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

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

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

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

2

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

2. 
   
3. 
   

5. 
   

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

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

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
   

\(\begin{aligned} 2\sin(4x+30)\sin(2x+10)&=\cos((4x+30)-(2x+10))-\cos((4x+30)+(2x+10))\\ &=\cos(4x+30-2x-10)-\cos(4x+30+2x+10)\\ &=\cos(2x+20)-\cos(6x+40)\\ &=\cos2(x+10)-\cos2(3x+20)\\ &=1-2\sin^2(x+10)-\left(1-2\sin^2(3x+20)\right)\\ &=1-2b^2-\left(1-2a^2\right)\\ &=1-2b^2-1+2a^2\\ &=\boxed{\boxed{2a^2-2b^2}} \end{aligned}\)

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
   

\(\begin{aligned} 2\sin(3x+60\degree)\sin(x+10\degree)&=\cos\left((3x+60\degree)-(x+10\degree)\right)-\cos\left((3x+60\degree)+(x+10\degree)\right)\\ &=\cos(2x+50\degree)-\cos(4x+70\degree)\\ &=\cos2(x+25\degree)-\cos2(2x+35\degree)\\ &=2\cos^2(x+25\degree)-1-\left(2\cos^2(2x+35\degree)-1\right)\\ &=2q^2-1-\left(2p^2-1\right)\\ &=2q^2-1-2p^2+1\\ &=2q^2-2p^2\\ &=2\left(q^2-p^2\right)\\ \sin(3x+60\degree)\sin(x+10\degree)&=\boxed{\boxed{q^2-p^2}} \end{aligned}\)

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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\(\eqalign{ \sin105^\circ+\sin15^\circ&=2\sin\left(\dfrac{105^\circ+15^\circ}2\right)\cos\left(\dfrac{105^\circ-15^\circ}2\right)\\ &=2\sin\left(\dfrac{120^\circ}2\right)\cos\left(\dfrac{90^\circ}2\right)\\ &=2\sin60^\circ\cos45^\circ\\ &=2\left(\dfrac12\sqrt3\right)\left(\dfrac12\sqrt2\right)\\ &=\boxed{\boxed{\dfrac12\sqrt6}} }\)

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

LIHAT JUGA:
Trigonometri