Math, asked by rajaerasezine, 1 year ago

sin^6(3pi/8)+sin^6(pi/8)+3cos^2(7pi/8)sin^2(7pi/8)

Answers

Answered by abhi178

we know,
( a +b)^3 -3ab (a +b)=a^3 + b^3

now ,
{sin^2 (3pi/8)}^3 +{sin^2 (pi/8)}^3 +3ccs^2 (7pi/8).sin^2 (7pi/8)

{sin^2 (3pi/8) + sin^2 (pi/8)}^3 -3sin^2 (3pi/8).sin^2 (pi/8){sin^2 (3pi/8)+sin^2 (pi/8) } + 3cos^2 (7pi/8).sin^2 (7pi/8)

we know,
sin (3pi/8 ) =sin (pi/2 -pi/8)=cos (pi/8)
sin (7pi/8) =sin (pi -pi/8) =sin (pi/8)
cos (7pi/8) =cos (pi - pi/8) = -cos (pi/8)
use this ,
{sin^2 (pi/8) +cos^2 (pi/8)}^3 -3sin^2 (pi/8).cos^2 (pi/8){sin^2 (pi/8)+ cos^2 (pi/8)} -3sin^2 (pi/8) cos^2 (pi/8)

1 - 3sin^2 (pi/8).cos^2 (pi/8) +3sin^1 (pi/8) cos^2 (pi/8)

1. (answer )

abhi178: now problem dear answer is same for both

rajaerasezine: 3sin^2 (pi/8).cos^2 (pi/8) +3sin^1 (pi/8) cos^2 (pi/8)=1

rajaerasezine: ?????

abhi178: no

rajaerasezine: so your answer is false ????

abhi178: how you read answer sincerely

abhi178: answer is correct

abhi178: how you think above =1 and no take I this in answer

rajaerasezine: ok thanks

abhi178: hmmm :-)

Answered by harshitha8

..... plzz mark as brainliest

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