Exercise Zone : Jumlah dan Selisih Trigonometri

Berikut ini adalah kumpulan soal mengenai jumlah dan selisih trigonometri tipe Standar. Jika ingin bertanya soal, silahkan gabung ke grup Facebook atau Telegram.

Tipe:

Standar SBMPTN HOTS

No 1

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

$\begin{array}{rl}\cos (x-y)& ={\displaystyle \frac{4}{5}}\\ \cos x\cos y+\sin x\sin y& ={\displaystyle \frac{4}{5}}\\ \cos x\cos y+{\displaystyle \frac{3}{10}}& ={\displaystyle \frac{4}{5}}\\ \cos x\cos y& ={\displaystyle \frac{4}{5}}-{\displaystyle \frac{3}{10}}\\ & ={\displaystyle \frac{5}{10}}\end{array}$

$\begin{array}{rl}\tan x\tan y& ={\displaystyle \frac{\sin x\sin y}{\cos x\cos y}}\\ & ={\displaystyle \frac{{\displaystyle \frac{3}{10}}}{{\displaystyle \frac{5}{10}}}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{3}{5}}}}}}\end{array}$

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

2

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

2. 
   
3. 
   

5. 
   

$\begin{array}{rl}\tan (A+5B)& =\tan (3A+4B-(2A-B))\\ & ={\displaystyle \frac{\tan (3A+4B)-\tan (2A-B)}{1+\tan (3A+4B)\tan (2A-B)}}\\ & ={\displaystyle \frac{6-2}{1+(6)(2)}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{4}{13}}}}}}\end{array}$

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

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

No. 4

Jika \cos(2x+35\degree)=p dan \cos(x+25\degree)=q maka \sin(3x+60\degree)\sin(x+10\degree)=

1. q^2-p^2
2. p^2-q^2
   
3. p^2+q^2
   

4. 1-2p^2q^2
5. 2p^2q^2-1
   

$\begin{array}{rl}2\sin (3x+60{}^{\circ })\sin (x+10{}^{\circ })& =\cos ((3x+60{}^{\circ })-(x+10{}^{\circ }))-\cos ((3x+60{}^{\circ })+(x+10{}^{\circ }))\\ & =\cos (2x+50{}^{\circ })-\cos (4x+70{}^{\circ })\\ & =\cos 2(x+25{}^{\circ })-\cos 2(2x+35{}^{\circ })\\ & =2{\cos }^2(x+25{}^{\circ })-1-(2{\cos }^2(2x+35{}^{\circ })-1)\\ & =2q^2-1-(2p^2-1)\\ & =2q^2-1-2p^2+1\\ & =2q^2-2p^2\\ & =2(q^2-p^2)\\ \sin (3x+60{}^{\circ })\sin (x+10{}^{\circ })& =\boxed{{\displaystyle \boxed{{\displaystyle q^2-p^2}}}}\end{array}$

Jadi, \sin(3x+60\degree)\sin(x+10\degree)=q^2-p^2.
JAWAB: A

No. 5

Tentukan nilai dari {\sin105\degree+\sin15\degree}

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$\begin{array}{rl}\sin {105}^{\circ }+\sin {15}^{\circ }& =2\sin \left({\displaystyle \frac{{105}^{\circ }+{15}^{\circ }}{2}}\right)\cos \left({\displaystyle \frac{{105}^{\circ }-{15}^{\circ }}{2}}\right)\\ & =2\sin \left({\displaystyle \frac{{120}^{\circ }}{2}}\right)\cos \left({\displaystyle \frac{{90}^{\circ }}{2}}\right)\\ & =2\sin {60}^{\circ }\cos {45}^{\circ }\\ & =2\left({\displaystyle \frac{1}{2}}\sqrt{3}\right)\left({\displaystyle \frac{1}{2}}\sqrt{2}\right)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{1}{2}}\sqrt{6}}}}}\end{array}$

Jadi, {\sin105\degree+\sin15\degree=\dfrac12\sqrt6}.

LIHAT JUGA:
Trigonometri