- සංයුක්ත ගණිතය I (ශුද්ධ ගණිතය) ප්‍රශ්න පත්‍රයේ A කොටසේ(කෙටි ප්‍රශ්න) 10 වැනි ගැටළුවෙහි B කොටසේ(රචනා ප්‍රශ්න) 17 වැනි ගැටළුවෙහි මෙම පාඩමේ අඩංගු සිද්ධාන්ත අඩංගු වේ.
- එකට එක සහ මතට වන ශ්‍රිතයක් වසම සහ සහවසම හුවමාරු කිරීමෙන් ලැබෙන ශ්‍රිතය ප්‍රතිලෝම ශ්‍රිතය වේ.
Sin-1(x ) ශ්‍රිතය
\mathrm{වසම}\;\lbrack-1,1\rbrack\;,\;\mathrm{පරාසය}\;\lbrack-\frac{\mathrm\pi}2,\frac{\mathrm\pi}2\rbrack
Cos -1(x ) ශ්‍රිතය
\mathrm{වසම}\;\lbrack-1,1\rbrack\;,\;\mathrm{පරාසය}\;\lbrack0,\mathrm\pi\rbrack
Tan -1(x ) ශ්‍රිතය
\mathrm{වසම}\;\lbrack R\rbrack\;,\;\mathrm{පරාසය}\;\lbrack-\frac{\mathrm\pi}2,\frac{\mathrm\pi}2\rbrack
සරල ගැටලු කිහිපයක් සලකා බලමු.
- \;Sin^{-1}\left(\cos\;\frac\pi3\right)\\=Sin^{-1}\left(\frac12\right)\\=\frac\pi6
2. \;Sin^{-1}\left(\sin\;\frac{5\pi}6\right)\\=Sin^{-1}\left[\sin\;\left(\pi-\frac\pi6\right)\right]\\=\frac\pi6
3. \cos^{-1}\left(\cos\;\frac{4\pi}3\right)\\\cos^{-1}\lbrack\cos(\mathrm\pi+\frac{\mathrm\pi}3)\rbrack\\=\frac{2\pi}3
4. Tan^{-1}\left(2\sin\frac{4\pi}3\right)\\=Tan^{-1}\left[2\sin\;\left(\pi+\frac\pi3\right)\right]\\=Tan^{-1}\left(2\sin\frac\pi3\right)\\=Tan^{-1}\left(\sqrt3\right)\\=\;\frac\pi3
5. \sin^{-1}\left(\sin^{-1\;}\left(\frac12\right)\right)\\=\frac12
6. \cos^{-1}\left(\cos^{-1\;}\left(-\frac1{\sqrt2}\right)\right)\\=\frac1{\sqrt2}
අවසානයට ත්‍රිකෝණමිතික ශ්‍රිතය ගැනීමේ දී ක්‍රියාව සහ ප්‍රතික්‍රියාව ලෙස සලකා මුල් අගයම ගතහැකි වුවද අවසානයට ප්‍රතිලෝම ශ්‍රිතය ගැනීමේදී මුල් අගය අනුරූප පරාසයට අයත් දැයි පරීක්ෂා කළ යුතුය.
උදාහරණ ගැටලුව 1
\begin{array}{l}2\cos\;^{-1}\left(\frac34\right)\;=\cos^{-1}\left(\frac18\right)\;\mathrm{බව}\;\mathrm{පෙන්වන්න}\\\end{array}.
\begin{array}{rcl}\mathrm A&=&\cos^{-1}\left(\frac34\right)\;\mathrm{යැයි}\;\mathrm{ගනිමු}.\\\;\mathrm{Cos}\;\mathrm A\;&=&\frac34>0\\;0&\leq&\mathrm A\leq\mathrm\pi\;\mathrm{වේ}.\mathrm{නමුත්}\;\;¾>0\;\mathrm{නිසා},\\\;0&<&\mathrm A<\frac{\mathrm\pi}2\;\;\;\;\mathrm{විය}\;\mathrm{යුතුය}.\;\\\mathrm Y&=&2\;\cos^{-1}\left(\frac34\right)\\\mathrm Y&=&2\mathrm A\;\\\mathrm{Cos}\;\mathrm Y\;&=&\;\mathrm{Cos}\;2\mathrm A\\&=&2Cos^2A-1\\&=&2\left(\frac9{16}\right)\;–\;1\\\;&=&\frac18\\\;0&<&\mathrm A<\frac{\mathrm\pi}2\;\;\mathrm{නිසා},0<2\mathrm A<\mathrm\pi\;\mathrm{වේ}.\mathrm{එනම්}\;\mathrm Y\;\mathrm{අදාල}\;\mathrm{පරාසයට}\;\mathrm{අයත්}\;\mathrm{වේ}.\;\mathrm{ඒනිසා},\;\\\;\;\mathrm Y&=&\mathrm{Cos}\;^{-1}\left(\frac18\right)\;\mathrm{වේ}.\\&&\end{array}
උදාහරණ ගැටලුව 2
\begin{array}{l}2\tan\;^{-1}\left(\frac12\right)\;=\cos^{-1}\left(\frac35\right)\;\mathrm{බව}\;\mathrm{පෙන්වන්න}\\\end{array}.
