Nilai

Akan dicari nilai $$\tan x^{\circ }$$ yang memenuhi persamaan $$\sin {\left(x+30\right)}^{\circ }=2\cos {\left(x-30\right)}^{\circ }$$

Ingat bahwa

$$\begin{gathered} \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} \\ \tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B} \\ \sin (A+B)=\sin A\cos B+\cos A\sin B \\ \cos (A-B)=\cos A\cos B+\sin A\sin B \end{gathered}$$ 

Diperhatikan

$$\begin{array}{rcl}\sin {\left(x+30\right)}^{\circ }& =& 2\cos {\left(x-30\right)}^{\circ }\\ \sin x^{\circ }\cos 30^{\circ }+\cos x^{\circ }\sin 30^{\circ }& =& 2\left(\cos x^{\circ }\cos 30^{\circ }+\sin x^{\circ }\sin 30^{\circ }\right)\\ \sin x^{\circ }\cdot \frac{1}{2}\sqrt{3}+\cos x^{\circ }\cdot \frac{1}{2}& =& 2\left(\cos x^{\circ }\cdot \frac{1}{2}\sqrt{3}+\sin x^{\circ }\cdot \frac{1}{2}\right)\\ \frac{1}{2}\left(\sqrt{3}\sin x^{\circ }+\cos x^{\circ }\right)& =& \sqrt{3}\cos x^{\circ }+\sin x^{\circ }\\ \sqrt{3}s\mathrm{in}{x}^{\circ }+\cos x^{\circ }& =& 2\sqrt{3}\cos x^{\circ }+2\sin x^{\circ }\\ \sqrt{3}\sin x^{\circ }-2\sin x^{\circ }& =& 2\sqrt{3}\cos x^{\circ }-\cos x^{\circ }\\ \left(\sqrt{3}-2\right)\sin x^{\circ }& =& \left(2\sqrt{3}-1\right)\cos x^{\circ }\\ \frac{\sin x^{\circ }}{\cos x^{\circ }}& =& \frac{2\sqrt{3}-1}{\sqrt{3}-2}\\ \tan x^{\circ }& =& \frac{2\sqrt{3}-1}{\sqrt{3}-2}\times \frac{\sqrt{3}+2}{\sqrt{3}+2}\\ \tan x^{\circ }& =& \frac{6+3\sqrt{3}-2}{3-4}\\ \tan x^{\circ }& =& \frac{4+3\sqrt{3}}{-1}\\ \tan x^{\circ }& =& -\left(4+3\sqrt{3}\right)\end{array}$$ 

Dengan demikian, diperoleh nilai $$\tan x^{\circ }$$ adalah $$-\left(4+3\sqrt{3}\right)$$

Oleh karena itu, jawaban yang benar yaitu D

Ingat,

Sudut dalam Segitiga

$$\mathrm{A}+\mathrm{B}+\mathrm{C}=180^{\circ }$$

Hubungan Tangen dengan Sinus dan Cosinus

$$\tan \mathrm{A}=\frac{\sin \mathrm{A}}{\cos \mathrm{A}}$$

Rumus Perbandingan Trigonometri untuk Selisih Dua Sudut (Cosinus)

$$\cos (\mathrm{A}-\mathrm{B})=\cos \mathrm{A}\cos \mathrm{B}+\sin \mathrm{A}\sin \mathrm{B}$$

Rumus Perbandingan Trigonometri untuk Penjumlahan Dua Sudut (Sinus)

$$\sin (\mathrm{A}+\mathrm{B})=\sin \mathrm{A}\cos \mathrm{B}+\cos \mathrm{A}\sin \mathrm{B}$$

Menentukan Akar-akar Persamaan Kuadrat: Rumus Kuadratik (abc)

$$\begin{gathered} a{x}^2+bx+c=0 \\ {x}_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \end{gathered}$$

Berdasarkan rumus tersebut, diperoleh sebagai berikut

Diketahui $$△\mathrm{ABC}$$ siku-siku di $$\mathrm{C}$$ , sehingga

$$\begin{array}{rcl}\mathrm{A}+\mathrm{B}+\mathrm{C}& =& 180^{\circ }\\ \mathrm{A}+\mathrm{B}+90^{\circ }& =& 180^{\circ }\\ \mathrm{A}+\mathrm{B}& =& 180^{\circ }-90^{\circ }\\ \mathrm{A}+\mathrm{B}& =& 90^{\circ }\end{array}$$

