Question

sa se calculeze valorile trigonometrie sinx,cosx,tgx pentru unghiurile ch măsura 3pi/8

Answer 1

Răspuns:

Explicație pas cu pas:

$\dfrac{3\pi }{8}=\dfrac{4\pi -\pi }{8} =\dfrac{4\pi }{8}-\dfrac{\pi }{8}=\dfrac{\pi }{2}- \dfrac{\pi }{8}\\sin(\dfrac{\pi }{2}- \dfrac{\pi }{8})=cos\dfrac{\pi }{8}=\sqrt{\frac{1+cos(2*\frac{\pi }{8} )}{2} } = \sqrt{\frac{1+cos(\frac{\pi }{4} )}{2} } =\sqrt{(1+ \frac{\sqrt{2} }{2} ):2}=\sqrt{\frac{2+\sqrt{2} }{4} }=\frac{\sqrt{2+\sqrt{2} } }{2}.$

$cos(\dfrac{\pi }{2}- \dfrac{\pi }{8})=sin\dfrac{\pi }{8}=\sqrt{\frac{1-cos(2*\frac{\pi }{8} )}{2} } = \sqrt{\frac{1-cos(\frac{\pi }{4} )}{2} } =\sqrt{(1- \frac{\sqrt{2} }{2} ):2}=\sqrt{\frac{2-\sqrt{2} }{4} }=\frac{\sqrt{2-\sqrt{2} } }{2}.$

$tg(\frac{\pi }{2}-\frac{\pi }{8})=ctg\frac{\pi }{8}=\dfrac{cos\frac{\pi }{8} }{sin\frac{\pi }{8} }=$

$=\frac{\sqrt{2+\sqrt{2} } }{2}: \frac{\sqrt{2-\sqrt{2} } }{2}=\sqrt{\frac{2+\sqrt{2} }{2-\sqrt{2} } }=\sqrt{\frac{(2+\sqrt{2})^{2} }{(2-\sqrt{2})(2+\sqrt{2}) } }=\frac{2+\sqrt{2} }{\sqrt{2} }=\frac{\sqrt{2}(2+\sqrt{2})}{2}$

$ctg(\frac{\pi }{2}-\frac{\pi }{8})=tg\frac{\pi }{8}=\frac{2}{\sqrt{2}(2+\sqrt{2})} =\frac{2\sqrt{2}(2-\sqrt{2})}{2(2+\sqrt{2})(2-\sqrt{2})}=\frac{\sqrt{2}(2-\sqrt{2})}{2}$