Trigonometri (04)

Perbandingan Trigonometri Sudut Berelasi

Kuadran Pertama

Diketahui sudut $\alpha$ terletak di kuadran pertama, maka 0$^0$ < $\alpha$ < 90$^0$

sin(90$^0 - \alpha$) = cos $\alpha$ cos(90$^0 - \alpha$) = sin $\alpha$ tan(90$^0 - \alpha$) = cotan $\alpha$

Kuadran Kedua

Diketahui sudut $\alpha$ terletak di kuadran kedua, maka 90$^0$ < $\alpha$ < 180$^0$

sin(90$^0 + \alpha$) = cos $\alpha$ cos(90$^0 + \alpha$) = $-$sin $\alpha$ tan(90$^0 + \alpha$) = $-$cotan $\alpha$

sin(180$^0 - \alpha$) = sin $\alpha$ cos(180$^0 - \alpha$) = $-$cos $\alpha$ tan(180$^0 - \alpha$) = $-$tan $\alpha$

Kuadran Ketiga

Diketahui sudut $\alpha$ terletak di kuadran ketiga, maka 180$^0$ < $\alpha$ < 270$^0$

sin(180$^0 + \alpha$) = $-$sin $\alpha$ cos(180$^0 + \alpha$) = $-$cos $\alpha$ tan(180$^0 + \alpha$) = tan $\alpha$

sin(270$^0 - \alpha$) = $-$cos $\alpha$ cos(270$^0 - \alpha$) = $-$sin $\alpha$ tan(270$^0 - \alpha$) = cotan $\alpha$

Kuadran Keempat

Diketahui sudut $\alpha$ terletak di kuadran keempat, maka 270$^0$ < $\alpha$ < 360$^0$

sin(270$^0 + \alpha$) = $-$cos $\alpha$ cos(270$^0 + \alpha$) = sin $\alpha$ tan(270$^0 + \alpha$) = $-$cotan $\alpha$

sin(360$^0 - \alpha$) = $-$sin $\alpha$ cos(360$^0 - \alpha$) = cos $\alpha$ tan(360$^0 - \alpha$) = $-$tan $\alpha$

Untuk sudut ($-\alpha$)

sin($-\alpha$) = $-$sin $\alpha$ cos($-\alpha$) = cos $\alpha$ tan($-\alpha$) = $-$tan $\alpha$

Untuk sudut lebih dari 360$^0$

sin(k.360$^0 + \alpha$) = sin $\alpha$ cos(k.360$^0 + \alpha$) = cos $\alpha$ tan(k.360$^0 + \alpha$) = tan $\alpha$ dimana k = 1, 2, ...

sin(k.360$^0 - \alpha$) = sin $\alpha$ cos(k.360$^0 - \alpha$) = cos $\alpha$ tan(k.360$^0 - \alpha$) = tan $\alpha$ dimana k = 2, 3, ...

Contoh :
Hitunglah nilai dari :
sin 240$^0$ = ....
Jawab :
Cara pertama :
sin 240$^0$ = sin (180$^0$ + 60$^0$)
sin 240$^0$ = $-$sin 60$^0$
sin 240$^0$ = $-\frac{1}{2}$$\sqrt{3}$
Cara kedua :
sin 240$^0$ = sin (270$^0 -$ 30$^0$)
sin 240$^0$ = $-$cos 30$^0$
sin 240$^0$ = $-\frac{1}{2}$$\sqrt{3}$

Hitunglah nilai dari :
cos 765$^0$ = ....
Jawab :
cos 765$^0$ = cos (2.360$^0$ + 45$^0$)
cos 765$^0$ = cos 45$^0$
cos 765$^0$ = $\frac{1}{2}$$\sqrt{2}$

Hitunglah nilai dari :
tan ($-$1200$^0$) = ....
Jawab :
tan ($-$1200$^0$) = $-$tan 1200$^0$
tan ($-$1200$^0$) = $-$tan (3.360$^0$ + 120$^0$)
tan ($-$1200$^0$) = $-$tan 120$^0$
tan ($-$1200$^0$) = $-$tan (180$^0 -$ 60$^0$)
tan ($-$1200$^0$) = tan 60$^0$
tan ($-$1200$^0$) = $\sqrt{3}$


Tugas :
Hitunglah nilai dari :
1.  cos 150$^0$ = ....
2.  tan 780$^0$ = ....
3.  sin ($-$1500$^0$) = ....