Question

Help with trigonometric identities....

Is anyone able to show that

(cost)^2(sint)^2+ (sint)^4+(cost)^2=1

Help would be greatly appreciated, thanks!

Answer 1

cos^2x (1-cos^2x) + (sin^2x)^2 + cos^2x

Answer 2

cos^2x - cos^4x + (1-cos^2x)^2 + cos^2x

cos^2x - cos^4x + cos^4x -2cos^2x+1

Answer 3

And so, you are left with 1.

Answer 4

cos^2*sin^2 + sin^4 + cos^2 = 1


(1-sin^2)*sin^2 + sin^4 + cos^2 = 1


sin^2 - sin^4 + sin^4 + cos^2 = 1


sin^2 + cos^2 = 1


1 = 1

Answer 5

Thank you saifoo.khan, but I am confused as to how cos^4x entered the equation...

Answer 6

Thank you so much jim! very helpful!

Answer 7

lol sef :)) missing some values in equations :p

Answer 8

you're welcome

Answer 9

(1-cos^2x)^2
FOIL here,

(1-cos^2 x)(1-cos^2 x)
1 - cos^2x - cos^2 x + cos^4x

Answer 10

@lgbasallote where?

Answer 11

cos^2x - cos^4x + (1-cos^2x)^2 + cos^2x

cos^2x - cos^4x + cos^4x -2cos^2x+1 <---missed the cos^2 x

Answer 12

Oh right. the last one.

Answer 13

Thanks everyone!