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Mohon bantuannya dengan caranya yaaaa

Berikut ini adalah pertanyaan dari farel1234567 pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Mohon bantuannya dengan caranya yaaaa
Mohon bantuannya dengan caranya yaaaa

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

\huge{\green{\mathfrak{1.}}}

Diketahui : \tan~\theta=-\frac{8}{15}dengan\sin~\theta > 0

Perhatikan gambar :

Pada gambar tersebut :

\tan~\theta=\frac{\text{BC}}{\text{AB}}

\to \frac{\text{BC}}{\text{AB}}=\frac{8}{15}

\to BC = 8 dan AB = 15

Dengan teorema Pythagoras :

AC = \sf \sqrt{15^2+8^2}=\sqrt{225+64}=\sqrt{289} = 17

Pada gambar tersebut :

\sin~\theta=\frac{\text{BC}}{\text{AC}}=\frac{8}{17}

Karena \sin~\theta > 0, dan diketahui \tan~\theta=-\frac{8}{15} < 0, maka : \cos~\theta=-\frac{15}{17}

\huge{\purple{\mathfrak{a.}}}

Sehingga :

\red{\sin~\theta\times \cos~\theta+\cos~\theta\times \sin~\theta}

=\left(\frac{8}{17}\right)\left(-\frac{15}{17}\right)+\left(-\frac{15}{17}\right)\left(\frac{8}{17}\right)

=-\frac{120}{289}-\frac{120}{289}~\red{\huge{=-\frac{240}{289}}}

\huge{\purple{\mathfrak{b.}}}

\cosec~\theta=\frac{1}{\sin~\theta}=\frac{1}{\frac{8}{17}}=\frac{17}{8}

\frac{\cosec~\theta}{\cot~\theta}=\frac{\frac{1}{\sin~\theta}}{\frac{\cos~\theta}{\sin~\theta}}=\frac{1}{\sin~\theta}\times \frac{\sin~\theta}{\cos~\theta}

\red{\frac{\cosec~\theta}{\cot~\theta}}=\frac{1}{\cos~\theta}=\frac{1}{-\frac{15}{17}}~\red{\huge{=-\frac{17}{15}}}

\\

\huge{\green{\mathfrak{2.}}}

\sin~(-300\degree)=\sin~-(360\degree-60\degree)=\sin~(60\degree-360\degree)

\boxed{\sin~(\alpha-\beta)=(\sin~\alpha\times \cos~\beta)-(\cos~\alpha\times \sin~\beta)}

Sehingga :

\sin~(-300\degree)=(\sin~60\degree\times \cos~360\degree)-(\cos~60\degree\times \sin~360\degree

\cos~(-300\degree)=\left(\frac{1}{2}\sqrt{3}\right)(1)+\left(\frac{1}{2}\right)(0)

\red{\huge{\sin~(-300\degree)=\frac{1}{2}\sqrt{3}}}

\\

\huge{\green{\mathfrak{3.}}}

\cos~(1.470\degree)=\cos~(4\times 360\degree+30\degree

\cos~(1.470\degree)=\cos~(360\degree+30\degree)

\boxed{\cos~(\alpha+\beta)=(\cos~\alpha\times \cos~\beta)-(\sin~\alpha\times \sin~\beta)}

Sehingga :

\cos~(1.470\degree)=(\cos~360\degree\times \cos~30\degree)-(\sin~360\degree\times \sin~30\degree)

\cos~(1.470\degree)=(1)\left(\frac{1}{2}\sqrt{3}\right)-(0)\left(\frac{1}{2}\right)

