SBMPTN Zone : Trigonometri

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

Tipe:

Standar SBMPTN HOTS

No.

Nilai minimum dari fungsi h(x)=\dfrac{1-\tan^2x}{2\sec^2x} untuk 0\leq x\leq2\pi adalah....

1. -3
2. -2
3. -1

4. -\dfrac12
5. 0

Dari Sherly

$\eqalign{ \dfrac{1-\tan^2x}{2\sec^2x}&=\dfrac{1-\dfrac{\sin^2x}{\cos^2x}}{\dfrac2{\cos^2x}}\\[4pt] &=\dfrac{\cos^2x-\sin^2x}2\\[4pt] &=\dfrac12\cos2x }$

Minimum = -\dfrac12

Jadi, Nilai minimum dari fungsi h(x)=\dfrac{1-\tan^2x}{2\sec^2x} untuk 0\leq x\leq2\pi adalah -\dfrac12. JAWAB: D

No.

Jika x-y=\dfrac12\pi maka \tan x adalah....

1. \dfrac{1+\tan y^2}{y}
2. -\dfrac{1-y^2}{\tan y}
3. \dfrac{\tan(1-y)}{(1+y)^2}

4. \dfrac{\tan y}{(1+y)^2}
5. -\dfrac1{\tan y}

x=\dfrac12\pi+y

\begin{aligned} \tan x&=\tan\left(\dfrac12\pi+y\right)\\ &=-\cot y\\ &=-\dfrac1{\tan y} \end{aligned}

Jadi, \tan x=-\dfrac1{\tan y}. JAWAB: E

No.

Jika \sin x+\cos x=2p, maka \sin^4x+\cos^4x= ....

1. 1-\left(2p^2-1\right)^2
2. 1+\dfrac12\left(2p^2-1\right)^2
3. 1-\dfrac12\left(2p^2-1\right)^2

4. 1+\dfrac12\left(4p^2-1\right)^2
5. 1-\dfrac12\left(4p^2-1\right)^2

Ganesha Operation

\begin{aligned} (\sin x+\cos x)^2&=\sin^2x+2\sin x\cos x+\cos^2x\\ (2p)^2&=2\sin x\cos x+1\\ \sin x\cos x&=\dfrac12\left(4p^2-1\right) \end{aligned}

\begin{aligned} \sin^4x+\cos^4x&=\left(\sin^2x+\cos^2x\right)^2-2\sin^2x\cos^2x\\ &=1^2-2\left(\sin x\cos x\right)^2\\ &=1-2\left(\dfrac12\left(4p^2-1\right)\right)^2\\ &=1-\dfrac12\left(4p^2-1\right)^2 \end{aligned}

Jadi, \sin^4x+\cos^4x=\dfrac12\left(4p^2-1\right). JAWAB: E

No.

Nilai yang ekivalen dengan \dfrac{\dfrac{\sin^2x+\cos^2x}{\tan^2x+\ctg^2x}}{\dfrac{\sin^2x-\cos^2x}{\tan^2x-\ctg^2x}} adalah

1. \dfrac2{2+\sin^22x}
2. \dfrac2{1+2\cos^22x}
3. \dfrac{2\sec^2x}{\sec^2x-1}

4. \dfrac{2\csc^2x}{\csc^2x+1}
5. \dfrac{2\sec^2x}{\sec^2x+1}

\begin{aligned} \dfrac{\dfrac{\sin^2x+\cos^2x}{\tan^2x+\ctg^2x}}{\dfrac{\sin^2x-\cos^2x}{\tan^2x-\ctg^2x}}&=\dfrac{\sin^2x+\cos^2x}{\tan^2x+\ctg^2x}\cdot\dfrac{\tan^2x-\ctg^2x}{\sin^2x-\cos^2x}\\[9pt] &=\dfrac1{\dfrac{\sin^2x}{\cos^2x}+\dfrac{\cos^2x}{\sin^2x}}\cdot\dfrac{\dfrac{\sin^2x}{\cos^2x}-\dfrac{\cos^2x}{\sin^2x}}{\sin^2x-\cos^2x}\\[9pt] &=\dfrac1{\dfrac{\sin^4x+\cos^4x}{\cos^2x\sin^2x}}\cdot\dfrac{\dfrac{\sin^4x-\cos^4x}{\cos^2x\sin^2x}}{\sin^2x-\cos^2x}\\[9pt] &=\dfrac{\cancel{\cos^2x\sin^2x}}{\sin^4x+\cos^4x}\cdot\dfrac{\sin^4x-\cos^4x}{\left(\cancel{\cos^2x\sin^2x}\right)\left(\sin^2x-\cos^2x\right)}\\[9pt] &=\dfrac1{\left(\sin^2x+\cos^2x\right)^2-2\sin^2x\cos^2x}\cdot\dfrac{\left(\sin^2x+\cos^2x\right)\left(\sin^2x-\cos^2x\right)}{\left(\sin^2x-\cos^2x\right)}\color{red}{\cdot\dfrac{\times2}{\times2}}\\[9pt] &=\dfrac2{2\left(1\right)^2-4\sin^2x\cos^2x}\cdot1\\[8pt] &=\dfrac2{2(1)-(2\sin x\cos x)^2}\\[8pt] &=\dfrac2{2-(\sin2x)^2}\\[8pt] &=\dfrac2{1+1-\sin^22x}\\[8pt] &=\dfrac2{1+\cos^22x}\\[8pt] &=\dfrac{\dfrac2{\cos^2x}}{\dfrac1{\cos^2x}+1}\\[8pt] &=\boxed{\boxed{\dfrac{2\sec^2x}{\sec^2x+1}}} \end{aligned}

Jadi, \dfrac{\dfrac{\sin^2x+\cos^2x}{\tan^2x+\ctg^2x}}{\dfrac{\sin^2x-\cos^2x}{\tan^2x-\ctg^2x}}=\dfrac{2\sec^2x}{\sec^2x+1}. JAWAB: E

No.

Diketahui \sin\alpha\cos\alpha=\dfrac8{25}. Nilai \dfrac1{\sin\alpha}-\dfrac1{\cos\alpha} adalah

1. \dfrac3{25}
2. \dfrac9{25}
3. \dfrac58

4. \dfrac35
5. \dfrac{15}8

\begin{aligned} \left(\cos\alpha-\sin\alpha\right)^2&=\cos^2\alpha+\sin^2\alpha-2\cos\alpha\sin\alpha\\ &1-2\left(\dfrac8{25}\right)\\ &=1-\dfrac{16}{25}\\ &=\dfrac9{25}\\ \cos\alpha-\sin\alpha&=\sqrt{\dfrac9{25}}\\ &=\boxed{\dfrac35} \end{aligned}

\begin{aligned} \dfrac1{\sin\alpha}-\dfrac1{\cos\alpha}&=\dfrac{\cos\alpha-\sin\alpha}{\sin\alpha\cos\alpha}\\ &=\dfrac{\dfrac35}{\dfrac8{25}}\\ &=\dfrac35\cdot\dfrac{25}8\\ &=\boxed{\boxed{\dfrac{15}8}} \end{aligned}

Jadi, \dfrac1{\sin\alpha}-\dfrac1{\cos\alpha}=\dfrac{15}8. JAWAB: E