Latihan Matematika Wajib Kelas X Fungsi Trigonometri dan Bilangan Real
# 7
Pilgan

Bentuk lain dari 2sin⁡2β−1sin⁡βcos⁡β\frac{2\sin^2\beta-1}{\sin\beta\cos\beta} adalah ....

A

tan⁡β−cotan⁡β\tan\beta-\operatorname{cotan}\beta

B

tan⁡β+cotan⁡β\tan\beta+\operatorname{cotan}\beta

C

sec⁡β−cosec⁡β\sec\beta-\operatorname{cosec}\beta

D

sec⁡β+cosec⁡β\sec\beta+\operatorname{cosec}\beta

E

sin⁡β−cos⁡β\sin\beta-\cos\beta

Pembahasan:

Perlu diingat identitas-identitas pada trigonometri berikut:

sin⁡2θ+cos⁡2θ=1⇔cos⁡2θ=1−sin⁡2θ\sin^2\theta+\cos^2\theta=1\Leftrightarrow\cos^2\theta=1-\sin^2\theta

tan⁡θ=sin⁡θcos⁡θ\tan\theta=\frac{\sin\theta}{\cos\theta}

cotan⁡θ=cos⁡θsin⁡θ\operatorname{cotan}\theta=\frac{\cos\theta}{\sin\theta}

Selain itu juga perlu diingat:

sifat komutatif dalam operasi penjumlahan dan perkalian, yaitu untuk sembarang bilangan aa dan bb berlaku

a+b=b+aa+b=b+a dan

a×b=b×aa\times b=b\times a

sifat distributif dalam operasi aljabar, yaitu untuk sembarang bilangan a, b,a,\ b, dan cc berlaku

(a+b).c=a.c+b.c\left(a+b\right).c=a.c+b.c

Pada soal diketahui ekspresi 2sin⁡2β−1sin⁡βcos⁡β\frac{2\sin^2\beta-1}{\sin\beta\cos\beta}. Berdasarkan identitas-identitas pada trigonometri yang disebutkan sebelumnya diperoleh

2sin⁡2β−1sin⁡βcos⁡β=sin⁡2β+sin⁡2β−1sin⁡βcos⁡β\frac{2\sin^2\beta-1}{\sin\beta\cos\beta}=\frac{\sin^2\beta+\sin^2\beta-1}{\sin\beta\cos\beta} berdasarkan sifat komutatif diperoleh

⇔2sin⁡2β−1sin⁡βcos⁡β=sin⁡2β−1+sin⁡2βsin⁡βcos⁡β\Leftrightarrow\frac{2\sin^2\beta-1}{\sin\beta\cos\beta}=\frac{\sin^2\beta-1+\sin^2\beta}{\sin\beta\cos\beta} berdasarkan sifat distributif diperoleh

⇔2sin⁡2β−1sin⁡βcos⁡β=sin⁡2β−1(1−sin⁡2β)sin⁡βcos⁡β\Leftrightarrow\frac{2\sin^2\beta-1}{\sin\beta\cos\beta}=\frac{\sin^2\beta-1\left(1-\sin^2\beta\right)}{\sin\beta\cos\beta}

⇔2sin⁡2β−1sin⁡βcos⁡β=sin⁡2β−1(cos⁡2β)sin⁡βcos⁡β\Leftrightarrow\frac{2\sin^2\beta-1}{\sin\beta\cos\beta}=\frac{\sin^2\beta-1\left(\cos^2\beta\right)}{\sin\beta\cos\beta}

⇔2sin⁡2β−1sin⁡βcos⁡β=sin⁡2β−cos⁡2βsin⁡βcos⁡β\Leftrightarrow\frac{2\sin^2\beta-1}{\sin\beta\cos\beta}=\frac{\sin^2\beta-\cos^2\beta}{\sin\beta\cos\beta}

⇔2sin⁡2β−1sin⁡βcos⁡β=sin⁡βsin⁡β−cos⁡βcos⁡βcos⁡βsin⁡β\Leftrightarrow\frac{2\sin^2\beta-1}{\sin\beta\cos\beta}=\frac{\sin\beta\sin\beta-\cos\beta\cos\beta}{\cos\beta\sin\beta}

⇔2sin⁡2β−1sin⁡βcos⁡β=sin⁡βcos⁡β−cos⁡βsin⁡β\Leftrightarrow\frac{2\sin^2\beta-1}{\sin\beta\cos\beta}=\frac{\sin\beta}{\cos\beta}-\frac{\cos\beta}{\sin\beta}

⇔2sin⁡2β−1sin⁡βcos⁡β=tan⁡β−cotan⁡β\Leftrightarrow\frac{2\sin^2\beta-1}{\sin\beta\cos\beta}=\tan\beta-\operatorname{cotan}\beta

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