\begin{array}{l}\begin{array}{rcl}\mathrm A&=&\tan^{-1}\left(\frac12\right)\;\mathrm{යැයි}\;\mathrm{ගනිමු}.\\\tan\;\mathrm A\;&=&\frac12\\-\frac{\mathrm\pi}2&<&\mathrm A<\frac{\mathrm\pi}2\;\mathrm{වේ}.\;\mathrm{නමුත්}\;\mathrm{Tan}\;\mathrm A\;\mathrm{ධන}\;\mathrm{නිසා}\;\;\begin{array}{rcl}\;0&<&\mathrm A<\frac{\mathrm\pi}2\;\;\end{array}\mathrm{වේ}.\;\\\mathrm Y&=&2\tan^{-1}\left(\frac12\right)\\\mathrm Y&=&2\mathrm A\;\\\mathrm{Cos}\;\mathrm Y\;&=&\;\mathrm{Cos}\;2\mathrm A\\&=&2Cos^2A-1\\&=&\dfrac{1-\tan^2A}{1+\tan^2A}\\\;&=&\dfrac{1-{\displaystyle\dfrac14}}{1+{\displaystyle\dfrac14}}\;\;=\;\dfrac35\\\;0&<&\mathrm A<\dfrac{\mathrm\pi}2\;\;\mathrm{නිසා},0<2\mathrm A<\mathrm\pi\;\mathrm{වේ}.\mathrm{එනිසා}\;\mathrm Y\;=\cos^{-1\;}\left(\dfrac35\right)\;\mathrm{වේ}.\;\;\\\;\;\mathrm{ඒනම්}\;,2\tan^{-1}\left(\dfrac12\right)&=&\mathrm{Cos}\;^{-1}\left(\dfrac35\right)\;\mathrm{වේ}.\end{array}\\\\\\\\\end{array}
උදාහරණ ගැටලුව 3
\begin{array}{l}Tan\;^{-1}\left(2\right)\;+Tan\;^{-1}\left(\dfrac13\right)\;=Tan\;^{-1}(7)\;\mathrm{බව}\;\mathrm{පෙන්වන්න}.\\\\\\\\\end{array}.
\begin{array}{l}\mathrm A=\mathrm{Tan}\;^{-1}(2)\;\\-\frac{\mathrm\pi}2<\mathrm A<\dfrac{\mathrm\pi}2\;\\\;\mathrm{Tan}\;\mathrm A\;=2\;\mathrm{ධන}\;\mathrm{නිසා},.\;\\\;0<\mathrm A<\dfrac{\mathrm\pi}2\\\mathrm B=\mathrm{Tan}\;^{-1}\left(\dfrac13\right)\\-\dfrac{\mathrm\pi}2<\mathrm A<\dfrac{\mathrm\pi}2\;\\\;\mathrm{TanB}\;=\dfrac13\;\mathrm{ධන}\;\mathrm{නිසා},.\;\\\;0<\mathrm B<\dfrac{\mathrm\pi}2\\\;Tan\;Y\;=\;Tan\;(A+B)\;\\=\dfrac{Tan\;A\;+\;TanB}{1–\;Tan\;A\;Tan\;B\;\;}\\=\dfrac{2+{\displaystyle\dfrac13}\;\;}{1–2(\dfrac13)\;\;}\\=7\\\mathrm{මෙහි}\;\mathrm A\;\mathrm{හා}\;\mathrm B\;\mathrm{හි}\;\mathrm{පරාස}\;\mathrm{පිළිබඳ}\;\mathrm{සලකා}\;\mathrm{බලමු}.\;\\0<A+B<\pi\;\;\mathrm{වේ}.\;\mathrm{එනම්}\;\;0<Y<\pi\;\mathrm{වේ}.\\\;\mathrm{නමුත්}\;Tan\;Y\;\mathrm{ධන}\;\;\;\;\mathrm{නිසා}\;\;0<Y<\pi/2\text{ වේ. }\\\text{ඒනිසා,}Y\;=\;Tan\;^{-1}(7)\;\mathrm{වේ}.\\\;\mathrm{එනම්},\;\;Tan\;^{-1}(2)\;+Tan\;^{-1}\left(\dfrac13\;\right)=Tan\;^{-1}(7)\;\mathrm{වේ}.\\\\\end{array}
උදාහරණ ගැටලුව 4
\begin{array}{l}\mathrm{Sin}\;^{-1}(–\mathrm x)\;=\;-\mathrm{Sin}\;^{-1}(\mathrm x)\;\mathrm{බව}\;\mathrm{පෙන්වන්න}.\\\\\end{array}.
