ত্রিকোণমিতিক সূত্রাবলি ও ত্রিভুজের সূত্রাবলী
sin(−1230∘)−cos{(2n+1)π+π3}=?\sin \left(-1230^{\circ}\right)-\cos \left\{(2 n+1) \pi+\frac{\pi}{3}\right\} =?sin(−1230∘)−cos{(2n+1)π+3π}=?
0.50.50.5
111
000
1.51.51.5
Solve:sin(−1230∘)−cos{(2n+1)π+π3}=−sin1230∘−cos{2nπ+(π+π3)}=−sin(3360∘+150∘)−cos(π+π3)=−sin150∘−(−cosπ3)=−sin(180∘−30∘)+cosπ3=−sin30∘+cosπ3=−12+12=0 (Ans.) \begin{aligned} & \sin \left(-1230^{\circ}\right)-\cos \left\{(2 n+1) \pi+\frac{\pi}{3}\right\} \\ = & -\sin 1230^{\circ}-\cos \left\{2 n \pi+\left(\pi+\frac{\pi}{3}\right)\right\} \\ = & -\sin \left(3360^{\circ}+150^{\circ}\right)-\cos \left(\pi+\frac{\pi}{3}\right) \\ = & -\sin 150^{\circ}-\left(-\cos \frac{\pi}{3}\right) \\ = & -\sin \left(180^{\circ}-30^{\circ}\right)+\cos \frac{\pi}{3} \\ = & -\sin 30^{\circ}+\cos \frac{\pi}{3}=-\frac{1}{2}+\frac{1}{2}=0 \text { (Ans.) } \end{aligned} =====sin(−1230∘)−cos{(2n+1)π+3π}−sin1230∘−cos{2nπ+(π+3π)}−sin(3360∘+150∘)−cos(π+3π)−sin150∘−(−cos3π)−sin(180∘−30∘)+cos3π−sin30∘+cos3π=−21+21=0 (Ans.)
কোন ত্রিভুজের তিন বাহুর দৈর্ঘ্য যথাক্রমে 3, 5, 7 cm হলে, বৃহত্তম কোণ কোনটি?
If cos3π9+sin3π18=m4(cosπ9+sinπ18) \cos^3\frac{\pi}{9}+ \sin^3\frac{\pi}{18} = \dfrac{m}{4} \left( \cos\frac{\pi}{9}+ \sin\frac{\pi}{18}\right)cos39π+sin318π=4m(cos9π+sin18π).Find mmm
দৃশ্যকল্প-১: △ABC \triangle \mathrm{ABC} △ABC এর A=75∘,B−C=15∘ \mathrm{A}=75^{\circ}, \mathrm{B}-\mathrm{C}=15^{\circ} A=75∘,B−C=15∘
