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Trigonometri Arsip | 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 …

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Problem – Tangen sudut rangkap

Problem:Buktikan bahwa $$begin{aligned}tan ntheta &=frac{displaystylebinom{n}{1} tan theta – binom{n}{3} tan^3theta+cdots}{displaystyle1-binom{n}{2} tan^2theta+cdots}end{aligned}$$ Solusi:Dengan menggunakan De Moivre’s Theorem $e^{(itheta)n}=e^{i(ntheta)}=cosleft(nthetaright)+isinleft(nthetaright)$. Sementara itu, $e^{(itheta)n}=left(costheta+isinthetaright)^n$. Sehingga bisa kita tuliskan$$begin{aligned}cosleft(nthetaright)+isinleft(nthetaright)&=sum_{k=0}^nbinom{n}{k}i^kcos^{n-k}thetasin^kthetaend{aligned}$$Tetapi kita tahu bahwa $i^2=-1$ dan $i^3=-i$. Sehingga bentuk di atas dapat disederhanakan menjadi $$begin{aligned}cosleft(nthetaright)+isinleft(nthetaright)&=sum_{k=0}^nbinom{n}{2k}(-1)^kcos^{n-2k}thetasin^{2k}theta+sum_{k=0}^nbinom{n}{2k+1}(i)^{2k+1}cos^{n-2k-1}thetasin^{2k+1}thetaend{aligned}$$Dengan memisahkan antara nilai Real dan Imaginer pada bentuk di atas, diperoleh:$$cos(ntheta)=sum_{k=0}^nbinom{n}{2k}(-1)^kcos^{n-2k}thetasin^{2k}theta$$$$sin(ntheta)=sum_{k=0}^nbinom{n}{2k+1}(-1)^kcos^{n-2k-1}thetasin^{2k+1}theta$$Selanjutnya, kita tahu bahwa $displaystyletan(ntheta)=frac{sin(ntheta)}{cos(ntheta)}$. Sehingga bentuk di atas …

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