SBMPTN Zone : Trigonometri

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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

$\begin{array}{rl}{\displaystyle \frac{1-{\tan }^{2}x}{2{\sec }^{2}x}}& ={\displaystyle \frac{1-{\displaystyle \frac{{\sin }^{2}x}{{\cos }^{2}x}}}{{\displaystyle \frac{2}{{\cos }^{2}x}}}}\\ & ={\displaystyle \frac{{\cos }^{2}x-{\sin }^{2}x}{2}}\\ & ={\displaystyle \frac{1}{2}}\cos 2x\end{array}$

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

$$\begin{array}{rl}\tan x& =\tan ({\displaystyle \frac{1}{2}}\pi +y)\\ & =-\cot y\\ & =-{\displaystyle \frac{1}{\tan y}}\end{array}$$

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

$$\begin{array}{rl}(\sin x+\cos x{)}^2& ={\sin }^{2}x+2\sin x\cos x+{\cos }^{2}x\\ (2p{)}^2& =2\sin x\cos x+1\\ \sin x\cos x& ={\displaystyle \frac{1}{2}}(4p^2-1)\end{array}$$

$$\begin{array}{rl}{\sin }^{4}x+{\cos }^{4}x& ={({\sin }^{2}x+{\cos }^{2}x)}^2-2{\sin }^{2}x{\cos }^{2}x\\ & =1^2-2{(\sin x\cos x)}^2\\ & =1-2{\left({\displaystyle \frac{1}{2}}(4p^2-1)\right)}^2\\ & =1-{\displaystyle \frac{1}{2}}{(4p^2-1)}^2\end{array}$$

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}

$$\begin{array}{rl}{\displaystyle \frac{{\displaystyle \frac{{\sin }^{2}x+{\cos }^{2}x}{{\tan }^{2}x+{\text{ctg}}^{2}x}}}{{\displaystyle \frac{{\sin }^{2}x-{\cos }^{2}x}{{\tan }^{2}x-{\text{ctg}}^{2}x}}}}& ={\displaystyle \frac{{\sin }^{2}x+{\cos }^{2}x}{{\tan }^{2}x+{\text{ctg}}^{2}x}}\cdot {\displaystyle \frac{{\tan }^{2}x-{\text{ctg}}^{2}x}{{\sin }^{2}x-{\cos }^{2}x}}\\ & ={\displaystyle \frac{1}{{\displaystyle \frac{{\sin }^{2}x}{{\cos }^{2}x}}+{\displaystyle \frac{{\cos }^{2}x}{{\sin }^{2}x}}}}\cdot {\displaystyle \frac{{\displaystyle \frac{{\sin }^{2}x}{{\cos }^{2}x}}-{\displaystyle \frac{{\cos }^{2}x}{{\sin }^{2}x}}}{{\sin }^{2}x-{\cos }^{2}x}}\\ & ={\displaystyle \frac{1}{{\displaystyle \frac{{\sin }^{4}x+{\cos }^{4}x}{{\cos }^{2}x{\sin }^{2}x}}}}\cdot {\displaystyle \frac{{\displaystyle \frac{{\sin }^{4}x-{\cos }^{4}x}{{\cos }^{2}x{\sin }^{2}x}}}{{\sin }^{2}x-{\cos }^{2}x}}\\ & ={\displaystyle \frac{\cancel{{\cos }^{2}x{\sin }^{2}x}}{{\sin }^{4}x+{\cos }^{4}x}}\cdot {\displaystyle \frac{{\sin }^{4}x-{\cos }^{4}x}{\left(\cancel{{\cos }^{2}x{\sin }^{2}x}\right)({\sin }^{2}x-{\cos }^{2}x)}}\\ & ={\displaystyle \frac{1}{{({\sin }^{2}x+{\cos }^{2}x)}^2-2{\sin }^{2}x{\cos }^{2}x}}\cdot {\displaystyle \frac{({\sin }^{2}x+{\cos }^{2}x)({\sin }^{2}x-{\cos }^{2}x)}{({\sin }^{2}x-{\cos }^{2}x)}}\cdot {\displaystyle \frac{\times 2}{\times 2}}\\ & ={\displaystyle \frac{2}{2{\left(1\right)}^2-4{\sin }^{2}x{\cos }^{2}x}}\cdot 1\\ & ={\displaystyle \frac{2}{2(1)-(2\sin x\cos x{)}^2}}\\ & ={\displaystyle \frac{2}{2-(\sin 2x{)}^2}}\\ & ={\displaystyle \frac{2}{1+1-{\sin }^{2}2x}}\\ & ={\displaystyle \frac{2}{1+{\cos }^{2}2x}}\\ & ={\displaystyle \frac{{\displaystyle \frac{2}{{\cos }^{2}x}}}{{\displaystyle \frac{1}{{\cos }^{2}x}}+1}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{2{\sec }^{2}x}{{\sec }^{2}x+1}}}}}}\end{array}$$

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

$$\begin{array}{rl}{(\cos \alpha -\sin \alpha )}^2& ={\cos }^2\alpha +{\sin }^2\alpha -2\cos \alpha \sin \alpha \\ & 1-2\left({\displaystyle \frac{8}{25}}\right)\\ & =1-{\displaystyle \frac{16}{25}}\\ & ={\displaystyle \frac{9}{25}}\\ \cos \alpha -\sin \alpha & =\sqrt{{\displaystyle \frac{9}{25}}}\\ & =\boxed{{\displaystyle {\displaystyle \frac{3}{5}}}}\end{array}$$

$$\begin{array}{rl}{\displaystyle \frac{1}{\sin \alpha }}-{\displaystyle \frac{1}{\cos \alpha }}& ={\displaystyle \frac{\cos \alpha -\sin \alpha }{\sin \alpha \cos \alpha }}\\ & ={\displaystyle \frac{{\displaystyle \frac{3}{5}}}{{\displaystyle \frac{8}{25}}}}\\ & ={\displaystyle \frac{3}{5}}\cdot {\displaystyle \frac{25}{8}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{15}{8}}}}}}\end{array}$$

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