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Pembuktian A sin x + B cos x = k cos (x - a)

 Penyelesaian persamaan $A \,sin\, x + B\, cos\, x$

 Persamaan $A\,sin \,x \, +\, B\, cos\,x$ dapat di ubah menjadi  $k cos (x-\alpha )$, sehingga $A \; sin x+ B \; cos x = k \, cos \, (x-\alpha )$

Pembuktian:

\(\begin{array}{rcl}A\, sin\,x +\,B\,cos \,x&=&k \, cos \, (x-\alpha )\\A\, sin\,x +\,B\,cos \,x&=&k \left (  cos\,x\,cos\alpha+sin\,x\,sin\,\alpha  \right )\\A\, sin\,x +\,B\,cos \,x&=&k\left (  cos\,x\,cos\alpha\right )+k \left ( sin\,x\,sin\,\alpha  \right)\\A\, sin\,x +\,B\,cos \,x&=&k \,cos\,x\,cos\,\alpha+k \,sin\,x\,sin\,\alpha\\\end{array}\)

Dari bentuk kesamaan tersebut maka:

$A\, sin\,x =k \,sin\,x\,sin\,\alpha$ diperoleh $A = k\, sin \,\alpha$ →$A^2 = k^2\, sin^2 \,\alpha$ .............................................. (1)

$B\, cos\,x =k \,cos\,x\,cos\,\alpha$ diperoleh $B = k\, cos \,\alpha$→$B^2 = k^2\, cos^2 \,\alpha$ .............................................. (2)

Dengan menjumlahkan (1) dan (2) diperoleh:

\(\begin{array}{rcl}A^2 + B^2 &=& k^2\, sin^2 \,\alpha + k^2 cos^2 \alpha\\A^2 + B^2 &=& k^2\left (sin^2 \,\alpha + cos^2 \alpha  \right )\\A^2 + B^2 &=& k^2\left (1 \right )\\A^2 + B^2 &=& k^2\\\sqrt{A^2 + B^2} &=&k\\\end{array}\)

Dengan membagikan (1) dan (2) diperoleh:

\(\frac{A}{B}=\frac{sin\,\alpha }{cos\, \alpha }=tan\,\alpha \)

Dengan demikian terbukti bahwa:

$A \; sin x+ B \; cos x = k \, cos \, (x-\alpha )$

Limit Tak Hingga Fungsi Aljabar

Limit tak Hingga Fungsi Aljabar

Bentuk 1: \(\displaystyle \lim_{ x\to \infty }\sqrt{ax^2+bx+c}-\sqrt{px^2+qx+r}=...\)

Penyelesaian:

\(\begin{array}{rcl}&=&\displaystyle \lim_{ x\to \infty}\sqrt{ax^2+bx+c}-\sqrt{px^2+qx+r}\times \left (\frac{\sqrt{ax^2+bx+c}+\sqrt{px^2+qx+r}}{\sqrt{ax^2+bx+c}+\sqrt{px^2+qx+r}}\right )\\&=&\displaystyle \lim_{x \to \infty }\frac{\left ( ax^2+bx+c \right )-\left ( px^2+qx+r \right )}{\sqrt{ax^2+bx+c}-\sqrt{px^2+qx+r}}\\&=&\displaystyle \lim_{x \to \infty }\frac{(a-p)x^2+(b-q)x+(c-r)}{\sqrt{ax^2+bx+c}+\sqrt{px^2+qx+r}} \end{array}\)

Bila $a = p$

\(\begin{array}{rcl}&=&\displaystyle \lim_{x \to \infty }\frac{(a-a)x^2+(b-q)x+(c-r)}{\sqrt{ax^2+bx+c}+\sqrt{ax^2+qx+r}} \\&=&\displaystyle \lim_{x \to \infty }\frac{(b-q)x+(c-r)}{\sqrt{ax^2+bx+c}+\sqrt{ax^2+qx+r}}\\&=&\displaystyle \lim_{x \to \infty }\frac{(b-q)}{2\sqrt{a}}\end{array}\)

Bila $a > p$

\(\begin{array}{rcl}&=&\displaystyle \lim_{x \to \infty }\frac{(a-p)x^2+(b-q)x+(c-r)}{\sqrt{ax^2+bx+c}+\sqrt{px^2+qx+r}} \\&=&\displaystyle \lim_{x \to \infty }\frac{kx^2+(b-q)x+(c-r)}{\sqrt{ax^2+bx+c}+\sqrt{px^2+qx+r}}\\&=&\infty\end{array}\)

Bila $a < p$

\(\begin{array}{rcl}&=&\displaystyle \lim_{x \to \infty }\frac{(a-p)x^2+(b-q)x+(c-r)}{\sqrt{ax^2+bx+c}+\sqrt{px^2+qx+r}} \\&=&\displaystyle \lim_{x \to \infty }\frac{-kx^2+(b-q)x+(c-r)}{\sqrt{ax^2+bx+c}+\sqrt{px^2+qx+r}}\\&=&-\infty\end{array}\)

Bentuk 2:\(\displaystyle \lim_{ x\to \infty }\sqrt[3]{ax^3+bx^2+cx+d}-\sqrt[3]{px^3+qx^2+rx+s}=...\)

Penyelesaian:

