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. ${2p^2q^2-1}$

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

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

---

No. 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

$\begin{array}{rl}2\cos \alpha \cdot \cos \beta & =\cos (\alpha +\beta )+\cos (\alpha -\beta )\\ 2\left({\displaystyle \frac{3}{8}}\right)& ={\displaystyle \frac{1}{2}}+\cos (\alpha -\beta )\\ {\displaystyle \frac{3}{4}}& ={\displaystyle \frac{1}{2}}+\cos (\alpha -\beta )\\ \cos (\alpha -\beta )& ={\displaystyle \frac{3}{4}}-{\displaystyle \frac{1}{2}}\\ & ={\displaystyle \frac{1}{4}}\end{array}$

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

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

Share this post