f(theta)=[[cos theta,sin theta],[-sin theta,cos theta]]_(" then ")

Simplify, costheta[[cos theta, sin theta],[-sin theta, cos theta]] + sin theta [[sin theta, -cos theta],[cos theta, sin theta]]

Simplify: cos theta[[cos theta , sin theta],[-sin theta , cos theta]]+sin theta[[sin theta, -cos theta],[cos theta ,sin theta]] .

If A = [[cos theta, sin theta],[-sin theta, cos theta]] , then what is A^(-1) ?

If F(theta)=[[cos theta,-sin theta, 0],[sin theta, cos theta, 0],[0,0,1]] , then show that F(theta) F (phi)= F(theta + phi) .

If A= [[cos theta,-sin theta],[sin theta,cos theta]] ,then Adj A is (a) [[cos theta,-sin theta],[cos theta,sin theta]] (b) [[1,0],[0,1]] (c) [[cos theta,sin theta],[-sin theta,cos theta]] (d) [[-1,0],[0,-1]]

[[ Simplify :cos thetacos theta,sin theta-sin theta,cos theta]]+sin theta[[sin theta,-cos thetacos theta,sin theta]]

Simplify cos theta[(cos theta,sin theta),(-sin theta,cos theta)]+sin theta[(sin theta,-cos theta),(cos theta,sin theta)]

cos theta/(1+sin theta)=(1-sin theta)/cos theta

The inverse of A=[[cos theta, sin theta],[-sin theta,cos theta]] is :

Evaluate |[cos theta, -sin theta], [sin theta, cos theta]|