【解析】【答案】 D 【解析】 ∵$$ \because \sin \alpha - \cos \beta = \frac { 1 } { 2 } , \cos \alpha - \sin \beta = \frac { 1 } { 3 } $$ ∴$$ ( \sin \alpha - \cos \beta ) ^ { 2 } = \sin ^ { 2 } \alpha + \cos ^ { 2 } \beta - 2 ...
{1 - \tan \alpha \ast \tan \beta }③利用这些公式可将某些不是特殊角的三角函数转化为特殊角的三角函数来求值,如:\tan 105^{{\circ} }=\tan (45^{{\circ} }+ 60^{{\circ} }) = \dfrac{\tan 45^{{\circ} } + \tan 60^{{\circ} }}{1 - \tan 45^{{\circ} }\ast \tan 60...
Step by step video & image solution for If x=r cos alpha cos beta,y=r cos alpha sin beta,z=r sin alpha then x^(2)+y^(2)+z^(2)= by Maths experts to help you in doubts & scoring excellent marks in Class 11 exams.Updated on:21/07/2023Class...
Step by step video, text & image solution for यदि cos(alpha + beta) = 3/5, sin(alpha - beta) = 5/13 तथा 0 lt alpha, beta lt pi/4, तब tan(2alpha) बराबर है by Maths experts to help you in doubts & scoring excellent marks in Class 12...
Answer to: Determine whether the below statement is true or false. cos(alpha - beta) = cos alpha cos beta + sin alpha sin beta a. True b. False By...
{1 + \tan \alpha \tan \beta }(1+ \tan \alpha \tan \beta \neq 0)利用这些公式可以将一些不是特殊角的三角函数转化为特殊角的三角函数来求值.如:\tan 105^{{\circ} }= \tan (45^{{\circ} }+ 60^{{\circ} })= \dfrac{\tan 45^{{\circ} } + \tan 60^{{\circ} }}{1 - \...
百度试题 结果1 题目已知sin\alpha -cos\beta =-\frac{2}{3}, cos\alpha +sin\beta =\frac{1}{3},求 sin(\alpha -\beta )的值。 相关知识点: 试题来源: 解析 本题考查了三角函数的计算。反馈 收藏
माना cos(alpha+beta)=(4)/(5),sin(alpha-beta)=(5)/(13) जहाँ 0lealpha,betale(pi)/(4) तो tan2alpha=
Step by step video & image solution for Let F(alpha)=[cosalpha-sinalpha0sinalphacosalpha0 0 0 1] and G(beta)=[cosbeta0sinbeta0 1 0-sinbeta0cosbeta] . Show that [F(alpha)]^(-1)=F(-alpha) (ii) [G(beta)]^(-1)=G(-beta) (iii) [F(alpha)G(beta)]^(-1)=G(-beta)F...
sin(α)+sin(β)=2sin(2α+β)cos(2α−β) Solve for α α∈C Solve for β β∈C