সংযুক্ত ও যৌগিক কোণের ত্রিকোণমিতিক অনুপাত
4(sin310∘+cos320∘)=?4\left(\sin ^{3} 10^{\circ}+\cos ^{3} 20^{\circ}\right) = ?4(sin310∘+cos320∘)=?
3(sin20∘+cos20∘) 3\left(\sin 20^{\circ}+\cos 20^{\circ}\right) 3(sin20∘+cos20∘)
3(sin10∘−cos20∘) 3\left(\sin 10^{\circ}-\cos 20^{\circ}\right) 3(sin10∘−cos20∘)
3(sin10∘+cos30∘) 3\left(\sin 10^{\circ}+\cos 30^{\circ}\right) 3(sin10∘+cos30∘)
3(sin10∘+cos20∘) 3\left(\sin 10^{\circ}+\cos 20^{\circ}\right) 3(sin10∘+cos20∘)
Solve:4(sin310∘+cos320∘)=3(sin10∘+cos20∘)=4sinsin310∘+0∘+4cos320∘)=3sin10∘−sin(3.10∘)+cos(3.20∘)+3cos20∘=3(sin10∘+cos20∘)−sin30∘+cos60∘=3(sin10∘+sin20∘)−12+12=3(sin10∘+cos20∘) \begin{array}{l} 4\left(\sin ^{3} 10^{\circ}+\cos ^{3} 20^{\circ}\right) \\ =3\left(\sin 10^{\circ}+\cos 20^{\circ}\right) \\ =\left.4 \sin \sin ^{3} 10^{\circ}+0^{\circ}+4 \cos ^{3} 20^{\circ}\right) \\ =3 \sin 10^{\circ}-\sin \left(3.10^{\circ}\right)+\cos \left(3.20^{\circ}\right) \\ +3 \cos 20^{\circ} \\ = 3\left(\sin 10^{\circ}+\cos 20^{\circ}\right)-\sin 30^{\circ}+\cos 60^{\circ} \\ = 3\left(\sin 10^{\circ}+\sin 20^{\circ}\right)-\frac{1}{2}+\frac{1}{2} \\ = 3\left(\sin 10^{\circ}+\cos 20^{\circ}\right) \end{array} 4(sin310∘+cos320∘)=3(sin10∘+cos20∘)=4sinsin310∘+0∘+4cos320∘)=3sin10∘−sin(3.10∘)+cos(3.20∘)+3cos20∘=3(sin10∘+cos20∘)−sin30∘+cos60∘=3(sin10∘+sin20∘)−21+21=3(sin10∘+cos20∘)
যদি tanθ=x \tan \theta=x tanθ=x হয়, তবে sin2θ \sin 2 \theta sin2θ এর মান কত?
tanθ tan3θ tan5θ tan7θ tan9θ tan11θ tan13θ=? θ=π/28
The number of real solutions of the equation sin(ex)=5x+5−x\sin \left( e ^ { x } \right) = 5 ^ { x } + 5 ^ { - x }sin(ex)=5x+5−x is ____________________.
If sinθ+cosθ=1\sin \theta +\cos \theta =1sinθ+cosθ=1, then the value of sin2θ\sin 2\thetasin2θ is equal to:
