Kumpulan soal dan pembahasan trigonometri analitika kelas 11

SMA KELAS 2 SMA MATERI SEKOLAH

21. Jika \(\sin α = p\) dan \(\sin β = q\), maka \(\sin(α+β)=…\)

A. \(p\sqrt{1-p^2} – q\sqrt{1-q^2}\)
B. \(p\sqrt{1-p^2} + q\sqrt{1-q^2}\)
C. \(q\sqrt{1-p^2} – p\sqrt{1-q^2}\)
D. \(p\sqrt{1-q^2} – q\sqrt{1-p^2}\)
E. \(p\sqrt{1-q^2} + q\sqrt{1-p^2}\)


Dengan menggunakan perbandingan pada segitiga siku-siku

\(\sin α = p\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\sin β=q\)
\(\cos α = \sqrt{1-p^2}\;\;\;\;\;\;\;\;\;\cos β=\sqrt{1-q^2}\)

\(\begin{align}
\sin(α+β)&=\sin α \cos β + \cos α \sin β\\
&=p\sqrt{1-q^2}+\sqrt{1-p^2}q\\
&=p\sqrt{1-q^2} + q\sqrt{1-p^2}\\
\end{align}\)


22. Diketahui A adalah sudut lancip dan \(\cos \frac{1}{2}A=\sqrt{\frac{x+1}{2x}}\), nilai \(\sin A\)

A. \(\sqrt{\frac{x^2-1}{x^2}}\)
B. \(\sqrt{\frac{x^2}{x^2+1}}\)
C. \(\sqrt{x^2-1}\)
D. \(\sqrt{x^2+1}\)
E. \(\sqrt{\frac{x^2+1}{x}}\)


Gunakan rumus \(\cos \frac{1}{2}A=±\sqrt{\frac{1+\cos A}{2}}\)

\(\cos \frac{1}{2}A=±\sqrt{\frac{1+\cos A}{2}}\)
karena A sudut lancip maka pilih yang bernilai positif
\(\cos \frac{1}{2}A=\sqrt{\frac{1+\cos A}{2}}\)
\(⇒\sqrt{\frac{x+1}{2x}}=\sqrt{\frac{1+\cos A}{2}}\)
\(⇒\frac{x+1}{2x}=\frac{1+\cos A}{2}\)
\(⇒\frac{x+1}{x}=1+\cos A\)
\(⇒\frac{x+1}{x}-1=\cos A\)
\(⇒\cos A =\frac{1}{x}\)

dengan bantuan perbandingan segitiga siku-siku

diperoleh \(\sin A = \frac{\sqrt{x^2-1}}{x}=\sqrt{\frac{x^2-1}{x^2}}\)


23. Nilai dari \(\frac{\cos 10°}{\cos 40° \cos 50°}\)

A. 3
B. 2
C. 1
D. 1/2
E. 1/4


gunakan rumus : \(\cos A \cos B = \frac{1}{2}(\cos (A+B)+\cos (A-B))\)

\(\frac{\cos 10°}{\cos 40° \cos 50°}\)
\(=\frac{\cos 10°}{\frac{1}{2}(\cos (40+50)° + \cos (40-50))°}\)
\(=\frac{\cos 10°}{\frac{1}{2}(\cos 90° + \cos (-10)°}\)
\(=\frac{\cos 10°}{\frac{1}{2}(0+ \cos 10°)}\)
\(=\frac{\cos 10°}{\frac{1}{2}\cos 10°}\)
\(=\frac{1}{\frac{1}{2}}\)
\(=2\)


24. Jika \(\tan 2x = 2\) untuk \(0<x<\frac{π}{2}\), maka nilai \(\tan x = …\)

A. \(\frac{1}{2}(\sqrt{5}-1)\)
B. \((\sqrt{5}-1)\)
C. \(\frac{1}{2}(\sqrt{5}+1)\)
D. \(2(\sqrt{5}-1)\)
E. \((\sqrt{5}+1)\)


Gunakan rumus \(\tan 2x=\frac{2\tan A}{1-\tan²A}\)

\(\tan 2x=\frac{2\tan A}{1-\tan²A}\)
\(⇒2=\frac{2\tan A}{1-\tan²A}\)
\(⇒2(1-\tan²A)=2\tan A\)
\(⇒1-\tan²A = \tan A\)
\(⇒\tan²A+\tan A – 1 = 0\)

misalkan \(\tan A = x\)

\(x^2 + x – 1 = 0\)

Dengan menggunakan rumus ABC

\(\begin{align}
x_{12}& = \frac{-b±\sqrt{b^2-4ac}}{2a}\\
&=\frac{-1±\sqrt{1^2 – 4(1)(-1)}}{2(1)}\\
&=\frac{-1±\sqrt{1+4}}{2}\\
&=\frac{-1±\sqrt{5}}{2}\\
&=\frac{1}{2}(-1±\sqrt{5})\\
\end{align}\)

karena \(0<x<\frac{π}{2}\) maka nilai x positif.

