HOTS Zone : Trigonometri

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Berikut ini adalah kumpulan soal mengenai Trigonometri. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.

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STANDARSNBTHOTS

No. 1

Misalkan
sin (sin (5x − 4y) ⋅ sin 4(x − y)) = a, dan
cos (cos (5x − 4y) ⋅ cos 4(x − y)) = b
dengan $0<y<x<{\displaystyle \frac{1}{20}}\pi$. Jika dinyatakan dalam a dan b maka cos (cos x) = ....

SUMBER

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}\sin (\sin (5x-4y)\cdot \sin 4(x-y))& =a\\ \sin (5x-4y)\cdot \sin (4x-4y)& ={\sin }^{-1}a& \text{color}red(1)\end{array}$

$\begin{array}{rl}\cos (\cos (5x-4y)\cdot \cos 4(x-y))& =b\\ \cos (5x-4y)\cdot \cos (4x-4y)& ={\cos }^{-1}b& \text{color}red(2)\end{array}$

(1) + (2)
$\begin{array}{rl}\sin (5x-4y)\cdot \sin (4x-4y)& ={\sin }^{-1}a\\ \cos (5x-4y)\cdot \cos (4x-4y)& ={\cos }^{-1}b& \text{color}red+\\ \cos (5x-4y)\cdot \cos (4x-4y)+\sin (5x-4y)\cdot \sin (4x-4y)& ={\sin }^{-1}a+{\cos }^{-1}b\\ \cos ((5x-4y)-(4x-4y))& ={\sin }^{-1}a+{\cos }^{-1}b\\ \cos x& ={\sin }^{-1}a+{\cos }^{-1}b\\ \cos (\cos x)& =\cos ({\sin }^{-1}a+{\cos }^{-1}b)\\ & =\cos ({\sin }^{-1}a)\cdot \cos ({\cos }^{-1}b)-\sin ({\sin }^{-1}a)\cdot \sin ({\cos }^{-1}b)\\ & =\sqrt{1-a^2}\text{cdotb}-a\cdot \sqrt{1-b^2}\end{array}$

Jadi,
JAWAB:

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No. 2

cos² 41° + cos² 43° + cos² 45° + cos² 47° + cos² 49° = ....

1. 1
2. 1,5
3. 2

4. 2,5
5. 5

Learncy.net

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}{\cos }^{2}41\text{degree}+{\cos }^{2}43\text{degree}+{\cos }^{2}45\text{degree}+{\cos }^{2}47\text{degree}+{\cos }^{2}49\text{degree}& ={\cos }^{2}41\text{degree}+{\cos }^{2}43\text{degree}+{\left({\displaystyle \frac{1}{2}}\sqrt{2}\right)}^2+{\sin }^{2}43\text{degree}+{\sin }^{2}41\text{degree}\\ & ={\cos }^{2}41\text{degree}+{\sin }^{2}41\text{degree}+{\cos }^{2}43\text{degree}+{\sin }^{2}43\text{degree}+{\displaystyle \frac{1}{4}}\cdot 2\\ & =1+1+{\displaystyle \frac{1}{2}}\\ & =2,5\end{array}$

Jadi, cos² 41° + cos² 43° + cos² 45° + cos² 47° + cos² 49° = 2,5.
JAWAB: D

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No. 3

Nilai (sin(15°) + sin(75°))² adalah ....

1. ${\displaystyle \frac{\sqrt{3}}{2}}$
2. ${\displaystyle \frac{3}{2}}$
3. $1+\sqrt{3}$

4. ${\displaystyle \frac{3}{4}}$
5. ${\displaystyle \frac{2}{3}}$

EDUVERSAL MATHEMATICS COMPETITION 2019

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}{(\sin \left(15°\right)+\sin \left(75°\right))}^2& ={\sin }^2\left(15°\right)+{\sin }^2\left(75°\right)+2\sin \left(15°\right)\sin \left(75°\right)\\ & ={\sin }^2\left(15°\right)+{\cos }^2\left(15°\right)+2\sin \left(15°\right)\cos \left(15°\right)\\ & =1+sin\left(30°\right)\\ & =1+{\displaystyle \frac{1}{2}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{3}{2}}}}}}\end{array}$

Jadi, ${(\sin \left(15°\right)+\sin \left(75°\right))}^2={\displaystyle \frac{3}{2}}$.
JAWAB: B

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No. 4

Nilai dari $\dfrac{3+\cot76°\cot16°}{\cot76°+\cot16°}$ adalah ....

