7. 1+cos2xsin2x​ sama dengan ....

Ingat kembali rumus penjumlahan dua sudut:

• $$\cos \left(\alpha +b\right)=\cos \alpha \cdot \cos \beta -\sin \alpha \cdot \sin \beta$$
• $$\cos 2\alpha =2\cdot {\cos }^2\alpha -1$$
• $$\sin 2\alpha =2\cdot \sin \alpha \cdot \cos \alpha$$

dan identitas trigonometri $${\sin }^2\alpha =1-{\cos }^2\alpha$$ sehingga dapat diperoleh rumus berikut:

$$\begin{array}{rcl}\cos 3\alpha & =& \cos \left(2\alpha +\alpha \right)\\ & =& \cos 2\alpha \cdot \cos \alpha -\sin 2\alpha \cdot \sin \alpha \\ & =& \left(2\cdot {\cos }^2\alpha -1\right)\cdot \mathrm{co}s\alpha -\left(2\cdot \sin \alpha \cdot \cos \alpha \right)\cdot \sin \alpha \\ & =& 2\cdot {\cos }^3\alpha -\mathrm{co}s\alpha -2\cdot {\sin }^2\alpha \cdot \cos \alpha \\ & =& 2\cdot {\cos }^3\alpha -\mathrm{co}s\alpha -2\cdot \left(1-{\cos }^2\alpha \right)\cdot \cos \alpha \\ & =& 2\cdot {\cos }^3\alpha -\mathrm{co}s\alpha -2\cdot \left(1-{\cos }^2\alpha \right)\cdot \cos \alpha \\ & =& 2\cdot {\cos }^3\alpha -\mathrm{co}s\alpha -2\cdot 1\cdot \cos \alpha +2\cdot {\cos }^2\alpha \cdot \cos \alpha \\ & =& 2\cdot {\cos }^3\alpha -\mathrm{co}s\alpha -2\cdot \cos \alpha +2\cdot {\cos }^3\alpha \\ & =& 2\cdot {\cos }^3\alpha +2\cdot {\cos }^3\alpha -\mathrm{co}s\alpha -2\cdot \cos \alpha \\ \cos 3\alpha & =& 4\cdot {\cos }^3\alpha -3\cdot \cos \alpha \end{array}$$ 

Oleh karena $$\cos x=\frac{1}{2}$$, maka dengan menggunakan rumus di atas, diperoleh:

 $$\begin{array}{rcl}\cos 3x& =& 4\cdot {\cos }^{3}x-3\cdot \cos x\\ & =& 4\times {\left(\frac{1}{2}\right)}^3-3\times \left(\frac{1}{2}\right)\\ & =& 4\times \frac{1^3}{2^3}-3\times \left(\frac{1}{2}\right)\\ & =& 4\times \frac{1}{8}-3\times \frac{1}{2}\\ & =& \frac{4\times 1}{8}-\frac{3\times 1}{2}\\ & =& \frac{4}{8}-\frac{3}{2}\\ & =& \frac{4÷4}{8÷4}-\frac{3}{2}\\ & =& \frac{1}{2}-\frac{3}{2}\\ & =& \frac{1-3}{2}\\ & =& \frac{-2}{2}\\ & =& -1\end{array}$$ 

Dengan demikian, diperoleh  $$\cos 3x$$ sama dengan $$-1$$.

Oleh karena itu, jawaban yang benar adalah E. 

Ingat bahwa :

Rumus identitas trigonometri

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

Rumus jumlah dan selisih dua sudut yaitu

$$\tan \left(A+B\right)=\frac{\tan A+\tan B}{1-\tan A\cdot \tan B}$$

Sudut rangkap pada sinus 

$$\begin{array}{rcl}\sin 2A& =& 2\sin A\cos A\\ \sin A& =& 2\sin \frac{1}{2}A\cos \frac{1}{2}A\end{array}$$

