Rozwiązane

Trygonometria
Zadanie
Oblicz wartość wyrażenia(alfa) + cos(alfa)



Trygonometria Zadanie Oblicz Wartość Wyrażeniaalfa Cosalfa class=

Odpowiedź :

[tex]\text{tg\,}\alpha=\dfrac{\sin\alpha}{\cos\alpha}\ ,\qquad \text{ctg\,}\alpha=\dfrac{\cos\alpha}{\sin\alpha}\\\\\alpha\in(0^o,90^o)\quad\implies\quad {\sin\alpha\,,\ \cos\alpha\,,\ \text{tg\,}\alpha\,,\ \text{ctg\,}\alpha>0[/tex]

[tex]\\\\ \text{tg\,}\alpha+\text{ctg\,}\alpha=4\\\\ \dfrac{\sin\alpha}{\cos\alpha}+\dfrac{\cos\alpha}{\sin\alpha}=4\\\\ \dfrac{\sin^2\alpha}{\cos\alpha\sin\alpha}+\dfrac{\cos^2\alpha}{\sin\alpha\cos\alpha}=4\\\\ \dfrac{1}{\sin\alpha\cos\alpha}=4\\\\\sin\alpha\cos\alpha=\dfrac14[/tex]

[tex]\sin\alpha+\cos\alpha=x\qquad\qquad\qquad \{\sin\alpha,\,\cos\alpha>0\ \ \implies\ x>0\} \\\\ (\sin\alpha+\cos\alpha)^2=x^2\\\\ \sin^2\alpha+2\sin\alpha\cos\alpha+\cos^2\alpha=x^2\\\\ 1+2\sin\alpha\cos\alpha=x^2\\\\ 1+2\cdot\frac14=x^2\\\\ 1\frac24=x^2\qquad\wedge\quad x>0\\\\ x=\sqrt{1\frac24}=\sqrt{\frac64}=\frac{\sqrt6}{2}\\\\ \boxed{\sin\alpha+\cos\alpha=\frac{\sqrt6}{2}}[/tex]