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Problem - Nilai Maksimum fungsi Trigonometri | Catatan Nurkholis

Problem – Nilai Maksimum fungsi Trigonometri

Problem:
Tentukan nilai maksimum dari
$$y=5- frac{15}{4cos x – 2sqrt{5}sin x +9}$$

Solusi:
Perhatikan bahwa $4cos x – 2sqrt{5}sin x$ dapat disederhanakan menjadi:
$$
begin{aligned}
4cos x – 2sqrt{5}sin x&=left(sqrt{16+20}right)left(cosalphacos x – sinalphasin xright)quad alpha=arctan{frac{sqrt{5}}{2}}\
&=6cos(x+alpha)
end{aligned}
$$
Sehingga soal di atas dapat disederhanakan menjadi:
$$y=5-frac{15}{6cos(x+alpha)+9}$$
Agar nilai $y$ maksimum, haruslah $displaystyle frac{15}{6cos(x+alpha)+9}$ minimum. Yang berarti bahwa nilai $cos(x+alpha)$ mencapai maksimum, yakni $= 1$. Sehingga Nilai maksimum $y$ adalah:
$$
begin{aligned}
y&=5-frac{15}{6cos(x+alpha)+9}\
&=5-frac{15}{6+9}\
&=5-frac{15}{15}\
&=5-1\
&=4
end{aligned}
$$
Jadi Nilai maksimum $y$ adalah $4$

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