If #cos(x) + cos^(2)x = 1#, then which of the following option is correct for, #sin^(12)x + 3sin^(10)x + 3sin^(8)x + sin^(6)x + 1#?

#A)0#
#B)1#
#C)2#
#D)3#

2 Answers
Mohan V.
Jan 11, 2018

The answer is 2

Explanation:

Given that,
# cos x + cos^2x = 1 ... ... . ( 1 ) #
#⇒ cos x = 1 − cos^2x #
#⇒ cos x = sin^2x#

When the given expression #sin^12x +3sin^10x + 3sin^8x +sin^6x+1#
#=(sin^2x)^6 +3sin^10x + 3(sin^2x)^4 +sin^6x+1 #....from (1)

#= (cosx)^6 +3sin^10x + 3(cosx)^4+sin^6x+1 #
#= cos^6x +sin^6x + 3cos^4x +3sin^10x +1#
#= (cos^2x +sin^2x)^3-3cos^2xsin^2x(cos^2x+sin^2x) + 3cos^4x +3(sin^2x)^5 +1#
#=1^3-3cos^2xcosx + 3cos^4x +3cos^5x+1 #
#=1-3cos^3x(1- cosx) +3cos^5x +1#
#=1-3cos^3x*cos^2x +3cos^5x +1#
#=1-3cos^5x +3cos^5x +1#
#=1+1#
#=2#

Hope this helps.

Ratnaker Mehta
Jan 11, 2018

# C)2#.

Explanation:

#cosx+cos^2x=1 rArr cosx=1-cos^2x=sin^2x......(star)#.

Now, #ul(sin^12x+3sin^10x+3sin^8x+sin^6x)+1#,

#=sin^6x{sin^6x+3sin^4x+3sin^2x+1}+1#,

#=(sin^2x)^3{(sin^2x+1)}^3+1#,

#={sin^2x(sin^2x+1)}^3+1#,

#={sin^4x+sin^2x}^3+1#,

#={(sin^2x)^2+sin^2x}^3+1#,

#={cos^2x+cosx}^3+1...........[because, (star)]#,

#=(1)^3+1................................................[because," Given]"#,

#=2,# as Respected Mohan V. has already derived!

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