sin2A=(4)/(5)" eon."tan A-sin2 pi*Aot(u sin110<=A<=(pi)/(4))1

If sin2A=(4)/(5) find the value of sin A(0

(ii) Find the value of sin^(2). (pi)/(4) + sin^(2). (3pi)/(4) + sin^(2). (5pi)/(4) + sin^(2). (7pi)/(4) .

sin ((pi) / (5)) sin ((2 pi) / (5)) sin ((3 pi) / (5)) sin ((4 pi) / (5)) =

Prove that : "sin"(pi)/(3) "tan"(pi)/(6) +"sin"(pi)/(2) "cos"(pi)/(3) =2 "sin"^(2)(pi)/(4) .

(1 + sin2A) / (cos2A) = (cos A + sin A) / (cos A-sin A) = tan ((pi) / (4) +1)

Prove that (1+sin2A)/(cos2A)=(cos A+sin A)/(cos A-sin A)=tan((pi)/(4)+A)

Prove that : (1+sin 2A)/(cos 2A) = (cos A + sin A)/(cos A - sin A) = tan (pi/4 + A)

To find the sum sin^(2) ""(2pi)/(7) + sin^(2)""(4pi)/(7) +sin^(2)""(8pi)/(7) , we follow the following method. Put 7theta = 2npi , where n is any integer. Then " " sin 4 theta = sin( 2npi - 3theta) = - sin 3theta This means that sin theta takes the values 0, pm sin (2pi//7), pmsin(2pi//7), pm sin(4pi//7), and pm sin (8pi//7) . From Eq. (i), we now get " " 2 sin 2 theta cos 2theta = 4 sin^(3) theta - 3 sin theta or 4 sin theta cos theta (1-2 sin^(2) theta)= sin theta ( 4sin ^(2) theta -3) Rejecting the value sin theta =0 , we get " " 4 cos theta (1-2 sin^(2) theta ) = 4 sin ^(2) theta - 3 or 16 cos^(2) theta (1-2 sin^(2) theta)^(2) = ( 4sin ^(2) theta -3)^(2) or 16(1-sin^(2) theta) (1-4 sin^(2) theta + 4 sin ^(4) theta) " " = 16 sin ^(4) theta - 24 sin ^(2) theta +9 or " " 64 sin^(6) theta - 112 sin^(4) theta - 56 sin^(2) theta -7 =0 This is cubic in sin^(2) theta with the roots sin^(2)( 2pi//7), sin^(2) (4pi//7), and sin^(2)(8pi//7) . The sum of these roots is " " sin^(2)""(2pi)/(7) + sin^(2)""(4pi)/(7) + sin ^(2)""(8pi)/(7) = (112)/(64) = (7)/(4) . The value of (tan^(2)""(pi)/(7) + tan^(2)""(2pi)/(7) + tan^(2)""(3pi)/(7))/(cot^(2)""(pi)/(7) + cot^(2)""(2pi)/(7) + cot^(2)""(3pi)/(7)) is