Prove that, cos32° sin20° + cos144° cos2° +sin68° cos56°=0
Prove that, cos32° sin20° + cos144° cos2° +sin68° cos56°=0 সমাধানঃ L.H.S= cos32° sin20° + cos144° cos2° +sin68° cos56° = $\frac{1}{2}$ …
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, cos32° sin20° + cos144° cos2° +sin68° cos56°=0 সমাধানঃ L.H.S= cos32° sin20° + cos144° cos2° +sin68° cos56° = $\frac{1}{2}$ …
Prove that,4 sin15° sin75°=$\sqrt{2}$ (cos 105° + sin75°) L.H.S: 4 sin15° sin75° = 2 (2 sin15° sin75°) = 2{cos(15°+75°)+cos(15°-75°)} =2 …
Prove that, cos $\frac{3\pi }{13}$ + cos $\frac{5\pi }{13}$- 2 cos $\frac{9\pi }{13}$ cos $\frac{12\pi }{13}$=0 Solution: L.H.S= cos $\frac{3\pi …
Prove that, sin10°+sin20°+sin40°+sin50°= sin70°+sin80° Solution: L.H.S = sin10°+sin20°+sin40°+sin50° = sin10°+sin50° +sin20°+sin40° = 2 sin$\frac{10° + 50°}{2}$ cos \frac{10° – 50°}{2} …
Prove that, sin19°+sin41°+sin83°=sin23°+sin37°+sin79° Solution: L.H.S: sin19°+sin41°+sin83° = 2 sin$\frac{19° + 41°}{2}$ cos $\frac{19° -41°}{2}$ +sin83° = 2 sin $\left(\frac{60°}{2}\right)$ cos …
Prove that, $\sin \left(\frac{2\pi }{3} + \theta \right) – sin\left(\frac{2\pi }{3} – \theta \right) + sin\theta$ = 0 $L\ldotp H\ldotp …
Prove that, cos (60° +A) + cos(60°-A) – cosA = 0 Solution: cos (60° +A) + cos(60°-A) – cosA = …
Prove that cos 80° – cos40° + $\sqrt{3}$ cos70° = 0 Solution: cos 80° – cos40° + $\sqrt{3}$ cos70° = …
Prove that sin10° +sin50° – sin70°= 0 Solution: sin10° +sin50° – sin70° = sin10° + 2 cos $\frac{50° + 70°}{2}$ …
ProveThat, $1 + \frac{\cos 105 + cos165}{\sin 105 + sin375}$ = 0 Solution: $L\ldotp H\ldotp S: 1 + \frac{\cos 105° …