解法1:原式= 12 (sin 10°sin 70°)sin 50° = 14 (cos 60°-cos 80°)sin 50° = 14( 12sin50°−cos80°sin50°) = 14[ 12sin50°− 12(sin130°−sin30°)] = 18(sin50°−sin50°+ 12) = 116 . 解法2:原式= 12 · 2cos10°sin10°sin50°sin70°2cos10° = 12 ·...
sin(-10°)<0,sin50°>0,故A错误;tan70°>1,sin70°<1,所以tan70°>sin70°故B错误;cos(-40°)=cos40°,cos310°=cos 50°,cos 40°>cos50°,故C错误;cos130°=cos50°,cos200°=cos 20°cos50°>cos 20°,故D正确。故答案选:D。反馈 收藏 ...
首先，利用三角函数的周期性和对称性，我们知道 sinθ = sin(180° - θ)。因此，我们可以将 sin50° 和 sin70° 分别转换为它们的补角形式，即 sin50° = sin(180° - 50°) = sin130°，和 sin70° = sin(180° - 70°) = sin110°。接下来，考虑到 sinθ * sin(180° - θ)...
The value of sin50∘sin130∘ is ……… View Solution Convert 25∘ in radian measure. View Solution If sin(θ+α)=aandsin(θ+β)=b, then prove that cos2(α−β)−4abcos(α−β)=1−2a2−2b2. View Solution Find distance between the points A(3,4)andB(−4,3)....
cos(10-70)-cos(10+70)]=2sin50[cos60-cos80]=sin50-2sin50cos80=sin50-[sin(50+80)+sin(50-80)]=sin50-sin130+sin30=sin50-sin50+sin30=1/24sin10sin50sin70=4cos20cos40cos80=4cos80(cos(20+40)+cos(40-20))/2=2cos80(cos60+cos20)=2cos80cos60+2cos80cos20=cos...
[解法1]直接用向量方法求解.由已知可得 (AB)=(OB)-(OA)=(cos70°-cos10° , sin70°≈sin10°) =(-2sin 40°sin 30°,2cos 40°sin 30°) =(-sin40°,cos40° ) =(-cos50°,sin50°) =(cos130°,sin130°) [解法2] 利用直线的斜率求解.由已知可得 A(cos10°,sin10°) B(...
sin 2 10°+sin 2 70°+sin 2 130°= 3 2 3 2 sin 2 20°+sin 2 80°+sin 2 140°= 3 2 3 2 sin 2 30°+sin 2 90°+sin 2 150°= 3 2 3 2 分析上述各式的共同特点,猜想出反映一般规律的等式,并对等式的正确性作出证明. 相关知识点: ...
=(2cos10-2sin10cos10)/cos20=2cos10(1-sin10)/cos20=2sin80(sin90-sin10)/cos20=2sin80(2cos50sin40)/cos20=8sin20cos50sin80=4(2sin80cos50)sin20=4(sin130-sin30)sin20=4(cos40-sin30)sin20=4cos40sin20-2sin20=2(sin60-sin(-20))-2sin20=2sin60+2sin20-2sin20=...
(v)cos130∘cos40∘l+sin130∘sin40∘=0 View Solution Prove that (i)sin(50∘+θ)cos(20∘+θ)−cos(50∘+θ)sin(20∘+θ)=12 (ii)cos(70∘+θ)cos(10∘+θ)+sin(70∘+θ)sin(10∘+θ)=12 View Solution
解:S=absinC/2=4sin10sin50sin70=2sin50[cos(10-70)-cos(10+70)]=2sin50[cos60-cos80]=sin50-2sin50cos80=sin50-[sin(50+80)+sin(50-80)]=sin50-sin130+sin30=sin50-sin50+sin30=1/2 解析看不懂?免费查看同类题视频解析查看解答 更多答案(1)...