Jadi nilai \(x=\tan A=\frac{1}{2}(-1+\sqrt{5})\)


25. Jika \(A+B+C=180°\) maka \(\sin (\frac{B+C}{2})\) sama dengan …

A. \(\cos (\frac{A}{2})\)
B. \(-\cos (\frac{A}{2})\)
C. \(\sin (\frac{A}{2})\)
D. \(-\sin (\frac{A}{2})\)
E. \(\sec (\frac{A}{2})\)


\(A+B+C=180°\)
\(B+C=180º-A\)

\(\begin{align}
\sin \left(\frac{B+C}{2}\right)&=\sin\left(\frac{180°-A}{2}\right)\\
&=\sin \left(90°-\frac{A}{2}\right)\\
&=\cos \left(\frac{A}{2}\right)\\
\end{align}\)

ket :\(\sin(90°-A) = \cos A, \cos(90°-A)=\sin A, 0<A<90°\)


26. Jika \(𝐴 + 𝐵 + 𝐶 = 180°\), maka \(\cos 𝐵 + \cos 𝐶\) sama dengan …

a. \(2 \cos (\frac{𝐴}{2}) \cos (\frac{𝐵−𝐶}{2})\)
b. \(2 \sin (\frac{𝐴}{2}) \cos (\frac{𝐵−𝐶}{2})\)
c. \(2 \cos (\frac{𝐶}{2}) \cos (\frac{𝐵−𝐶}{2})\)
d. \(2 \cos (\frac{𝐵}{2}) \cos (\frac{𝐶}{2})\)
e. \(2 \sin (\frac{𝐵}{2}) \sin (\frac{𝐶}{2})\)


\(\begin{align}
\cos 𝐵 + \cos 𝐶 &= 2 \cos \frac{1}{2}(𝐵 + 𝐶) \cos\frac{1}{2}(𝐵 − 𝐶)\\
&= 2 \cos \frac{1}{2}(180 − 𝐴) \cos (\frac{𝐵−𝐶}{2})\\
&= 2 \cos(90 −\frac{𝐴}{2}) \cos (\frac{𝐵−𝐶}{2})\\
&= 2 \sin (\frac{𝐴}{2}) \cos (\frac{𝐵−𝐶}{2})\\
\end{align}\)


27. Nilai dari \(\left(\cos \frac{𝜋}{7}\cos\frac{2𝜋}{7}\cos\frac{4𝜋}{7}\right)\) adalah …

A. \(-\frac{1}{8}\)
B. \(-\frac{1}{4}\)
C. \(0\)
D. \(\frac{1}{2}\)
E. \(\frac{1}{3}\)


\(\left(\cos \frac{𝜋}{7}\cos\frac{2𝜋}{7}\cos\frac{4𝜋}{7}\right)\)
Kali dengan \(\frac{2\sin\frac{𝜋}{7}}{2 \sin\frac{𝜋}{7}}\), diperoleh

\(\frac{2 \sin\frac{𝜋}{7}\cos\frac{𝜋}{7}\cos\frac{2𝜋}{7}\cos\frac{4𝜋}{7}}{2 \sin\frac{𝜋}{7}}\)
Lanjutkan dengan bantuan rumus \(\sin 2𝐴 = 2 \sin 𝐴 \cos 𝐴\)
\(=\frac{\sin\frac{2𝜋}{7}\cos\frac{2𝜋}{7}\cos\frac{4𝜋}{7}}{2 \sin\frac{𝜋}{7}}\)
\(=\frac{2 \sin\frac{2𝜋}{7}\cos\frac{2𝜋}{7}\cos\frac{4𝜋}{7}}{4 \sin\frac{𝜋}{7}}\)
\(=\frac{\sin\frac{4𝜋}{7}\cos\frac{4𝜋}{7}}{4 \sin\frac{𝜋}{7}}\)
\(=\frac{2 \sin\frac{4𝜋}{7}\cos\frac{4𝜋}{7}}{8 \sin\frac{𝜋}{7}}\)
\(=\frac{\sin\frac{8𝜋}{7}}{8 \sin\frac{𝜋}{7}}\)
\(=\frac{\sin (𝜋 +\frac{𝜋}{7})}{8 \sin\frac{𝜋}{7}}\)
\(= −\frac{\sin (\frac{𝜋}{7})}{8\sin\frac{𝜋}{7}}\)
\(= −\frac{1}{8}\)


