Question 965164
(a) If cos^2(29∘)−sin^2(29∘)=cosA∘, then 

we have,

cos^2(theta)−sin^2(theta∘)=cos2(theta)∘

therefore,

A=58∘

(b) If cos^2(2x)−sin^2(2x)=cosB,

cos^2(theta)−sin^2(theta∘)=cos2(theta)∘

therefore,cos(4x)=cosB

or 4x=B