cos2θ=2cos2θ−1=1−2sin2θ\cos 2\theta = 2\cos^2 \theta - 1 = 1 - 2\sin^2 \thetacos2θ=2cos2θ−1=1−2sin2θ 这个公式表示一个角的余弦值的两倍角可以由该角的余弦值或正弦值的平方来表示。它在三角函数的计算和证明中非常有用,特别是在处理与角度加倍或减半有...
costhetacos2thetacos3theta=1/4 03:55 The general solution of the equation sin^4x + cos^4x= sinx cos x is 02:48 Solve the equation (sinx+cosx)^(1+sin2x)=2, when 0lt=xlt=pi 05:57 Solve the inequation 2^(1/(cos^2 x))sqrt(y^2-y+1/2)<= 1 03:41 The equation 2cos^2x...
This cos 2 theta calculator will teach you all there is about the double angle formula for the cosine function.
解析 $\cos {2\theta }=(cos)^2θ -(sin)^2θ =1-2{sin}^{2}\theta $$=2{cos}^{2}\theta -1$ 综上所述,答案是:${cos}^{2}\theta -{sin}^{2}\theta $,$1-2{sin}^{2}\theta $,$2{cos}^{2}\theta -1$反馈 收藏
(sintheta+sin2theta)/(1+costheta+cos2theta)= View Solution Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Raj...
解:\tan\theta =3,则\cos 2\theta =\frac{\cos^2\theta -\sin^2\theta}{\cos^2\theta +\sin^2\theta }=\frac{1-tan^2\theta}{1+tan^2\theta } =\frac{1-9}{1+9} =-\frac{4}{5},所以C选项是正确的. 由条件利用角三角函数的基本关系,二倍角的余弦公式,求得\cos 2\theta 的值....
∫cos(u)2ducos(u)2 由于1212对于uu是常数,所以将1212移到积分外。 12∫cos(u)du12∫cos(u)du cos(u)对uu的积分为sin(u)sin(u)。 12(sin(u)+C)12(sin(u)+C) 化简。 12sin(u)+C12sin(u)+C 使用2θ2θ替换所有出现的uu。 12sin(2θ)+C12sin(2θ)+C...
ρ^2=cos2θ的图像是双纽线。解：本题利用了卡西尼卵形线和正弦螺线等曲线的特殊性质求解。P^2=cos2θ是极坐标中纽扣曲线的方程 。求解方法为把它化为直角坐标系：r^4=(cos^2θ-sin^2θ)r^2 (x^2+y^2)^2=x^2-y^2 即 x^4+y^4+2(x^2y^2)-x^2+y^2=0。又因为x，y正负皆...
【题目】如果$$ \cos \theta = 1 $$,那么cos2θ为( ) A、2 B、1 C、-1 D、0 相关知识点: 试题来源: 解析 【解析】 由题意可得,$$ \cos 2 \theta = 2 \cos ^ { 2 } \theta - 1 = 2 \times 1 - 1 = 1 $$ 综上所述,答案选择:B 反馈 收藏 ...
NCERT solutions for CBSE and other state boards is a key requirement for students. Doubtnut helps with homework, doubts and solutions to all the questions. It has helped students get under AIR 100 in NEET & IIT JEE. Get PDF and video solutions of IIT-JEE Mains & Advanced previous year pap...