সৃজনশীলঃ ‘f(x)=sin⁡x.…’ চট্টগ্রাম কলেজ ২০২৬ – বিপরীত ত্রিকোনমিতিক ফাংশন ও ত্রিকোণমিতিক সমীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

দেওয়া আছে, $f\left(x\right)=\sin x$ প্রদত্ত সমীকরণ,

$\left\{f\right(2\theta ){\}}^2-\{f\left(\theta \right){\}}^2=\frac{1}{3}\left\{f\right(3\theta \left)\right\}$
বা, ${\sin }^{2}2\theta -{\sin }^2\theta =\frac{1}{3}\sin 3\theta$
বা, $(2\sin \theta \cdot \cos \theta {)}^2-{\sin }^2\theta =\frac{1}{3}(3\sin \theta -4{\sin }^3\theta )$
বা, $4{\sin }^2\theta \cdot {\cos }^2\theta -{\sin }^2\theta =\frac{1}{3}\sin \theta (3-4{\sin }^2\theta )$

বা, ${\sin }^2\theta (4{\cos }^2\theta -1)=\frac{1}{3}\sin \theta (3-4{\sin }^2\theta )$
বা, $3{\sin }^2\theta \{4(1-{\sin }^2\theta )-1)\}=\sin \theta (3-4{\sin }^2\theta )$
বা, $3{\sin }^2\theta (3-4{\sin }^2\theta )-\sin \theta (3-4{\sin }^2\theta )=0$
বা, $\sin \theta (3-4{\sin }^2\theta )(3\sin \theta -1)=0$

হয়, $\sin \theta =0$
$\therefore \theta =2n\pi$

অথবা, $3-4{\sin }^2\theta =0$
বা, ${\sin }^2\theta =\frac{3}{4}$
বা, $\sin \theta =\pm \frac{\sqrt{3}}{2}=\pm \sin \frac{\pi }{3}$
$\therefore \theta =n\pi \pm (-1{)}^n\frac{\pi }{3}$

অথবা, $3\sin \theta -1=0$
বা, $\sin \theta =\frac{1}{3}$
$\therefore \theta =n\pi +(-1{)}^n{\sin }^{-1}\left(\frac{1}{3}\right)$

নির্ণেয় সমাধান : $\theta =2n\pi ,n\pi \pm (-1{)}^n\frac{\pi }{3},n\pi +(-1{)}^n{\sin }^{-1}\left(\frac{1}{3}\right)$.

দেওয়া আছে, $f\left(x\right)=\sin x$

$\therefore f^{-1}\left(x\right)={\sin }^{-1}x$
এবং $f^{-1}\left(x\right)+f^{-1}\left(y\right)+f^{-1}\left(z\right)=\pi$
বা, ${\sin }^{-1}x+{\sin }^{-1}y+{\sin }^{-1}z=\pi$
বা, ${\sin }^{-1}x+{\sin }^{-1}y=\pi -{\sin }^{-1}z$
বা, ${\sin }^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2})=\pi -{\sin }^{-1}z$
বা, $\sin {\sin }^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2})=\sin (\pi -{\sin }^{-1}z)$
বা, $x\sqrt{1-y^2}+y\sqrt{1-x^2}=\sin ({\sin }^{-1}z)$
বা, $x\sqrt{1-y^2}+y\sqrt{1-x^2}=z$
বা, ${\left(x\sqrt{1-y^2}\right)}^2={(z-y\sqrt{1-x^2})}^2$
বা, $x^2(1-y^2)=z^2-2yz\sqrt{1-x^2}+{\left(y\sqrt{1-x^2}\right)}^2$
বা, $x^2-x^2{y}^2=z^2+y^2-x^2{y}^2-2yz\sqrt{1-x^2}$.
বা, ${(x^2-y^2-z^2)}^2={(-2yz\sqrt{1-x^2})}^2$
বা, $x^4+y^4+z^4-2x^2{y}^2-2x^2{z}^2+2y^2{z}^2=4y^2{z}^2(1-x^2)$
বা, $x^4+y^4+z^4=4y^2{z}^2-4x^2{y}^2{z}^2+2x^2{y}^2+2x^2{z}^2-2y^2{z}^2$
$\therefore x^4+y^4+z^4+4x^2{y}^2{z}^2=2(x^2{y}^2+y^2{z}^2+z^2{x}^2)$.
(দেখানো হলো)