Let E (θ) = [cos θ&sin θ -sin θ &cos θ ] then E(α) E(β) is equal to

Question

Let E (θ) = [cos θ&sin θ -sin θ &cos θ ] then E(α) E(β) is equal to (A)  E((α+β/2))  (B)  E(α β)  (C)  E(α+β)  (D)  E(α-β)

Tardigrade Admin · Accepted Answer

Given, E (θ)=[ beginmatrixcos θ&sin θ -sin θ&cos θ endmatrix] Now, E(α)E(β)=[ beginmatrixcos α&sin α -sin α&cos α endmatrix] [ beginmatrixcos β&sin β -sin β&cos β endmatrix] =[ beginmatrixcos α cos β-sin α sin β&cos α sin β+sin α cos β -(sin α cos β+cos α sin β)&-sin α sin β+cos α cos β endmatrix] =[ beginmatrixcos(α+β)&sin(α+β) -sin(α+β)&cos(α+β) endmatrix]=E(α+β)

Q. Let $E (θ) = [cos θ - s in θ s in θ cos θ]$ then $E (α) E (β)$ is equal to

$E (\frac{α + β}{2})$

$E (α β)$

$E (α + β)$

$E (α - β)$

Q. Let E(θ)=[cosθ−sinθ​sinθcosθ​] then E(α)E(β) is equal to

E(2α+β​)

E(αβ)

E(α+β)

E(α−β)

Given, E(θ)=[cosθ−sinθ​sinθcosθ​] Now, E(α)E(β)=[cosα−sinα​sinαcosα​][cosβ−sinβ​sinβcosβ​] =[cosαcosβ−sinαsinβ−(sinαcosβ+cosαsinβ)​cosαsinβ+sinαcosβ−sinαsinβ+cosαcosβ​] =[cos(α+β)−sin(α+β)​sin(α+β)cos(α+β)​]=E(α+β)

Q. Let $E (θ) = [cos θ - s in θ s in θ cos θ]$ then $E (α) E (β)$ is equal to

$E (\frac{α + β}{2})$

$E (α β)$

$E (α + β)$

$E (α - β)$