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# Mohon bantuannya kak# Asal Lapor​

Berikut ini adalah pertanyaan dari Agnesmonica91 pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

# Mohon bantuannya kak
# Asal Lapor​
# Mohon bantuannya kak# Asal Lapor​

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

\purple{\huge{1.}}

Terdapat identitas trigononetri :

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

\cos~\text{A}=\sqrt{1-\sin^2~\text{A}}=\sqrt{1-\left(\frac{4}{5}\right)^2}=\sqrt{1-\frac{16}{25}}=\sqrt{\frac{9}{25}}=\frac{3}{5}

\cos~\text{B}=\sqrt{1-\sin^2~\text{B}}=\sqrt{1-\left(\frac{8}{10}\right)^2}=\sqrt{1-\frac{64}{100}}=\sqrt{\frac{36}{100}}=\frac{6}{10}

Terdapat identitas trigonometri :

\boxed{\boxed{\sin~(x+y)=\sin~x.\cos~y+\cos~x.\sin~y}}

\sin~(\text{A}+\text{B})=\sin~\text{A}.\cos~\text{B}+\cos~\text{A}.\sin~\text{B}=\left(\frac{4}{5}\right).\left(\frac{6}{10}\right)+\left(\frac{3}{5}\right).\left(\frac{8}{10}\right)=\frac{24}{50}+\frac{24}{50}=\frac{48}{50}

\red{\huge{\sin~(\text{A}+\text{B})=\frac{24}{25}}}

\\

\purple{\huge{2.}}

Terdapat identitas trigonometri :

\boxed{\boxed{\cos~(x+y)=\cos~x.\cos~y-\sin~x.\sin~y}}

\cos~(\text{A}+\text{B})=\cos~\text{A}.\cos~\text{B}-\sin~\text{A}.\sin~\text{B}=\left(\frac{3}{5}\right).\left(\frac{6}{10}\right)-\left(\frac{4}{5}\right).\left(\frac{8}{10}\right)=\frac{18}{50}-\frac{32}{50}=-\frac{14}{50}

\red{\huge{\cos~(\text{A}+\text{B})=-\frac{7}{25}}}

\\

\purple{\huge{3.}}

\sin~\alpha=\sqrt{1-\cos^2~\alpha}=\sqrt{1-\left(\frac{5}{13}\right)^2}=\sqrt{1-\frac{25}{169}}=\sqrt{\frac{144}{169}}=\frac{12}{13}

\cos~2\alpha=\cos~(\alpha+\alpha)=\cos~\alpha.\cos~\alpha-\sin~\alpha.\sin~\alpha=\cos^2~\alpha-\sin^2~\alpha

\cos~2\alpha=\left(\frac{5}{13}\right)^2-\left(\frac{12}{13}\right)^2=\frac{25}{169}-\frac{144}{169}

\red{\huge{\cos~2\alpha=-\frac{119}{169}}}

\\

\purple{\huge{4.}}

Terdapat identitas trigonometri :

\boxed{\boxed{\begin{array}{ccc}\cos~x+~\cos~y=\\2.\cos~\frac{1}{2}(x+y).\cos~\frac{1}{2}(x-y)\end{array}}}

\cos~75\degree+\cos~15\degree=2.\cos~\frac{1}{2}(75\degree+15\degree).\cos~\frac{1}{2}(75\degree-15\degree)=2.\cos~45\degree.\cos~30\degree=2.\left(\frac{1}{2}\sqrt{2}\right).\left(\frac{1}{2}\sqrt{3}\right)

\red{\huge{\begin{array}{ccc}\cos~75\degree+\cos~15\degree\\=\frac{1}{2}\sqrt{6}\end{array}}}

\\

\purple{\huge{5.}}

Terdapat identitas trigonometri :

\boxed{\boxed{\begin{array}{ccc}\sin~x.\cos~y=\\\frac{1}{2}.\left(\sin~(x+y)+\sin~(x-y)\right)\end{array}}}

\sin~105\degree.\cos~15\degree=\frac{1}{2}.\left(\sin~(105\degree+15\degree)+\sin~(105\degree-15\degree)\right)=\frac{1}{2}.\left(\sin~120\degree+\sin~90\degree\right)=\frac{1}{2}.\left(\frac{1}{2}\sqrt{3}+1\right)

\red{\huge{\begin{array}{ccc}\sin~105\degree.\cos~15\degree\\=\frac{1}{4}\sqrt{3}+\frac{1}{2}\end{array}}}

\\

\purple{\huge{6.}}

\cos~\alpha=\sqrt{1-\sin^2~\alpha}=\sqrt{1-\left(\frac{5}{13}\right)^2}=\sqrt{1-\frac{25}{169}}=\sqrt{\frac{144}{169}}=\frac{12}{13}

\sin~2\alpha=\sin~(\alpha+\alpha)=\sin~\alpha.\cos~\alpha+\cos~\alpha.\sin~\alpha=2.\sin~\alpha.\cos~\alpha

\sin~2\alpha=2.\left(\frac{5}{13}\right).\left(\frac{12}{13}\right)

\red{\huge{\sin~2\alpha=\frac{120}{169}}}

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Last Update: Mon, 12 Jul 21
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