`cosA.cos2A.cos4A.cos8A.cos16A.cos32A`=`sin(64A)/(64sinA) `

Prove that 4sinA.cosA.cos2A=sin4A

Prove that 4sinA.cosA.cos2A=sin4A

Prove that cos A cos 2A cos 4A cos 8A= (sin 16A)/(16 sin A) .

If A is not an integral multiple of (pi) , prove that cos A cos 2A cos 4A cos 8A =(sin 16A)/(16 sin A) Hence deduce that cos. (2pi)/(15). Cos. (4pi)/(15) .cos. (8pi)/(18). Cos. (16pi)/(15)=(1)/(16)

(cos3A+sin3A)/(cosA-sinA)=

Solve sinA.cos(A-B)-cosA.sin(A-B)

(cos5A+cos3A+cosA)/(sin5A-sin3A+sinA)=

(sinA + sin3A) / (cosA + cos3A) = tanalpha, (cosA-cos2A) / (sinA + sin2A) = tanbeta, (sinA + sin3A + sin5A + sin7A) / (cosA + cos3A + cos5A + cos7A) = tan gamma , (cosA-cos5A-cos9A + cos13A) / (sinA-sin5A + sin9A-sin13A) = tandelta, then (a) alpha = 4beta (b) gamma = 2alpha (c) gamma = delta (d) none

Prove that (cosA+cos3A+cos5A+cos7A)/(sinA+sin3A+sin5A+sin7A)=cot4A

Prove that (1+sin2A)/(cos2A)=(cosA+sinA)/(cosA-sinA)=tan((pi)/(4)+A)