1. tan 16°
2. tan 46°

3. cot 16°
4. cot 44°

5. cot 46°

KPATN 2024 Jenjang Guru

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}{\displaystyle \frac{3+\cot 76°\cot 16°}{\cot 76°+\cot 16°}}& ={\displaystyle \frac{3+{\displaystyle \frac{\cos 76°}{\sin 76°}}{\displaystyle \frac{\cos 16°}{\sin 16°}}}{{\displaystyle \frac{\cos 76°}{\sin 76°}}+{\displaystyle \frac{\cos 16°}{\sin 16°}}}}\text{color}red\cdot {\displaystyle \frac{\sin 76°\sin 16°}{\sin 76°\sin 16°}}\\ & ={\displaystyle \frac{3\sin 76°\sin 16°+\cos 76°\cos 16°}{\sin 16°\cos 76°+\cos 16°\sin 76°}}\\ & ={\displaystyle \frac{2\sin 76°\sin 16°+\sin 76°\sin 16°+\cos 76°\cos 16°}{\sin (16°+76°)}}\\ & ={\displaystyle \frac{\cos (76°-16°)-\cos (76°+16°)+\cos (76°-16°)}{\sin (16°+76°)}}\\ & ={\displaystyle \frac{\cos 60°-\cos 92°+\cos 60°}{\sin 92°}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{2}}-(1-2{\sin }^{2}46°)+{\displaystyle \frac{1}{2}}}{2\sin 46°\cos 46°}}\\ & ={\displaystyle \frac{2{\sin }^{2}46°}{2\sin 46°\cos 46°}}\\ & ={\displaystyle \frac{\sin 46°}{\cos 46°}}\\ & =\text{color}blue\boxed{{\displaystyle \boxed{{\displaystyle \text{color}black\tan 46°}}}}\end{array}$

Jadi, nilai dari $\dfrac{3+\cot76°\cot16°}{\cot76°+\cot16°}$ adalah tan 46°.
JAWAB: B

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No. 5

Jika $\sin^{-1}x+\sin^{-1}y+\sin^{-1}z=\dfrac{3\pi}2$, maka nilai dari $x^{2023}+y^{2023}+z^{2023}+\dfrac{27}{x^{2023}+y^{2023}+z^{2023}}$ adalah ....

1. 0
2. 3

3. 6
4. 9

5. 12

KPATN 2024 Jenjang Guru

ALTERNATIF PENYELESAIAN

Perhatikan bahwa $-\dfrac{\pi}2\leq\sin^{-1}\alpha\leq\dfrac{\pi}2$, sehingga $-\dfrac{3\pi}2\leq\sin^{-1}x+\sin^{-1}y+\sin^{-1}z\leq\dfrac{3\pi}2$. Karena $\sin^{-1}x+\sin^{-1}y+\sin^{-1}z=\dfrac{3\pi}2$, maka $\sin^{-1}x=\sin^{-1}y=\sin^{-1}z=\dfrac{\pi}2$. Didapat
$x=y=z=\sin\dfrac{\pi}2=1$

$\begin{array}{rl}{x}^{2023}+y^{2023}+z^{2023}+{\displaystyle \frac{27}{x^{2023}+y^{2023}+z^{2023}}}& =1^{2023}+1^{2023}+1^{2023}+{\displaystyle \frac{27}{1^{2023}+1^{2023}+1^{2023}}}\\ & =3+{\displaystyle \frac{27}{3}}\\ & =3+9\\ & =\text{color}blue\boxed{{\displaystyle \boxed{{\displaystyle \text{color}black12}}}}\end{array}$

Jadi, nilai dari $x^{2023}+y^{2023}+z^{2023}+\dfrac{27}{x^{2023}+y^{2023}+z^{2023}}$ adalah 12.
JAWAB: E

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No. 6

Jika $\sqrt{\sin^4x+4\cos^2x}-\sqrt{\cos^4x+4\sin^2x}=\dfrac13$, maka nilai 6 cos 2x adalah ....

Evaluasi GMOM Februari 2024 tingkat SMA

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}\sqrt{{\sin }^{4}x+4{\cos }^{2}x}-\sqrt{{\cos }^{4}x+4{\sin }^{2}x}& ={\displaystyle \frac{1}{3}}\\ \sqrt{{\sin }^{4}x+4(1-{\sin }^{2}x)}-\sqrt{{\cos }^{4}x+4(1-{\cos }^{2}x)}& ={\displaystyle \frac{1}{3}}\\ \sqrt{{\sin }^{4}x-4{\sin }^{2}x+4}-\sqrt{{\cos }^{4}x-4{\cos }^{2}x+4}& ={\displaystyle \frac{1}{3}}\\ \sqrt{{({\sin }^{2}x-2)}^2}-\sqrt{{({\cos }^{2}x-2)}^2}& ={\displaystyle \frac{1}{3}}\\ |{\sin }^{2}x-2|-|{\cos }^{2}x-2|& ={\displaystyle \frac{1}{3}}\end{array}$
Perhatikan bahwa sin² x ≤ 1, dan cos² x ≤ 1, maka sin² x − 2 dan cos² x − 2 bernilai negatif, sehingga

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$\begin{array}{rl}|{\sin }^{2}x-2|-|{\cos }^{2}x-2|& ={\displaystyle \frac{1}{3}}\\ 2-{\sin }^{2}x-(2-{\cos }^{2}x)& ={\displaystyle \frac{1}{3}}\\ 2-{\sin }^{2}x-2+{\cos }^{2}x& ={\displaystyle \frac{1}{3}}\\ {\cos }^{2}x-{\sin }^{2}x& ={\displaystyle \frac{1}{3}}\\ \cos 2x& ={\displaystyle \frac{1}{3}}\\ 6\cos 2x& =\text{color}blue\boxed{{\displaystyle \boxed{{\displaystyle \text{color}black2}}}}\end{array}$

Jadi, nilai 6 cos 2x adalah 2.

