Question #b545b

1 Answer
Sunayan S
Dec 3, 2017

See the answer below...

Explanation:

#color(red)(sinx+cosx=a# -:Given condition

Now,

#color(red)(sin^6x+cos^6x#
#=(sin^2x)^3+(cos^2x)^3#
#=(sin^2x+cos^2x){(sin^2x)^2-sin^2xcdotcos^2x+(cos^2x)^2}#
#=(sin^2x)^2-sin^2xcdotcos^2x+(cos^2x)^2#[As #color(red)(sin^2x+cos^2x=1#]
#=(sin^2x)^2+(cos^2x)^2-sin^2xcdotcos^2x#
#=(sin^2x+cos^2x)^2-2cdot sin^2xcdotcos^2x-sin^2xcdot cos^2x#
#=1-3cdot sin^2xcdotcos^2x#

Let's come to the given condition:

#color(red)(sinx+cosx=a#
#=>(sinx+cosx)^2=a^2#
#=>sin^2x+cos^2x+2cdot sinxcdotcosx=a^2#
#=>1+2cdot sinxcdotcosx=a^2#
#=sinxcdotcosx=(a^2-1)/2#

Come to the 6 square part:

continued......
#=1-3cdot sin^2xcdotcos^2x#
#=1-3cdot((a^2-1)/2)^2#

Hence,the answer is #{1-3cdot((a^2-1)/2)^2}#

Simplify it if you need...
Hope it helps...
Thank you...

Related questions
Impact of this question
218 views around the world
You can reuse this answer
Creative Commons License