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{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

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