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No. 7

Jika $\sin A=\sqrt{2pq}$ dan $\tan A=\dfrac{2pq}{p-q}$, maka p² + q² = ....

Evaluasi GMOM September 2024

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}\tan A& ={\displaystyle \frac{2pq}{p-q}}\\ {\displaystyle \frac{\sin A}{\cos A}}& ={\displaystyle \frac{2pq}{p-q}}\\ {\displaystyle \frac{\sqrt{2pq}}{\cos A}}& ={\displaystyle \frac{2pq}{p-q}}\\ \cos A& =p-q\end{array}$

$\begin{array}{rl}{\sin }^{2}A+{\cos }^{2}A& =1\\ {\left(\sqrt{2pq}\right)}^2+(p-q{)}^2& =1\\ 2pq+p^2-2pq+q^2& =1\\ p^2+q^2& =\text{color}blue\boxed{{\displaystyle \boxed{{\displaystyle \text{color}black1}}}}\end{array}$

Jadi, p² + q² = 1.

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No. 8

Nilai dari

cos(56°)⋅cos(2⋅56°)⋅cos(2²⋅56°)⋯cos(2²³56°)

adalah ....

1. 2⁻²⁴
2. 2⁻²³
3. 2²⁴

4. 2²³
5. 2⁻²⁸

Penyisihan LMNas UGM ke-28

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}\sin 2a& =2\sin a\cos a\\ \cos a& ={\displaystyle \frac{\sin 2a}{2\sin a}}\end{array}$

$\begin{array}{rl}\cos \left(56°\right)\cdot \cos (2\cdot 56°)\cdot \cos (2^2\cdot 56°)\cdots \cos (2^{23}\cdot 56°)& ={\displaystyle \frac{\sin (2\cdot 56°)}{2\sin \left(56°\right)}}\cdot {\displaystyle \frac{\sin (2^2\cdot 56°)}{2\sin (2\cdot 56°)}}\cdot {\displaystyle \frac{\sin (2^3\cdot 56°)}{2\sin (2^2\cdot 56°)}}\cdots {\displaystyle \frac{\sin (2^{24}\cdot 56°)}{2\sin (2^{23}\cdot 56°)}}\\ & ={\displaystyle \frac{\sin (2^{24}\cdot 56°)}{2^{24}\sin \left(56°\right)}}\end{array}$
Kita buktikan bahwa 2²⁴⋅56° = 56° + k⋅360°, atau 2²⁴⋅56 = 56 mod 360.
$\begin{array}{rl}{2}^{24}\cdot 56\mathrm{mod}360& \equiv 8(2^{24}\cdot 7\mathrm{mod}45)\mathrm{mod}360\\ & \equiv 8\left(7\mathrm{mod}45\right)\mathrm{mod}360& \text{color}red\varphi (45)=24\\ & \equiv 56\mathrm{mod}360\end{array}$

$\begin{array}{rl}{\displaystyle \frac{\sin (2^{24}\cdot 56°)}{2^{24}\sin \left(56°\right)}}& ={\displaystyle \frac{\sin \left(56°\right)}{2^{24}\sin \left(56°\right)}}\\ & ={\displaystyle \frac{1}{2^{24}}}\\ & =\text{color}blue\boxed{{\displaystyle \boxed{{\displaystyle \text{color}black{2}^{-24}}}}}\end{array}$

Jadi, cos(56°)⋅cos(2⋅56°)⋅cos(2²⋅56°)⋯cos(2²³56°) = 2⁻²⁴.
JAWAB: A

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No. 9

Jika sudut lancip θ memenuhi persamaan $\sin \theta ={\displaystyle \frac{3}{5}}$, manakah dari pilihan berikut ini yang benar?

1. 0° < θ < 30°
2. 30° < θ < 45°
3. 45° < θ < 60°

4. 60° < θ < 75°
5. 75° < θ < 90°

Penyisihan OGM 10 Tingkat SMA

ALTERNATIF PENYELESAIAN

Perhatikan bahwa jika 0° < α < θ < β < 90°, maka 0 < sin α < sin θ < sin β < 1.
Perhatikan tabel nilai sin berikut ini:

$\theta$	0°	30°	45°	60°	90°
$\sin \theta$	0	${\displaystyle \frac{1}{2}}=0,5$	${\displaystyle \frac{\sqrt{2}}{2}}=0,7$	${\displaystyle \frac{\sqrt{3}}{2}}=0,86$	1

Karena $\sin \theta ={\displaystyle \frac{3}{5}}=0,6$, maka $\theta$ berada di antara 30° dan 45°.

Jadi, 30° < θ < 45°.
JAWAB: B

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