\begin{array}{rcl}\mathrm{Tan}\;^{-1}(2\mathrm x)\;&=&\mathrm A\;\mathrm{හා}\;\;\mathrm{Tan}\;^{-1}(3\mathrm x)\;=\mathrm B\;\mathrm{ලෙස}\;\mathrm{ගනිමු}.\\-\frac{\mathrm\pi}2&<&\mathrm A<\frac{\mathrm\pi}2\;\\Tan\;\mathrm A\;&=&\;2\mathrm x-\\\frac{\mathrm\pi}2&<&\mathrm B<\frac{\mathrm\pi}2\\Tan\;\mathrm B\;&=&\;3\mathrm x\\\mathrm A+\mathrm B&=&\;\frac{\mathrm\pi}4\\Tan(\mathrm A+\mathrm B)&=&\mathrm{Tan}\;\frac{\mathrm\pi}4\;\\\mathrm{Tan}\;\mathrm A\;+\mathrm{Tan}\;\mathrm B1–\;\mathrm{Tan}\;\mathrm A\;\mathrm{Tan}\;\mathrm B\;&=&\;1\\;2\mathrm x+3\mathrm x1–2\mathrm x.3\mathrm x&=&1\\5\mathrm x&=&1–6\mathrm x^2\;\\6\mathrm x^2+5\mathrm x–1&=&0\\(6\mathrm x–1)(\mathrm x+1)&=&0\\\mathrm X&=&\frac16\;\mathrm{හෝ}\;\mathrm x=(–1)\end{array}
උදාහරණ ගැටලුව 5
\begin{array}{rcl}\mathrm{Sin}\;^{-1}(\mathrm x)\;+\;\mathrm{Cos}\;^{-1}(\mathrm x)\;&=&\;\frac{\mathrm\pi}2\;\mathrm{බව}\;\mathrm{පෙන්වන්න}.\\&&\\&&\end{array}.
\begin{array}{rcl}Sin^{-1}(x)\;&=&\;A\;\mathrm{යැයි}\;\mathrm{ගනිමු}.\\-\frac\pi2&\leq&A\leq\frac\pi2\;\\\frac\pi2&\geq&(–A)\geq–\frac\pi2\\&&\mathrm{දෙපසට}\;\mathrm ම\;\frac{\mathrm\pi}2\;\mathrm{ක්}\;\mathrm{එකතු}\;\mathrm{කරමු}.\\\pi&\geq&(\frac{\mathrm\pi}2–A)\geq0\;\;\;\;\;\;\;\;\;\;\;\;\\Sin\;A\;&=&\;X\;\\Cos\;\left(\frac{\mathrm\pi}2–A\right)\;\;&=&\;X\\\;Cos^{\;-1}\left[\cos\left(\frac{\mathrm\pi}2–A\right)\right]&=&Cos^{\;-1}(x)\\\frac{\mathrm\pi}2–A\;&=&\;Cos^{-1}(x)\;\\\frac{\mathrm\pi}2&=&\;A\;+\;Cos^{-1}(x)\;\\&&\text{එමනිසා,}\;\\Sin\;^{-1}(x)\;+\;Cos\;^{-1}(x)\;&=&\;\frac{\mathrm\pi}2\;\mathrm{වේ}.\end{array}
මෙම සෘජුකෝණාස්‍රය තුළ එක සමාන සුදු පැහැ සමචතුරස්‍ර 7ක් තිබෙනවා. ඔබට ලැබෙන අභියෝගය වන්නේ එක් සුදු පැහැ සමචතුරස්‍රයක වර්ගඵලය සෙවීමයි.
ප්‍රතිලෝම ත්‍රිකෝණමිතික සමීකරණ විසඳීම.