\red{\huge{\cos~(1.470\degree)=\frac{1}{2}\sqrt{3}}}

\\

\huge{\green{\mathfrak{4.}}}

\frac{\sin~x}{1~+~\cos~x}+\frac{\sin~x}{1~-~\cos~x}

=\frac{\sin~x(1~-~\cos~x)+\sin~x(1~+~\cos~x)}{(1~+~\cos~x)(1~-~\cos~x)}

=\frac{\sin~x-\sin~x~\cos~x+\sin~x+\sin~x~\cos~x}{1-\cos^2~x}

\boxed{\cos^2~x+\sin^2~x=1}

=\frac{2~\sin~x}{\sin^2~x}

\red{\huge{\frac{\sin~x}{1+\cos~x}+\frac{\sin~x}{1-\cos~x}=\frac{2}{\sin~x}}}

[tex]\huge{\green{\mathfrak{1.}}}[/tex]Diketahui : [tex]\tan~\theta=-\frac{8}{15}[/tex] dengan [tex]\sin~\theta > 0[/tex]Perhatikan gambar :Pada gambar tersebut :[tex]\tan~\theta=\frac{\text{BC}}{\text{AB}}[/tex][tex]\to \frac{\text{BC}}{\text{AB}}=\frac{8}{15}[/tex][tex]\to[/tex] BC = 8 dan AB = 15Dengan teorema Pythagoras :AC = [tex]\sf \sqrt{15^2+8^2}=\sqrt{225+64}=\sqrt{289}[/tex] = 17Pada gambar tersebut :[tex]\sin~\theta=\frac{\text{BC}}{\text{AC}}=\frac{8}{17}[/tex]Karena [tex]\sin~\theta > 0[/tex], dan diketahui [tex]\tan~\theta=-\frac{8}{15} < 0[/tex], maka : [tex]\cos~\theta=-\frac{15}{17}[/tex][tex]\huge{\purple{\mathfrak{a.}}}[/tex]Sehingga :[tex]\red{\sin~\theta\times \cos~\theta+\cos~\theta\times \sin~\theta}[/tex][tex]=\left(\frac{8}{17}\right)\left(-\frac{15}{17}\right)+\left(-\frac{15}{17}\right)\left(\frac{8}{17}\right)[/tex][tex]=-\frac{120}{289}-\frac{120}{289}~\red{\huge{=-\frac{240}{289}}}[/tex][tex]\huge{\purple{\mathfrak{b.}}}[/tex][tex]\cosec~\theta=\frac{1}{\sin~\theta}=\frac{1}{\frac{8}{17}}=\frac{17}{8}[/tex][tex]\frac{\cosec~\theta}{\cot~\theta}=\frac{\frac{1}{\sin~\theta}}{\frac{\cos~\theta}{\sin~\theta}}[/tex][tex]=\frac{1}{\sin~\theta}\times \frac{\sin~\theta}{\cos~\theta}[/tex][tex]\red{\frac{\cosec~\theta}{\cot~\theta}}=\frac{1}{\cos~\theta}=\frac{1}{-\frac{15}{17}}~[/tex][tex]\red{\huge{=-\frac{17}{15}}}[/tex][tex]\\[/tex][tex]\huge{\green{\mathfrak{2.}}}[/tex][tex]\sin~(-300\degree)=\sin~-(360\degree-60\degree)[/tex][tex]=\sin~(60\degree-360\degree)[/tex][tex]\boxed{\sin~(\alpha-\beta)=(\sin~\alpha\times \cos~\beta)-(\cos~\alpha\times \sin~\beta)}[/tex]Sehingga :[tex]\sin~(-300\degree)=(\sin~60\degree\times \cos~360\degree)-(\cos~60\degree\times \sin~360\degree[/tex][tex]\cos~(-300\degree)=\left(\frac{1}{2}\sqrt{3}\right)(1)+\left(\frac{1}{2}\right)(0)[/tex][tex]\red{\huge{\sin~(-300\degree)=\frac{1}{2}\sqrt{3}}}[/tex][tex]\\[/tex][tex]\huge{\green{\mathfrak{3.}}}[/tex][tex]\cos~(1.470\degree)=\cos~(4\times 360\degree+30\degree[/tex][tex]\cos~(1.470\degree)=\cos~(360\degree+30\degree)[/tex][tex]\boxed{\cos~(\alpha+\beta)=(\cos~\alpha\times \cos~\beta)-(\sin~\alpha\times \sin~\beta)}[/tex]Sehingga :[tex]\cos~(1.470\degree)=(\cos~360\degree\times \cos~30\degree)-(\sin~360\degree\times \sin~30\degree)[/tex][tex]\cos~(1.470\degree)=(1)\left(\frac{1}{2}\sqrt{3}\right)-(0)\left(\frac{1}{2}\right)[/tex][tex]\red{\huge{\cos~(1.470\degree)=\frac{1}{2}\sqrt{3}}}[/tex][tex]\\[/tex][tex]\huge{\green{\mathfrak{4.}}}[/tex][tex]\frac{\sin~x}{1~+~\cos~x}+\frac{\sin~x}{1~-~\cos~x}[/tex][tex]=\frac{\sin~x(1~-~\cos~x)+\sin~x(1~+~\cos~x)}{(1~+~\cos~x)(1~-~\cos~x)}[/tex][tex]=\frac{\sin~x-\sin~x~\cos~x+\sin~x+\sin~x~\cos~x}{1-\cos^2~x}[/tex][tex]\boxed{\cos^2~x+\sin^2~x=1}[/tex][tex]=\frac{2~\sin~x}{\sin^2~x}[/tex][tex]\red{\huge{\frac{\sin~x}{1+\cos~x}+\frac{\sin~x}{1-\cos~x}=\frac{2}{\sin~x}}}[/tex]

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Last Update: Fri, 09 Jul 21
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