Trig identity

Question

Trig identity

anonymous · Answer

$$\large 1-(\sin^6x+cox^6x)=3\sin^2xcos^2x$$

anonymous · Answer

$$\large 1-(x^6+6x^5y+10x^4y^2+20x^3y^3+10x^3y^4+6xy^5+y^6)$$

hartnn · Answer

$\large a^3+b^3=(a+b)(a^2+ab+b^2)$
write 
$\sin^6x=(\sin^2x)^3$

hartnn · Answer

same for cos

anonymous · Answer

like this ? $$1-(\sin^3x)^3(\cos^3x)^3$$

hartnn · Answer

$\large 1-(\sin^6x+cox^6x)=1-((\sin^2x)^3+(\cos^2x)^3)$

anonymous · Answer

doesnt that equal 5

hartnn · Answer

no, how ?

hartnn · Answer

2*3=6

anonymous · Answer

OH yeah sorry i was adding

hartnn · Answer

now apply that formula

anonymous · Answer

i don't know what to do

hartnn · Answer

$\large (\sin^2x)^3+(\cos^2x)^3=(..+..)(..+..+..)$
$\large a^3+b^3=(a+b)(a^2-ab+b^2)$
sorry, i made a typo before.

hartnn · Answer

$\large (\sin^2x)^3+(\cos^2x)^3=(..+..)(..-..+..)$

anonymous · Answer

$$\large (\sin^2x+\cos^2x) $$

anonymous · Answer

do i multiply that with Pascal's triangle

hartnn · Answer

were u not able to fill this ? $\large (\sin^2x)^3+(\cos^2x)^3=(..+..)(..-..+..)$ 
also , standard identity $sin^2x+cos^2x= ??$

hartnn · Answer

$\large a^3+b^3=(a+b)(a^2-ab+b^2) \ a=\sin^2x,b=\cos^2x$

anonymous · Answer

ohh so $$(1)(\sin^2x)^2-(\sin^2xcos^2x)+(\cos^2x)$$

hartnn · Answer

$(1)((\sin^2x)^2-(\sin^2x\cos^2x)+(\cos^2x)^2)$
now use 
$\large (a^2+b^2)=(a+b)^2-2ab$

hartnn · Answer

$\large a=\sin^2x,b=\cos^2x$

anonymous · Answer

$$(\sin^2x+\cos^2)^2-2\sin^2xcos^2x$$

hartnn · Answer

yes, and that simplifies to ?

anonymous · Answer

$$1-2\sin^2xcos^2x$$

anonymous · Answer

but wont it be 1-1 ... because of the initial one in the question

hartnn · Answer

$\large 1-(\sin^6x+cox^6x)=1-(1-2\sin^2x\cos^2x-\sin^2x\cos^2x)=?$

hartnn · Answer

yeah, 1 cancels out, what remains ?

anonymous · Answer

$$2\sin^2xcos^2x-\sin^2xcos^2x)$$

hartnn · Answer

$1-(1-2\sin^2x\cos^2x-\sin^2x\cos^2x)=1-1+2\sin^2x\cos^2x+\sin^2x\cos^2x=?$

hartnn · Answer

$2\sin^2xcos^2x+\sin^2xcos^2x$
any doubts ?

anonymous · Answer

nope i dont think so but how does that equal the right

hartnn · Answer

because 2+1=3 :P

anonymous · Answer

i am so confused :(

hartnn · Answer

$2\sin^2xcos^2x+\sin^2xcos^2x=(2+1)\sin^2xcos^2x=3\sin^2xcos^2x$

anonymous · Answer

OH haha i see it

anonymous · Answer

thankyou

hartnn · Answer

welcome ^_^