Sudut rangkap pada cosinus

$$\begin{array}{rcl}\cos 2A& =& 2{\cos }^{2}A-1\end{array}$$

Sehingga 

$$\begin{array}{rcl}\tan \left(\frac{\pi }{4}+\frac{\theta }{2}\right)& =& \frac{\tan \frac{\pi }{4}+\tan \frac{\theta }{2}}{1-\tan \frac{\pi }{4}\tan \frac{\theta }{2}}\\ & =& \frac{1+\tan \frac{\theta }{2}}{1-\tan \frac{\theta }{2}}\\ & =& \frac{1+\frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}}{1-\frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}}\\ & =& \frac{\cos \frac{1}{2}\theta +\sin \frac{1}{2}\theta }{\cos \frac{1}{2}\theta }\cdot \frac{\cos \frac{1}{2}\theta }{\cos \frac{1}{2}\theta -\sin \frac{1}{2}\theta }\\ & =& \frac{{\cos }^2\frac{1}{2}\theta +\sin \frac{1}{2}\theta \cos \frac{1}{2}\theta }{{\cos }^2\frac{1}{2}\theta -\sin \frac{1}{2}\theta \cos \frac{1}{2}\theta }\cdot \frac{2}{2}\\ & =& \frac{2{\cos }^2\frac{1}{2}\theta +2\sin \frac{1}{2}\theta \cos \frac{1}{2}\theta }{2{\cos }^2\frac{1}{2}\theta -2\sin \frac{1}{2}\theta \cos \frac{1}{2}\theta }\\ & =& \frac{\left(1+2{\cos }^2\frac{1}{2}\theta -1\right)+\sin \theta }{\left(1+2{\cos }^2\frac{1}{2}\theta -1\right)-\sin \theta }\\ & =& \frac{1+\cos \theta +\sin \theta }{1+\cos \theta -\sin \theta }\left(\mathrm{terbukti}\right)\end{array}$$

Dengan demikian terbukti bahwa $$\tan \left(\frac{\pi }{4}+\frac{\theta }{2}\right)=\frac{1+\cos \theta +\sin \theta }{1+\cos \theta -\sin \theta }$$

 

Ingat sifat trigonometri berikut.

$$\begin{array}{rcl}\sin 2\alpha & =& 2\sin \alpha \cos \alpha \\ \cos 2\alpha & =& {\cos }^2\alpha -{\sin }^2\alpha \\ \sin \left(\alpha +\beta \right)& =& \sin \alpha \cos \beta +\sin \beta \cos \alpha \\ \cos (\alpha +\beta )& =& \cos \alpha \cos \beta -\sin \alpha \sin \beta \end{array}$$

Misal: $$p=2\sin 105^{\circ }{\cos }^{3}105^{\circ }-2{\sin }^{3}105^{\circ }\cos 105^{\circ }$$, maka diperoleh:

$$\begin{array}{rcl}p& =& 2\sin 105^{\circ }{\cos }^{3}105^{\circ }-2{\sin }^{3}105^{\circ }\cos 105^{\circ }\\ & =& 2\sin 105^{\circ }\cos 105^{\circ }{\cos }^{2}105^{\circ }-2\sin 105^{\circ }\cos 105^{\circ }{\sin }^{2}105^{\circ }\\ & =& \sin 2(105^{\circ }){\cos }^{2}105^{\circ }-\sin 2(105^{\circ }){\sin }^{2}105^{\circ }\\ & =& \sin 210^{\circ }{\cos }^{2}105^{\circ }-\sin 210^{\circ }{\sin }^{2}105^{\circ }\\ & =& \sin 210^{\circ }\left({\cos }^{2}105^{\circ }-{\sin }^{2}105^{\circ }\right)\\ & =& \sin 210^{\circ }\left(\cos 2(105^{\circ })\right)\\ & =& \sin 210^{\circ }\cos 210^{\circ }\end{array}$$ 

Selanjutnya dapat ditentukan nilai dari $$\sin 210^{\circ }\mathrm{dan}\cos 210^{\circ }$$ sebagai berikut.

$$\begin{array}{rcl}\sin 210^{\circ }& =& \sin (180^{\circ }+30^{\circ })\\ & =& \sin 180^{\circ }\cos 30^{\circ }+\sin 30^{\circ }\cos 180^{\circ }\\ & =& 0\cdot \frac{1}{2}\sqrt{3}+\frac{1}{2}\cdot \left(-1\right)\\ & =& -\frac{1}{2}\\ \cos 210^{\circ }& =& \cos (180^{\circ }+30^{\circ })\\ & =& \cos 180^{\circ }\cos 30^{\circ }-\sin 180^{\circ }\sin 30^{\circ }\\ & =& \left(-1\right)\cdot \frac{1}{2}\sqrt{3}-0\cdot \left(\frac{1}{2}\right)\\ & =& -\frac{1}{2}\sqrt{3}\end{array}$$ 