28. Nilai dari \(\left(\sin \frac{𝜋}{14}\sin\frac{3𝜋}{14}\sin\frac{9𝜋}{14}\right)\) adalah …

A. \(\frac{1}{16}\)
B. \(\frac{1}{8}\)
C. \(\frac{1}{4}\)
D. \(\frac{1}{2}\)
E. \(1\)



29. Nilai dari \((\tan 9° − \tan 27° − \tan 63° + \tan 81°)\)

A. 1
B. 2
C. 3
D. 4
E. 5


Rumus bantuan untuk menyelesaiakan soal di atas
\(\tan(90 − 𝐴) = \cot 𝐴\)
\(\tan 𝐴 =\frac{\sin 𝐴}{\cos 𝐴}\) dan \(\cot 𝐴 =\frac{\cos 𝐴}{\sin 𝐴}\)
Identitas \(\sin^2 𝐴 + \cos^2 𝐴 = 1\)
\(\sin 2𝐴 = 2 \sin 𝐴 \cos 𝐴\)
\(\sin 𝐴 − \sin 𝐵 = 2 \cos\frac{1}{2}(𝐴 + 𝐵) \sin\frac{1}{2}(𝐴 − 𝐵)\)

\(\begin{align}
(\tan 9° − \tan 27° − \tan 63° + \tan 81°)&=\tan 81° + \tan 9° − (\tan 27° + \tan 63°)\\
&= \tan(90 − 9)° + \tan 9° − (\tan 27° + \tan(90 − 27)°\\
&= \cot 9° + \tan 9° − (\tan 27° + \cot 27°)\\
&=\frac{\cos 9°}{\sin 9°}+\frac{\sin 9°}{\cos 9°}− \left(\frac{\cos 27°}{\sin 27°}+\frac{\sin 27°}{\cos 27°}\right)\\
&=\frac{\sin^2 9°+\cos^2 9°}{\sin 9° \cos 9°}− (\frac{\sin^2 27°+\cos^2 27°}{\sin 27° \cos 27°)}\\
&=\frac{1}{\sin 9° \cos 9°}−\frac{1}{\sin 27° \cos 27°}\\
&=\frac{2}{2\sin 9° \cos 9°}−\frac{2}{2 \sin 27° \cos 27°}\\
&=\frac{2}{\sin 18°}−\frac{2}{\sin 54°}\\
&=\frac{2(\sin 54°−\sin 18°)}{\sin 18° \sin 54°}\\
&=\frac{2(2 \cos 36° \sin 18°)}{\sin 18° \sin(90−36)°}\\
&=\frac{4 \cos 36°}{\cos 36°}\\
&= 4\\
\end{align}\)


30. Nilai dari \((\sin 50° − \sin 70° + \sin 10°)\) sama dengan …

A. \(\frac{1}{2}\)
B. 0
C. 1
D. 2
E. 3


Gunakan rumus \(\sin 𝐴 − \sin 𝐵 = 2 \cos\frac{1}{2}(𝐴 + 𝐵) \sin\frac{1}{2}(𝐴 − 𝐵)\)

\(\begin{align}
(\sin 50° − \sin 70° + \sin 10°)&= (\sin 50° − \sin 70°) + \sin 10°\\
&= 2 \cos 60° \sin(−10°) + \sin 10°\\
&= 2 (\frac{1}{2}) (− \sin 10°) + \sin 10°\\
&= − \sin 10° + \sin 10°\\
&= 0\\
\end{align}\)


31. Nilai dari \(\left(\cos \frac{𝜋}{5}\cos\frac{2𝜋}{5}\cos\frac{4𝜋}{5}\cos\frac{8𝜋}{5}\right)\) adalah …

A. \(-\frac{1}{16}\)
B. \(-\frac{1}{8}\)
C. \(0\)
D. \(\frac{1}{16}\)
E. \(\frac{1}{8}\)


\(\left(\cos \frac{𝜋}{5}\cos\frac{2𝜋}{5}\cos\frac{4𝜋}{5}\cos\frac{8𝜋}{5}\right)\)

kalikan dengan \(\frac{2\sin\frac{π}{5}}{2\sin\frac{π}{5}}\), diperoleh

\(\begin{align}
\cos \frac{𝜋}{5}\cos\frac{2𝜋}{5}\cos\frac{4𝜋}{5}\cos\frac{8𝜋}{5}&=\frac{2\sin\frac{π}{5}\cos \frac{𝜋}{5}\cos\frac{2𝜋}{5}\cos\frac{4𝜋}{5}\cos\frac{8𝜋}{5}}{2\sin\frac{π}{5}}\\
&=\frac{2\sin\frac{2π}{5}\cos\frac{2𝜋}{5}\cos\frac{4𝜋}{5}\cos\frac{8𝜋}{5}}{4\sin\frac{π}{5}}\\
&=\frac{2\sin\frac{4π}{5}\cos\frac{4𝜋}{5}\cos\frac{8𝜋}{5}}{8\sin\frac{π}{5}}\\
&=\frac{2\sin\frac{8π}{5}\cos\frac{8𝜋}{5}}{16\sin\frac{π}{5}}\\
&=\frac{\sin\frac{16π}{5}}{16\sin\frac{π}{5}}\\
&=\frac{\sin(3π+\frac{π}{5})}{16\sin\frac{π}{5}}\\
&=\frac{\sin\frac{π}{5}}{16\sin\frac{π}{5}}\\
&=\frac{1}{16}\\
\end{align}\)



 

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