ত্রিকোণমিতিক সূত্রাবলি ও ত্রিভুজের সূত্রাবলী
cos(36∘−θ)cos(36∘+θ)+cos(54∘+θ)cos(54∘−θ)=? \cos \left(36^{\circ}-\theta\right) \cos \left(36^{\circ}+\theta\right)+ \cos \left(54^{\circ}+\theta\right) \cos \left(54^{\circ}-\theta\right)= ? cos(36∘−θ)cos(36∘+θ)+cos(54∘+θ)cos(54∘−θ)=?
cosθ\cos \thetacosθ
cos4θ\cos 4 \thetacos4θ
cos3θ\cos 3 \thetacos3θ
cos2θ\cos 2 \thetacos2θ
Solve:cos(36∘−θ)cos(36∘+θ)+cos(54∘+θ)cos(54∘−θ)=12(cos72∘+cos2θ)+12(cos108∘+cos2θ)=12{cos(90∘−18∘)+cos2θ}+12{cos(90∘+18∘)+cos2θ}=12(cos2θ+cos18∘) \begin{array}{l} \cos \left(36^{\circ}-\theta\right) \cos \left(36^{\circ}+\theta\right)+ \\ \cos \left(54^{\circ}+\theta\right) \cos \left(54^{\circ}-\theta\right) \\ =\frac{1}{2}\left(\cos 72^{\circ}+\cos 2 \theta\right)+\frac{1}{2}\left(\cos 108^{\circ}+\cos 2 \theta\right) \\ =\frac{1}{2}\left\{\cos \left(90^{\circ}-18^{\circ}\right)+\cos 2 \theta\right\}+ \\ \frac{1}{2}\left\{\cos \left(90^{\circ}+18^{\circ}\right)+\cos 2 \theta\right\} \\ =\frac{1}{2}\left(\cos 2 \theta+\cos 18^{\circ}\right) \end{array} cos(36∘−θ)cos(36∘+θ)+cos(54∘+θ)cos(54∘−θ)=21(cos72∘+cos2θ)+21(cos108∘+cos2θ)=21{cos(90∘−18∘)+cos2θ}+21{cos(90∘+18∘)+cos2θ}=21(cos2θ+cos18∘)
+12(cos2θ−cos18∘)=12(cos2θ+cos18∘+cos2θ−cos18∘)=12⋅2cos2θ=cos2θ= R.H.S. (Proved) \begin{array}{r} +\frac{1}{2}\left(\cos 2 \theta-\cos 18^{\circ}\right) \\ =\frac{1}{2}\left(\cos 2 \theta+\cos 18^{\circ}+\cos 2 \theta-\cos 18^{\circ}\right) \\ =\frac{1}{2} \cdot 2 \cos 2 \theta=\cos 2 \theta=\text { R.H.S. (Proved) } \end{array} +21(cos2θ−cos18∘)=21(cos2θ+cos18∘+cos2θ−cos18∘)=21⋅2cos2θ=cos2θ= R.H.S. (Proved)
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∘