Sehingga diperoleh nilai $$p$$ sebagai berikut.

$$\begin{array}{rcl}p& =& \sin 210^{\circ }\cos 210^{\circ }\\ & =& -\frac{1}{2}\cdot \left(-\frac{1}{2}\sqrt{3}\right)\\ & =& \frac{1}{4}\sqrt{3}\end{array}$$  

Jadi, jawaban yang benar adalah C.

Ingat kembali:

• rumus sinus untuk sudut ganda: $$\sin 2\alpha =2\cdot \sin \alpha \cdot \cos \alpha$$
• rumus cosinus untuk sudut ganda: $$\cos 2\alpha ={\cos }^2\alpha -{\sin }^2\alpha$$ 
• identitas trigonometri: $${\sin }^2\alpha +{\cos }^2\alpha =1$$ 

Oleh karena $$\sin A+\sin 2A=x$$ dan $$\cos A+\cos 2A=y$$, maka dapat diperoleh:

$$\begin{array}{rcl}{x}^2& =& {\left(\sin A+\sin 2A\right)}^2\\ & =& {\sin }^{2}A+2\cdot \sin A\cdot \sin 2A+{\sin }^{2}2A\\ & =& {\sin }^{2}A+2\cdot \sin A\cdot \left(2\cdot \sin A\cdot \cos A\right)+{\sin }^{2}2A\\ & =& {\sin }^{2}A+2\cdot 2\cdot \sin A\cdot \sin A\cdot \cos A+{\sin }^{2}2A\\ & =& {\sin }^{2}A+4\cdot {\sin }^{2}A\cdot \cos A+{\sin }^{2}2A\\ & & \\ y^2& =& {\left(\cos A+\cos 2A\right)}^2\\ & =& {\cos }^{2}A+2\cdot \cos A\cdot \cos 2A+{\cos }^{2}2A\\ & =& {\cos }^{2}A+2\cdot \cos A\cdot \left({\cos }^{2}A-{\sin }^{2}A\right)+{\cos }^{2}2A\\ & =& {\cos }^{2}A+2\cdot {\cos }^{3}A-2\cdot \cos A\cdot {\sin }^{2}A+{\cos }^{2}2A\\ & & \\ x^2+y^2& =& {\sin }^{2}A+4\cdot {\sin }^{2}A\cdot \cos A+{\sin }^{2}2A\\ & & +{\cos }^{2}A+2\cdot {\cos }^{3}A-2\cdot \cos A\cdot {\sin }^{2}A+{\cos }^{2}2A\\ & =& +4\cdot {\sin }^{2}A\cdot \cos A+2\cdot {\cos }^{3}A-2\cdot \cos A\cdot {\sin }^{2}A\\ & & \left({\sin }^{2}A+{\cos }^{2}A\right)+\left({\sin }^{2}2A+{\cos }^{2}2A\right)\\ & =& 4\cdot {\sin }^{2}A\cdot \cos A-2\cdot {\sin }^{2}A\cdot \cos A+2\cdot {\cos }^{3}A+1+1\\ & =& 2\cdot {\sin }^{2}A\cdot \cos A+2\cdot {\cos }^{3}A+2\\ & =& 2\cdot \cos A\cdot \left({\sin }^{2}A+{\cos }^{2}A\right)+2\\ & =& 2\cdot \cos A\cdot 1+2\\ & =& 2\cdot \cos A+2\end{array}$$   

