can you solve this for me?

Question

Member since: June 07, 2007
Total points: 83 (Level 1)
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niks
S Pls help me solve this?
given that cosA+cos^2A=1
prove that
sin^12A+3sin^10A+3sin^8A+
sin^6A+2sin^4A+2sin^2A-2=
1

Answer 1

we have cos A = sin ^2 A

sin ^12 A + 3 sin ^10A = sin ^10A(3+ cos A)
add 3 sin ^8A to get 3 sin ^8A + 3 sin ^10 A + cos A sin ^10 A
= sin ^8A(3+ 3 sin ^2A + cos A sin ^2 A)
= sin ^8A( 3 + 3 cos A + cos^2 A)
= sin ^8A( 3 + 3 cos A + 1- sin ^2 A)
= sin ^8A (4 + 2 cos A)
add sin ^6A to get
= sin ^6A((4+ 2 cos A) cos A + 1)
= sin ^6A ( 4 cos A + 2 cos ^2 A + 1)
= sin ^6A( 4 cos A + 2 - cos A + 1)
= 3 sin ^6A( 1+ cos A)
add 2 sin ^4A
to get = sin ^4A(3+ 3 cos A) cos A + 2 sin ^4A
there by reducing you get the result

Answer 2

Remember that
sin^12A = (sin^2A)^6

for sin^2A substitute (1 - cos^A) because sin^2A+cos^2A = 1

then look for cosA+cos^2A and substitute 1 for that.

This should give you the answer
(U may use other trigonometric ids)

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sin^12A+3sin^10A+3sin^8A+sin^6A+2sin^4A+2sin^2A-2
= (1 - con^2A)^6 + 3((1 - con^2A)^5 ..... etc