6.7.5 Fungsi Trigonometri, SPM Praktis (Kertas 1)

6.7.5 Fungsi Trigonometri, SPM Praktis (Kertas 1)
Soalan 11:
Buktikan identiti \(\frac{kos^2x}{1-\sin x}=1+\sin x\)
Penyelesaian:
\(\begin{array}{l}\text{Sebelah kiri}\\=\frac{kos^2x}{1-\sin x}\\=\frac{1-\sin^2x}{1-\sin x}\leftarrow\boxed{\sin^2x+kos^2x=1}\\=\frac{{(1+\sin x)}{(1-\sin x)}}{1-\sin x}\\=1+\sin x\\=\text{Sebelah kanan}\end{array}\)  
Soalan 12:
Buktikan identiti \(\sin^2x-kos^2x=\frac{\tan^2x-1}{\tan^2x+1}\)   
Penyelesaian:
\(\begin{aligned} &\text { Sebelah kanan } \frac{\tan ^{2} x-1}{\tan ^{2} x+1}\\ &=\frac{\frac{\sin ^{2} x}{\operatorname{los} x}-1}{\frac{\sin ^{2} x}{\operatorname{kos}^{2} x}+1} \leftarrow \tan x=\frac{\sin x}{\operatorname{los} x}\\ &=\frac{\frac{\sin ^{2} x-\operatorname{kos}^{2} x}{\operatorname{kos} 2 x}}{\frac{\sin ^{2} x+\cos ^{2} x}{\operatorname{kos}^{2} x}}\\ &=\frac{\sin ^{2} x-\cos ^{2} x}{\sin ^{2} x+\cos ^{2} x}\\ &=\sin ^{2} x-\operatorname{sos}^{2} x \leftarrow \quad \sin ^{2} x+\operatorname{kos}^{2} x=1\\ &=\text { Sebelah kiri } \end{aligned}\)
Soalan 13:
Buktikan identiti tan2 θ– sin2 θ = tan2θ sin2 θ
Penyelesaian:
\(\begin{array}{l}\text{Sebelah kiri}\\=\tan^2\theta-\sin^2\theta\\=\frac{\sin^2\theta}{kos^2\theta}-\sin^2\theta\\=\frac{\sin^2\theta-\sin^2\theta kos^2\theta}{kos^2\theta}\\=\frac{\sin^2\theta{(1-kos^2\theta)}}{kos^2\theta}\\=\frac{\sin^2\theta\sin^2\theta}{kos^2\theta}\\={(\frac{\sin^2\theta}{kos^2\theta})}{(\sin^2\theta)}\\=\tan^2\theta\sin^2\theta\\=\text{Sebelah kanan}\end{array}\)
Soalan 14:
Buktikan identiti kosek2 θ (sek2 θ – tan2 θ) – 1 = kot2 θ
Penyelesaian:
Sebelah kiri,
kosek2 θ (sek2θ – tan2 θ) – 1
= kosek2 θ (1) – 1  ← (tan2 θ + 1 = sek2θ , sek2 θ – tan2θ  = 1)
= kosek2 θ – 1
= kot2 θ  ←(1 + kot2 θ = kosek2 θ , kosek2 θ – 1 = kot2 θ  )
= Sebelah kanan

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