6.7.6 Fungsi Trigonometri, SPM Praktis (Kertas 1)

6.7.6 Fungsi Trigonometri, SPM Praktis (Kertas 1)
Soalan 15:
Buktikan identiti \(\frac2{kos2A+1}=sek^2A\)
Penyelesaian:
\(\begin{array}{l}\text{Sebelah kiri}\\\text{=}\frac2{kos2A+1}\\=\frac2{{(2kos^2A-1)}+1}\leftarrow\boxed{kos2A=2kos^2A-1}\\=\frac2{2kos^2A}=\frac1{kos^2A}\\=sek^2A\\=\text{Sebelah kanan}\end{array}\)
Soalan 16:
Buktikan identiti \(\frac{2\tan A}{2-sek^2A}=\tan2A\)
Penyelesaian:
\(\begin{array}{l}\text{Sebelah kiri}\\\text{=}\frac{2\tan A}{2-sec^2A}\\=\frac{2\tan A}{2-{(\tan^2A+1)}}\leftarrow\boxed{\tan^2A+1=sec^2A}\\=\frac{2\tan A}{1-\tan^2A}\\=\tan2A\\=\text{Sebelah kanan}\end{array}\)
Soalan 17:
Buktikan identiti tan x + kot x = 2 kosek 2x
Penyelesaian:
Sebelah kiri,
tan x + kot x
\(\begin{array}{l}=\frac{\sin x}{kosx}+\frac{kosx}{\sin x}\\=\frac{\sin^2x+kos^2x}{kosx\sin x}\\=\frac1{kosx\sin x}\leftarrow\boxed{\sin^2x+kos^2x=1}\\=\frac1{\frac12\sin2x}\leftarrow\boxed{\begin{array}{l}\sin2x=2\sin xkosx\\\frac12\sin2x=\sin xkosx\end{array}}\\=\frac2{\sin2x}\\=2{(\frac1{\sin2x})}\\=2kosek2x\\=\text{Sebelah kanan}\end{array}\)
Soalan 18:
Buktikan identiti \(\frac{kosx-\sin2x}{kos2x+\sin x-1}=\frac1{\tan x}\)
Penyelesaian:
\(\begin{array}{l}\text{Sebelah kiri}\\\frac{kosx-\sin2x}{kos2x+\sin x-1}\\=\frac{kosx-2\sin xkosx}{{(1-2\sin^2x)}+\sin x-1}\leftarrow\boxed{kos2x=1-2\sin^2x}\\=\frac{kosx{(1-2\sin x)}}{\sin x-2\sin^2x}\\=\frac{kosx{(1-2\sin x)}}{\sin x{(1-2\sin x)}}\\=\frac{kosx}{\sin x}\\=kotx\\=\frac1{\tan x}\\\text{Sebelah kanan}\end{array}\)

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