sehingga diperoleh 

 $$\begin{array}{rcl}\left(x^2+y^2\right)\left(x^2+y^2-3\right)& =& \left(2\cdot \cos A+2\right)\left(2\cdot \cos A+2-3\right)\\ & =& \left(2\cdot \cos A+2\right)\left(2\cdot \cos A+2-3\right)\\ & =& \left(2\cdot \cos A+2\right)\left(2\cdot \cos A-1\right)\\ & =& 4\cdot {\cos }^{2}A-2\cdot \cos A+4\cdot \cos A-2\\ & =& 4\cdot {\cos }^{2}A+2\cdot \cos A-2\\ & =& 2\cdot \left(2\cdot {\cos }^{2}A+\cos A-1\right)\\ & =& 2\cdot \left(2\cdot {\cos }^{2}A+\cos A-\left({\sin }^{2}A+{\cos }^{2}A\right)\right)\\ & =& 2\cdot \left(2\cdot {\cos }^{2}A+\cos A-{\sin }^{2}A-{\cos }^{2}A\right)\\ & =& 2\cdot \left(\cos A+2\cdot {\cos }^{2}A-{\cos }^{2}A-{\sin }^{2}A\right)\\ & =& 2\cdot \left(\cos A+{\cos }^{2}A-{\sin }^{2}A\right)\\ & =& 2\cdot \left(\cos A+\cos 2A\right)\\ & =& 2\cdot y\\ & =& 2y\end{array}$$    

Dengan demikian, telah ditunjukan bahwa $$\left(x^2+y^2\right)\left(x^2+y^2-3\right)=2y$$. 

Ingat,

Sudut Rangkap (Sinus)

$$\sin 2\mathrm{A}=2\sin \mathrm{A}\cos \mathrm{A}$$

Sudut Rangkap (Cosinus)

$$\cos 2\mathrm{A}=1-2{\sin }^2\mathrm{A}$$

Rumus Perbandingan Sisi Trigonometri

$$\begin{gathered} \sin \mathrm{A}=\frac{\mathrm{depan}}{\mathrm{miring}} \\ \cos \mathrm{A}=\frac{\mathrm{samping}}{\mathrm{miring}} \end{gathered}$$

Berdasarkan rumus tersebut, diperoleh sebagai berikut

Diketahui $$\sin (\mathrm{x}+15^{\circ })=m$$ dengan $$0^{\circ }\le x\le 15^{\circ }$$ sehingga sisi depan $$m$$ dan sisi miring $$1$$

Menentukan sisi samping dengan teorema pythagoras

$$\sqrt{1^2-m^2}=\sqrt{1-m^2}$$

Menghitung $$\cos \left(x+15^{\circ }\right)$$

$$\cos \left(\mathrm{x}+15^{\circ }\right)=\frac{\mathrm{samping}}{\mathrm{miring}}=\frac{\sqrt{1-m^2}}{1}=\sqrt{1-m^2}$$

►Menentukan ekspresi $$\sin \left(2x+60^{\circ }\right)$$ dalam $$m$$

Diketahui $$\sin (\mathrm{x}+15^{\circ })=m$$ dan $$\cos \left(\mathrm{x}+15^{\circ }\right)=\sqrt{1-m^2}$$

Misalkan $$\mathrm{A}=x+15^{\circ }$$

sehingga $$2x+60^{\circ }=2\left(x+15^{\circ }\right)+30^{\circ }=2\mathrm{A}+30^{\circ }$$

$$\begin{array}{rcl}\sin \left(2\mathrm{x}+60^{\circ }\right)& =& \sin \left(2\mathrm{A}+30^{\circ }\right)\\ & =& \sin 2\mathrm{A}\cos 30^{\circ }+\cos 2\mathrm{A}\sin 30^{\circ }\\ & =& \left(2\sin \mathrm{A}\cos \mathrm{A}\right)\cos 30^{\circ }+\left(1-2{\sin }^2\mathrm{A}\right)\sin 30^{\circ }\\ & =& 2\cdot m\cdot \sqrt{1-m^2}\cdot \frac{1}{2}\sqrt{3}+\left(1-2\cdot m^2\right)\frac{1}{2}\\ & =& \sqrt{3}m\sqrt{1-m^2}+\frac{1}{2}-m^2\\ & =& -m^2+\sqrt{3}m\sqrt{1-m^2}+\frac{1}{2}\end{array}$$

Dengan demikian, ekspresi $$\sin \left(2x+60^{\circ }\right)$$ dalam $$m$$ adalah $$\begin{array}{rcl}& & -m^2+\sqrt{3}m\sqrt{1-m^2}+\frac{1}{2}\end{array}$$.