SBMPTN 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.

---

No. 1

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. 2

Jika {\cos\alpha\cdot\cos\beta=\dfrac38} dan {\cos (\alpha+\beta) = \dfrac12} maka \tan (\alpha-\beta) =

1. \dfrac14
2. \dfrac12\sqrt3
3. \dfrac13\sqrt5

4. \sqrt{15}
5. \dfrac12\sqrt{15}

Ganesha Operation

\(\eqalign{ 2\cos\alpha\cdot\cos\beta&=\cos (\alpha+\beta)+\cos (\alpha-\beta)\\ 2\left(\dfrac38\right)&=\dfrac12+\cos (\alpha-\beta)\\ \dfrac34&=\dfrac12+\cos (\alpha-\beta)\\ \cos (\alpha-\beta)&= \dfrac34-\dfrac12\\ &=\dfrac14 }\)

{\tan\left(\alpha-\beta\right)=\dfrac{\sqrt{15}}1=\boxed{\boxed{\sqrt{15}}}}

Jadi, \tan (\alpha-\beta) =\sqrt{15}.
JAWAB: D

Share this post