Exercise Zone : Jumlah dan Selisih Trigonometri

Berikut ini adalah kumpulan soal mengenai jumlah dan selisih trigonometri tingkat dasar. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.

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No. 1

Diketahui ${\cos(x-y)=\dfrac45}$ dan ${\sin x\sin y=\dfrac3{10}}$. Nilai $\tan x\tan y$ adalah ....

$\begin{aligned}\cos(x-y)&=\dfrac45\\\cos x\cos y+\sin x\sin y&=\dfrac45\\\cos x\cos y+\dfrac3{10}&=\dfrac45\\\cos x\cos y&=\dfrac45-\dfrac3{10}\\&=\dfrac5{10}\end{aligned}$

$\begin{aligned}\tan x\tan y&=\dfrac{\sin x\sin y}{\cos x\cos y}\\&=\dfrac{\dfrac3{10}}{\dfrac5{10}}\\&=\boxed{\boxed{\dfrac35}}\end{aligned}$

Jadi, $\tan x\tan y=\dfrac35$.

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No. 2

Diketahui ${\tan(3A+4B)=6}$ dan ${\tan(2A-B)=2}$. Nilai dari $\tan(A+5B)=$ ....

1. $-\dfrac8{11}$
2. $8$
3. $4$

4. $\dfrac8{11}$
5. $\dfrac4{13}$

$\begin{aligned}\tan(A+5B)&=\tan\left(3A+4B-(2A-B)\right)\\&=\dfrac{\tan(3A+4B)-\tan(2A-B)}{1+\tan(3A+4B)\tan(2A-B)}\\[8pt]&=\dfrac{6-2}{1+(6)(2)}\\&=\boxed{\boxed{\dfrac4{13}}}\end{aligned}$

Jadi, Nilai dari $\tan(A+5B)=\dfrac4{13}$.
JAWAB: E

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No. 3

Jika ${\sin(3x+20)=a}$ dan ${\sin(x+10)=b}$ maka ${2\sin(4x+30)\sin(2x+10)=}$

1. ${1-2a^2b^2}$
2. ${a^2-b^2}$
   
3. ${b^2-a^2}$
   

4. ${2a^2-2b^2}$
5. ${2b^2-2a^2}$
   

$$\begin{array}{rl}2\sin (4x+30)\sin (2x+10)& =\cos ((4x+30)-(2x+10))-\cos ((4x+30)+(2x+10))\\ & =\cos (4x+30-2x-10)-\cos (4x+30+2x+10)\\ & =\cos (2x+20)-\cos (6x+40)\\ & =\cos 2(x+10)-\cos 2(3x+20)\\ & =1-2{\sin }^2(x+10)-(1-2{\sin }^2(3x+20))\\ & =1-2b^2-(1-2a^2)\\ & =1-2b^2-1+2a^2\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 2a^2-2b^2}}}}\end{array}$$

Jadi, $2\sin(4x+30)\sin(2x+10)=2a^2-2b^2$.
JAWAB: D

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