Question

6.26)
Proszę oo wytłumaczenie może być na kartce zdj byle by zrozumieć
Dam najjjj

Answer 1

Verified answer

[tex]\huge\boxed{\begin{array}{llll}a)&\alpha=60^\circ,&b)&\alpha=30^\circ\\c)&\alpha=45^\circ,&d)&\alpha=30^\circ\end{array}}[/tex]

Funkcje trygonometryczne kątów ostrych

Jeżeli kąt jest ostry, to wszystkie funkcje trygonometryczne tego kąta są dodatnie.

[tex]\underline{\bold{D: sin\alpha, cos\alpha, tg\alpha, ctg\alpha > 0}}[/tex]

Przypomnijmy podstawowe wzory trygonometryczne:

[tex]\boxed{\begin{array}{ll}\text{Jedynka trygonometryczna:}&sin^2\alpha+cos^2\alpha=1\\\\\text{Wzory na tangens: }&tg\alpha=\dfrac{sin\alpha}{cos\alpha}\\\\&tg\alpha\cdot ctg\alpha=1\end{array}}[/tex]

a)

[tex]6sin^2\alpha=5-cos\alpha\\\\6(1-cos^2\alpha)=5-cos\alpha\\\\6-6cos^2\alpha=5-cos\alpha\\\\-6cos^2\alpha+cos\alpha+6-5=0\\\\-6cos^2\alpha+cos\alpha+1=0\\\\\Delta=1^2-4\cdot (-6)\cdot 1=1+24=25\\\\\sqrt{\Delta}=\sqrt{25}=5\\\\cos\alpha_1=\dfrac{-1-5}{2\cdot (-6)}=\dfrac{-6}{-12}=\dfrac12\\\\cos\alpha_2=\dfrac{-1+5}{2\cdot (-6)}=\dfrac4{-12}=-\dfrac13 \notin D[/tex]

Wartość kąta α odczytujemy z tablic wartości funkcji trygonometrycznych:

[tex]\boxed{\bold{cos\alpha=\dfrac12 \Rightarrow \alpha=60^\circ}}[/tex]

b)

[tex]2cos^2\alpha=5sin\alpha-1\\\\2(1-sin^2\alpha)=5sin\alpha-1\\\\2-2sin^2\alpha=5sin\alpha-1\\\\-2sin^2\alpha-5sin\alpha+2+1=0\\\\-2sin^2\alpha-5sin\alpha+3=0\\\\\Delta=(-5)^2-4\cdot (-2)\cdot 3=25+24=49\\\\\sqrt{\Delta}=\sqrt{49}=7\\\\sin\alpha_1=\dfrac{5-7}{2\cdot (-2)}=\dfrac{-2}{-4}=\dfrac12\\\\sin\alpha_2=\dfrac{5+7}{2\cdot (-2)}=\dfrac{12}{-4}=-3 \notin D\\\\\boxed{\bold{sin\alpha=\dfrac12 \Rightarrow \alpha=30^\circ}}[/tex]

c)

[tex]tg\alpha+ctg\alpha=2\\\\tg\alpha+\dfrac{1}{tg\alpha}=2\\\\\dfrac{tg^2\alpha}{tg\alpha}+\dfrac1{tg\alpha}=2\\\\\dfrac{tg^2\alpha+1}{tg\alpha}=2\\\\tg^2\alpha+1=2tg\alpha\\\\tg^2\alpha-2tg\alpha+1=0\\\\\Delta=(-2)^2-4\cdot 1\cdot 1=4-4=0\\\\tg\alpha=\dfrac{2}{2\cdot 1}=\dfrac22=1\\\\\boxed{\bold{tg\alpha=1 \Rightarrow \alpha=45^\circ}}[/tex]

d)

[tex]3tg^2\alpha+2\sqrt3tg\alpha=3\\\\3tg^2\alpha+2\sqrt3tg\alpha-3=0\\\\\Delta=(2\sqrt3)^2-4\cdot 3\cdot (-3)=12+36=48\\\\\sqrt{\Delta}=\sqrt{48}=4\sqrt3\\\\tg\alpha_1=\dfrac{-2\sqrt3-4\sqrt3}{2\cdot 3}=\dfrac{-6\sqrt3}6=-\sqrt3 \notin D\\\\tg\alpha_2=\dfrac{-2\sqrt3+4\sqrt3}{2\cdot 3}=\dfrac{2\sqrt3}6=\dfrac{\sqrt3}3\\\\\boxed{\bold{tg\alpha=\dfrac{\sqrt3}3\Rightarrow \alpha=30^\circ}}[/tex]