\\sf tg(a+a)=\dfrac{tg(a)+tg(a)}{1-tg(a)\cdot tg(a)}\\\boxed{\boxed{\boxed{\sf tg(2a)=\dfrac{2tg(a)}{1-tg^2(a)}}}}\end{array}}$