Diperoleh $$\mathrm{A}+\mathrm{B}=90^{\circ }$$, sehingga

$$\begin{array}{rcl}\cos \mathrm{B}& =& \cos \left(90^{\circ }-\mathrm{A}\right)\\ & =& \cos 90^{\circ }\cos \mathrm{A}+\sin 90^{\circ }\sin \mathrm{A}\\ & =& 0+1\cdot \sin \mathrm{A}\\ & =& \sin \mathrm{A}\end{array}$$

Diketahui $$2\tan \mathrm{A}=\sin \mathrm{B}$$ , sehingga

$$\begin{array}{rcl}2\tan \mathrm{A}& =& \sin \mathrm{B}\\ 2\frac{\sin \mathrm{A}}{\cos \mathrm{A}}& =& \sin \mathrm{B}\\ 2\sin \mathrm{A}& =& \cos \mathrm{A}\sin \mathrm{B}\\ \cos \mathrm{A}\sin \mathrm{B}& =& 2\sin \mathrm{A}\end{array}$$

► Menentukan nilai $$\sin \mathrm{A}$$

$$\begin{array}{rcl}\sin (\mathrm{A}+\mathrm{B})& =& \sin \mathrm{A}\cos \mathrm{B}+\cos \mathrm{A}\sin \mathrm{B}\\ \sin \left(90^{\circ }\right)& =& \sin \mathrm{A}\sin \mathrm{A}+2\sin \mathrm{A}\\ 1& =& {\sin }^2\mathrm{A}+2\sin \mathrm{A}\\ 0& =& {\sin }^2\mathrm{A}+2\sin \mathrm{A}-1\end{array}$$

Dengan rumus kuadratik (abc) diperoleh akar-akar dari persamaan kuadrat

$$\begin{array}{rcl}{\sin }^2\mathrm{A}+2\sin \mathrm{A}-1& =& 0\\ \sin \mathrm{A}& =& \frac{-2\pm \sqrt{2^2-4\left(1\right)\left(-1\right)}}{2\left(1\right)}\\ & =& \frac{-2\pm \sqrt{4+4}}{2}\\ & =& \frac{-2\pm \sqrt{8}}{2}\\ & =& \frac{-2\pm 2\sqrt{2}}{2}\\ & =& -1\pm \sqrt{2}\end{array}$$

Dengan demikian, nilai $$\sin \mathrm{A}$$ adalah $$-1+\sqrt{2}$$ atau $$-1-\sqrt{2}$$. 

Konsep:

$$\begin{array}{rcl}\sin (A+B)& =& \sin A\cos B+\cos A\sin B\\ \cos (A-B)& =& \cos A\cos B+\sin A\sin B\\ \frac{\sin A}{\cos A}& =& \tan A\end{array}$$

Penyelesaian:

$$\begin{array}{rcl}\sin (A+B)\cos C& =& 2\sin A\cos (B-C)\\ \left(\sin A\cos B+\cos A\sin B\right)\cos C& =& 2\sin A\left(\cos B\cos C+\sin B\sin C\right)\\ \sin A\cos B\cos C+\cos A\sin B\cos C& =& 2\sin A\cos B\cos C-2\sin A\sin B\sin C\\ \sin A\cos B\cos C-2\sin A\cos B\cos C& =& -2\sin A\sin B\sin C-\cos A\sin B\cos C\\ \sin A\cos B(\cos C-2\cos C)& =& -2\sin A\sin B\sin C-\cos A\sin B\cos C\\ -\sin A\cos B\cos C+\cos A\sin B\cos C& =& -2\sin A\sin B\sin C\\ \sin A\cos B\cos C-\cos A\sin B\cos C& =& 2\sin A\sin B\sin C\\ \cos C(\sin A\cos B-\cos A\sin B)& =& 2\sin A\sin B\sin C\\ -\cos C(\sin B\cos A-\cos B\sin A)& =& 2\sin A\sin B\sin C\\ \frac{(\sin B\cos A-\cos B\sin A)}{2\sin A\sin B}& =& \frac{\sin C}{-\cos C}\\ \frac{\sin (B-A)}{2\sin A\sin B}& =& -\tan C\\ -\tan C& =& \frac{\sin (B-A)}{2\sin A\sin B}\end{array}$$

Jadi, tidak terbukti bahwa $$\tan C=\frac{\sin (B-A)}{2\sin A\sin B}$$.

Jawaban: A