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$Yuk kita temukan perbandingan trigonometri pada kuadran II, menggunakan sumbu koordinat! Cerminkan segitiga di Kuadran I Ke Kuadran II sehingga : {:[x^(')=-x],[y^(')=y],[r^(')=r]:} Perhatikan segitiga merah di Kuadran 2, aplikasikan perbandingan trigonometri pada segitiga siku-siku {:[sin(180^(@)-alpha)=(dots..)/(dots..)=-dots dots.],[cos(180^(@)-alpha)=(x^('))/(r^('))=(-x)/(r)=-cos alpha],[tan(180^(@)-alpha)=(dots..)/(dots..)=-=dots..]:}$

Yuk kita temukan perbandingan trigonometri pada kuadran II, menggunakan sumbu koordinat! $\newline$ Cerminkan segitiga di Kuadran I Ke Kuadran II sehingga : $\newline$ $\begin{aligned} x^{\prime} & =-x \\ y^{\prime} & =y \\ r^{\prime} & =r \end{aligned}$ $\newline$ Perhatikan segitiga merah di Kuadran $2$ , aplikasikan perbandingan trigonometri pada segitiga siku-siku $\newline$ $\begin{array}{l} \sin \left(180^{\circ}-\alpha\right)=\frac{\ldots . .}{\ldots . .}=-\ldots \ldots . \\ \cos \left(180^{\circ}-\alpha\right)=\frac{x^{\prime}}{r^{\prime}}=\frac{-x}{r}=-\cos \alpha \\ \tan \left(180^{\circ}-\alpha\right)=\frac{\ldots . .}{\ldots . .}=-=\ldots . . \end{array}$

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Algebra 1

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Q. Yuk kita temukan perbandingan trigonometri pada kuadran II, menggunakan sumbu koordinat!

\newline

Cerminkan segitiga di Kuadran I Ke Kuadran II sehingga :

\newline

\begin{aligned} x^{\prime} & =-x \\ y^{\prime} & =y \\ r^{\prime} & =r \end{aligned}

\newline

Perhatikan segitiga merah di Kuadran

2

, aplikasikan perbandingan trigonometri pada segitiga siku-siku

\newline

\begin{array}{l} \sin \left(180^{\circ}-\alpha\right)=\frac{\ldots . .}{\ldots . .}=-\ldots \ldots . \\ \cos \left(180^{\circ}-\alpha\right)=\frac{x^{\prime}}{r^{\prime}}=\frac{-x}{r}=-\cos \alpha \\ \tan \left(180^{\circ}-\alpha\right)=\frac{\ldots . .}{\ldots . .}=-=\ldots . . \end{array}

Reflect Triangle: Reflect the triangle from Quadrant I to Quadrant II: \begin{align*} x^{\prime} &= -x, \ y^{\prime} &= y, \ r^{\prime} &= r \end{align*}
Calculate $\sin(180°-\alpha):$ For $\sin(180°-\alpha)$ , use the y-coordinate and the radius: $\newline$ $\sin(180°-\alpha) = \frac{y'}{r'} = \frac{y}{r}$ $\newline$ Since we're in Quadrant II, sin is positive, so: $\newline$ $\sin(180°-\alpha) = \sin \alpha$
Calculate $\cos(180°-\alpha)$ : For $\cos(180°-\alpha)$ , use the x-coordinate and the radius: $\newline$ $\cos(180°-\alpha) = \frac{x'}{r'} = \frac{-x}{r}$ $\newline$ Since $\cos$ is negative in Quadrant II, we have: $\newline$ $\cos(180°-\alpha) = -\cos \alpha$
Calculate $\tan(180°-\alpha):$ For $\tan(180°-\alpha)$ , use the y-coordinate and the x-coordinate: $\newline$ $\tan(180°-\alpha) = \frac{y^{'} }{x^{'} } = \frac{y}{-x}$ $\newline$ Since $\tan$ is positive in Quadrant II, we have: $\newline$ $\tan(180°-\alpha) = -\tan \alpha$