Prove that, cosα cos(120+α) cos(240+α)= $\frac{1}{4}$ cos3α
Prove that, cosα cos(120+α) cos(240+α)= $\frac{1}{4}$ cos3α $cos\alpha cos(120 + \alpha )cos(240 + \alpha )$ = $\frac{1}{4}cos3\alpha$ Solution: cosα cos(120°+α) …
SN Dey Class 11 Transformations of Sums and Product (যোগফল ও গুণফলের রূপান্তর)
All Math Problems of SN Dey Class 11 Transformations of Sums and Product are Available Here. সৌরেন্দ্রনাথ দে ক্লাস 11 ছায়া প্রকাশনী বইয়ের যোগফল ও গুণফলের রূপান্তর অধ্যায়ের সমাধান
Prove that, 4 sinθ sin $\left(\frac{\pi }{3} + \theta \right)$ sin $\left(\frac{\pi }{3} – \theta \right)$ =sin3θ
4 sinθ sin $\left(\frac{\pi }{3} + \theta \right)$ sin $\left(\frac{\pi }{3} – \theta \right)$ =sin3θ Prove that, 4 sinθ sin(π/3+θ ) …
Prove that,cosθ cos(60°-θ) cos(60°+θ)= $\frac{1}{4}$ cos 3θ
Prove that,cosθ cos(60°-θ) cos(60°+θ)= $\frac{1}{4}$ cos 3θ Solution: cos θ cos(60°– θ) cos(60°+ θ) = $\frac{1}{2}$ cos θ {2 cos(60°- θ) …
Prove that, cos40° cos100° cos160°= $\frac{1}{8}$
Prove that, cos40° cos100° cos160°= $\frac{1}{8}$ Solution: cos40° cos100° cos160° = $\frac{1}{2}$ (2 cos100° cos40°) cos160° = $\frac{1}{2}$ (cos 140° …
Prove that, tan20° tan40° tan80° = $\sqrt{3}$
সমাধানঃ Prove that, tan20° tan40° tan80° = $\sqrt{3}$ L.H.S: $\tan 20°tan40°tan80°$ = $\frac{\sin 20°sin40°sin80°}{\cos 20°cos40°cos80°}$ = $\frac{2\sin 20°sin40°sin80°}{2\cos 20°cos40°cos80°}$ = …
Prove that, 8 sin20° sin40° sin80° = $\sqrt{3}$
Prove that, 8 sin20° sin40° sin80° = $\sqrt{3}$ Solution: 8 sin20° sin40° sin80° = 4(2 sin20° sin40°) sin80° =4 {cos(20°+40°)-cos(20°-40°)} …
Prove that, cos20° cos40° cos80°=$\frac{1}{8}$
Prove that, cos20° cos40° cos80°=$\frac{1}{8}$ Solution: cos20° cos40° cos80° = $\frac{1}{2}$ (2 cos20° cos40°) cos80° = $\frac{1}{2}$ {cos (20°+40°) + …
Prove that, cos24° + cos55° +cos125° + cos204°+ cos300°=$\frac{1}{2}$
Prove that, cos24° + cos55° +cos125° + cos204°+ cos300°=$\frac{1}{2}$ Solution: L.H.S: cos24° + cos55° +cos125° + cos204°+ cos300° = cos24° …
Prove that, cos10° cos20° +sin45° cos145° +sin55° cos245°=0
Prove that, cos10° cos20° +sin45° cos145° +sin55° cos245°=0 Solution: L.H.S =cos10° cos20° +sin45° cos145° +sin55° cos245° = $\frac{1}{2}$ ( 2cos10° …
Prove that, cos306° +cos234° +cos162° +cos18°=0
Prove that, cos306° +cos234° +cos162° +cos18°=0 Solution: L.H.S = (cos306° +cos234°)+(cos162° +cos18°) = 2 cos $\frac{306° + 234°}{2}$ cos $\frac{306° …