උදාහරණ ගැටලුව 6
\begin{array}{rcl}Tan^{\;-1}(2x)\;+\;Tan\;^{-1}(3x)\;&=&\;\frac{\pi\;}4\text{විසඳන්න.}\\&&\end{array} \begin{array}{rcl}\mathrm{Tan}\;^{-1}(2\mathrm x)\;&=&\mathrm A\;\mathrm{හා}\;\;\mathrm{Tan}\;^{-1}(3\mathrm x)\;=\mathrm B\;\mathrm{ලෙස}\;\mathrm{ගනිමු}.\\-\frac{\mathrm\pi}2&<&\mathrm A<\frac{\mathrm\pi}2\;\\\mathrm{Tan}\;\mathrm A\;&=&\;2\mathrm x\\-\frac{\mathrm\pi}2&<&\mathrm B<\frac{\mathrm\pi}2\\\mathrm{Tan}\;\mathrm B\;&=&\;3\mathrm x\\\;\;\mathrm A+\mathrm B&=&\;\frac{\mathrm\pi}4\\\mathrm{Tan}(\mathrm A+\mathrm B)&=&\mathrm{Tan}\;\frac{\mathrm\pi}4\;\\\frac{\mathrm{Tan}\;\mathrm A\;+\mathrm{Tan}\;\mathrm B}{1–\;\mathrm{Tan}\;\mathrm A\;\mathrm{Tan}\;\mathrm B}\;&=&\;1\\\frac{\;2\mathrm x+3\mathrm x}{1–2\mathrm x.3\mathrm x}&=&1\\\;5\mathrm x&=&1–6\mathrm x^2\;\\6\mathrm x^2+5\mathrm x–1&=&0\\\;(6\mathrm x–1)(\mathrm x+1)&=&0\\\;\mathrm X&=&\frac16\;\mathrm{හෝ}\;\mathrm x=(–1)\end{array}දැන් මෙම ලැබුණු පිළිතුරු පළමු සමීකරණය තෘප්ත කරන්නේ දැයි පරීක්ෂා කළ යුතුයි.
එසේ පරීක්ෂා කිරීමෙන් පසු x = (–1) පිළිගත නොහැකි බව පෙනේ.ඒනිසා,මෙහි පිළිතුර ,x= 1/6 වේ.
උදාහරණ ගැටලුව 7
Tan\;^{-1}\sqrt{x+2}\;\;+\;Tan\;^{-1}\sqrt{\frac1{2x+1}\;}\;=\;\frac\pi2\;\;\;\;\mathrm{විසඳන්න}.
\begin{array}{rcl}\mathrm{Tan}\;^{-1}\sqrt{\mathrm x+2\;}&=&\;\;\mathrm A\;\;\mathrm ද,\;\\-\frac{\mathrm\pi}2&<&\mathrm A<\frac{\mathrm\pi}2\;\\\;\mathrm{Tan}\;\mathrm A\;&=&\;\sqrt{\mathrm x+2\;}\\\mathrm{Tan}\;^{-1}\sqrt{\frac1{2\mathrm x+1}}\;\;&=&\;\mathrm B\;\;\;\mathrm ද\;\mathrm{ලෙස}\;\mathrm{ගනිමු}.\;\\-\frac{\mathrm\pi}2&<&\mathrm A<\frac{\mathrm\pi}2\;\;\\\mathrm{Tan}\;\mathrm B\;&=&\;\sqrt{\frac1{2\mathrm x+1}}\;\;\;\;\;\;\;⤇\;①\\\;\mathrm A\;+\;\mathrm B\;&=&\;\frac{\mathrm\pi}2\\\mathrm A\;&=&\;\frac{\mathrm\pi}2\;–\;\mathrm B\;\\\mathrm{Tan}\;\mathrm A\;&=&\;\mathrm{Tan}(\frac{\mathrm\pi}2\;–\;\mathrm B)\;\\\;\mathrm{Tan}\;\mathrm A\;&=&\;\mathrm{CotB}\;\\\;\therefore\;\;\mathrm{CotB}\;&=&\;\sqrt{\mathrm x+2}\;\;\;⤇\;②\\&&\;①\;\;\mathrm{හා}\;②\;\mathrm{න්},\\\;\;\frac1{\sqrt{2\mathrm x+1}}&=&\;\frac{\;1}{\mathrm x+2}\;\;\;\;\\2\mathrm x\;+\;1\;\;&=&\;\mathrm x\;+\;2\;\\\mathrm x\;&=&\;1\;\\&&\mathrm{මෙම}\;\mathrm{විසදුම}\;\mathrm{සමීකරණය}\;\mathrm{තෘප්ත}\;\mathrm{කරයි}.\end{array}පසුගිය විභාග ගැටලුවක් සලකා බලමු
2018 AL / I කොටස / (17) C
\begin{array}{rcl}2Tan\;^{-1}\left(\frac13\right)\;+\;Tan\;^{-1}\frac43&=&\;\frac\pi2\;බ\mathrm ව\;\mathrm{පෙන්වන්න}\end{array}
“A mathematical theory is not to be considered complete until you have made it so clear that you can explain it to the first man whom you meet on the street.”
-David Hilbert –