\(\begin{array}{rcl}&=&\displaystyle \lim_{ x\to \infty}\sqrt[3]{ax^3+bx^2+cx+d}-\sqrt[3]{px^3+qx^2+rx+s}\times \left ( \frac{\sqrt[3]{\left ( ax^3+bx^2+cx+d \right )^{2}}+\sqrt[3]{\left ( ax^3+bx^2+cx+d \right ){\left ( px^3+qx^2+rx+s \right )}}+\sqrt[3]{\left ( px^3+qx^2+rx+s \right )^{2}}}{\sqrt[3]{\left ( ax^3+bx^2+cx+d \right )^{2}}+\sqrt[3]{\left ( ax^3+bx^2+cx+d \right )\left ( px^3+qx^2+rx+s \right )}+\sqrt[3]{\left ( px^3+qx^2+rx+s \right )^{2}}} \right )\\&=&\displaystyle \lim_{ x\to \infty}\frac{\left ( ax^3+bx^2+cx+d \right )-\left ( px^3+qx^2+rx+s \right )}{\sqrt[3]{(ax^3+bx^2+cx+d)^{2}}+\sqrt[3]{(ax^3+bx^2+cx+d)(px^3+qx^2+rx+s)}+\sqrt[3]{(px^3+qx^2+rx+s)^{2}}}\\&=&\displaystyle \lim_{ x\to \infty}\frac{(a-p)x^3+(b-q)x^2+(c-r)x+(d-s) }{\sqrt[3]{(ax^3+bx^2+cx+d)^{2}}+\sqrt[3]{(ax^3+bx^2+cx+d)(px^3+qx^2+rx+s)}+\sqrt[3]{(px^3+qx^2+rx+s)^{2}}} \end{array}\)

Bila $a = p$

\(\begin{array}{rcl}&=&\displaystyle \lim_{ x\to \infty}\frac{(a-a)x^3+(b-q)x^2+(c-r)x+(d-s) }{\sqrt[3]{(ax^3+bx^2+cx+d)^{2}}+\sqrt[3]{(ax^3+bx^2+cx+d)(ax^3+qx^2+rx+s)}+\sqrt[3]{(ax^3+qx^2+rx+s)^{2}}}\\&=&\displaystyle \lim_{ x\to \infty}\frac{(b-q)x^2+(c-r)x+(d-s) }{\sqrt[3]{(ax^3)^{2}}+\sqrt[3]{(ax^3)(ax^3)}+\sqrt[3]{(ax^3)^{2}}}\\&=&\displaystyle \lim_{ x\to \infty}\frac{(b-q)x^2+(c-r)x+(d-s) }{\sqrt[3]{(a^2x^6)}+\sqrt[3]{(a^2x^6)}+\sqrt[3]{(a^2x^6)}}\\&=&\displaystyle \lim_{ x\to \infty}\frac{(b-q)x^2+(c-r)x+(d-s) }{\sqrt[3]{(a)^{2}}x^2+\sqrt[3]{(a)^{2}}x^2+\sqrt[3]{(a)^{2}}x^2} \\&=&\displaystyle \lim_{ x\to \infty}\frac{(b-q)x^2+(c-r)x+(d-s) }{3\sqrt[3]{a^{2}}x^2}\\&=&\frac{b-q }{3\sqrt[3]{a^{2}}} \end{array}\)

Bila $a > p$

\(\begin{array}{rcl}&=&\displaystyle \lim_{ x\to \infty}\frac{(a-p)x^3+(b-q)x^2+(c-r)x+(d-s) }{\sqrt[3]{(ax^3+bx^2+cx+d)^{2}}+\sqrt[3]{(ax^3+bx^2+cx+d)(px^3+qx^2+rx+s)}+\sqrt[3]{(px^3+qx^2+rx+s)^{2}}}\\&=&\displaystyle \lim_{ x\to \infty}\frac{kx^3+(b-q)x^2+(c-r)x+(d-s) }{\sqrt[3]{(ax^3+bx^2+cx+d)^{2}}+\sqrt[3]{(ax^3+bx^2+cx+d)(ax^3+qx^2+rx+s)}+\sqrt[3]{(ax^3+qx^2+rx+s)^{2}}}\\&=&\displaystyle \lim_{ x\to \infty}\frac{kx^3+(b-q)x^2+(c-r)x+(d-s) }{lx^2}\\&=&\displaystyle \lim_{ x\to \infty}\frac{kx^3 }{lx^2}\\&=&\infty \end{array}\)

Bila $a < p$

\(\begin{array}{rcl}&=&\displaystyle \lim_{ x\to \infty}\frac{(a-p)x^3+(b-q)x^2+(c-r)x+(d-s) }{\sqrt[3]{(ax^3+bx^2+cx+d)^{2}}+\sqrt[3]{(ax^3+bx^2+cx+d)(px^3+qx^2+rx+s)}+\sqrt[3]{(px^3+qx^2+rx+s)^{2}}}\\&=&\displaystyle \lim_{ x\to \infty}\frac{-kx^3+(b-q)x^2+(c-r)x+(d-s) }{\sqrt[3]{(ax^3+bx^2+cx+d)^{2}}+\sqrt[3]{(ax^3+bx^2+cx+d)(ax^3+qx^2+rx+s)}+\sqrt[3]{(ax^3+qx^2+rx+s)^{2}}}\\&=&\displaystyle \lim_{ x\to \infty}\frac{-kx^3+(b-q)x^2+(c-r)x+(d-s) }{lx^2}\\&=&\displaystyle \lim_{ x\to \infty}\frac{-kx^3 }{lx^2}\\&=&-\infty \end{array}\)

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