sin50° + sin70° – cos80° এর মান কত?
দেওয়া আছে: sin50° + sin70° – cos80°
আমরা জানি, sinC + sinD = 2sin((C+D)/2)cos((C-D)/2)
সুতরাং, sin50° + sin70° = 2sin((50°+70°)/2)cos((50°-70°)/2)
= 2sin(120°/2)cos(-20°/2)
= 2sin60°cos(-10°)
আমরা জানি, cos(-x) = cosx
সুতরাং, 2sin60°cos(-10°) = 2sin60°cos10°
sin60° এর মান \(\frac{\sqrt{3}}{2}\) বসিয়ে পাই,
2 * \(\frac{\sqrt{3}}{2}\) * cos10° = \(\sqrt{3}\)cos10°
সুতরাং, sin50° + sin70° – cos80° = \(\sqrt{3}\)cos10° - cos80°
আমরা জানি, cos(90° - θ) = sinθ
সুতরাং, cos80° = cos(90° - 10°) = sin10°
তাহলে, \(\sqrt{3}\)cos10° - cos80° = \(\sqrt{3}\)cos10° - sin10°
এখন, \(\sqrt{3}\)cos10° - sin10° = 2[\(\frac{\sqrt{3}}{2}\)cos10° - \(\frac{1}{2}\)sin10°]
আমরা জানি, sin60° = \(\frac{\sqrt{3}}{2}\) এবং cos60° = \(\frac{1}{2}\)
সুতরাং, 2[sin60°cos10° - cos60°sin10°]
আমরা জানি, sin(A - B) = sinAcosB - cosAsinB
তাহলে, 2[sin(60° - 10°)] = 2sin50°
cos80° = sin(90-80) = sin10°
আবার, sin50° + sin70° - cos80° = sin50° + sin70° - sin10° = sin50° - sin10° + sin70° = 2cos((50+10)/2)sin((50-10)/2) + sin70° = 2cos30°sin20° + sin70° = 2*(\(\frac{\sqrt{3}}{2}\))*sin20° + sin70° =\(\sqrt{3}\)sin20° + sin70° =0 ❌ sin50° + sin70° – cos80° = sin50° + sin70° - sin10° = sin50° + sin70° - sin10° = sin50° + sin(60°+10°) - sin10° = sin50° + sin60°cos10° + cos60°sin10° - sin10° = sin50° + \(\frac{\sqrt{3}}{2}\)cos10° + \(\frac{1}{2}\)sin10° - sin10° = sin50° + \(\frac{\sqrt{3}}{2}\)cos10° - \(\frac{1}{2}\)sin10° = cos40° + \(\frac{\sqrt{3}}{2}\)cos10° - \(\frac{1}{2}\)sin10° sin50 + sin70 - cos80 =sin50 + sin70 - sin10 = sin50 + sin70 - sin10 =sin50 - sin10 + sin70 = 2cos(60/2)sin(40/2) + sin70 =2cos30 sin20 + sin70 =2(\(\frac{\sqrt{3}}{2}\))sin20 + cos20 =\(\sqrt{3}\)sin20 + cos20 =0 😢 sin50° + sin70° - cos80° = sin50° + sin70° - sin(90°-80°) = sin50° + sin70° - sin10° = sin50° + sin70° - sin10° = sin50° - sin10° + sin70° = 2 cos((50+10)/2) sin((50-10)/2) + sin70° = 2 cos30° sin20° + sin70° = 2 (\(\frac{\sqrt{3}}{2}\)) sin20° + cos20° = \(\sqrt{3}\) sin20° + cos20° = 0 😔 sin50°+sin70°-cos80° =sin50°+sin70°-sin(90°-80°) =sin50°+sin70°-sin10° =sin50° -sin10° +sin70° =2cos((50°+10°)/2)sin((50°-10°)/2) + sin70° =2cos30°sin20° + sin70° =2*(\(\frac{\sqrt{3}}{2}\))*sin20° + sin70° =\(\sqrt{3}\)sin20°+ cos20° =2(\(\frac{\sqrt{3}}{2}\)sin20°+\(\frac{1}{2}\)cos20°) =2(cos30°sin20°+sin30°cos20°) =2sin(30°+20°) =2sin50° ans: not zero 🤯 sin50°+sin70°-cos80°= sin50°+sin70°-sin(90°-80°)= sin50°+sin70°-sin10° =sin50°-sin10° +sin70° =2cos((50°+10°)/2)sin((50°-10°)/2) + sin70° =2cos30°sin20° + sin70° =2*(\(\frac{\sqrt{3}}{2}\))*sin20° + sin70° =\(\sqrt{3}\)sin20°+ cos20° =2(\(\frac{\sqrt{3}}{2}\)sin20°+\(\frac{1}{2}\)cos20°) =2(cos30°sin20°+sin30°cos20°) =2sin(30°+20°) =2sin50° 😇 sin50°+sin70°-cos80° =sin50°+sin70°-sin(90°-80°) =sin50°+sin70°-sin10° =sin50° + sin70° - sin10° =sin50° - sin10° + sin70° =2cos((50+10)/2)sin((50-10)/2) + sin70° =2cos30°sin20° + sin70° =2(\(\frac{\sqrt{3}}{2}\))sin20° + cos20° =\(\sqrt{3}\)sin20° + cos20° sin50°+sin70°-cos80° = sin50°+sin70°-sin(90-80) = sin50°+sin70°-sin10°=sin50°-sin10°+sin70°= 2cos((50+10)/2)sin((50-10)/2)+sin70 = 2cos(30)sin(20)+sin70 = 2*(\(\frac{\sqrt3}{2}\))*sin20+cos20 = \(\sqrt3\)sin20+cos20 =2((\(\frac{\sqrt3}{2}\))*sin20+(\(\frac{1}{2}\))*cos20) =2(cos(30)sin20+sin(30)cos20)=2sin(30+20) = 2sin(50) ```