Question

Prove that:-

Answer 1

?

Answer 2

\[3\left[ \sin ^{4}\left( \frac{ 3\pi }{ 2 } -\alpha\right)+\sin ^{4}\left( 3\pi+\alpha \right) \right]-2\left[ \sin ^{6}\left( \frac{ \pi }{ 2 }+\alpha \right)+\sin ^{6}\left( 5\pi -\alpha\right) \right] = 1\]

Answer 3

sorry for being late

Answer 4

identities!

whats sin (x+pi/2) =..
sin (pi -x) =...
sin (pi+x) = ..
?

Answer 5

yeah, i knw those steps but i'm getting stuck at the end of the problem :(
could you plz solve it?

Answer 6

post the last step that you're sure is correct in your work

Answer 7

wait...

Answer 8

are you there?

Answer 9

yes

Answer 10

check if it is correct...

Answer 11

yes, thats correct.

Answer 12

then what am i going to do in order to proceed further?

Answer 13

i can think of next step as using this
\(a^2+b^2 = (a+b)^2 -2ab\)
as we know that sin^2 +cos^2 =1

Answer 14

then the first part is coming
\[3\left( 1-2\sin ^{2}\alpha \cos ^{2}\alpha \right)\]

Answer 15

ohh, i got a better, shorter trick!

lets try to bring everything in terms of sin^2 and sin^4 !

Answer 16

for
s^6 +c^6
use (a^3+b^3) formula,
know that formula ?

Answer 17

yeah...ok, i'm trying it out wid dat formula

Answer 18

right,
then where ever you see cos^2
just replace it by sin^2

Answer 19

i mean by 1-sin^2
:P

Answer 20

same goes for sin^4 +cos^4 term :)

Answer 21

oh goddd!!! i'm not getting it! after which step should follow that which you are talking about? :(

Answer 22

*should i follow

Answer 23

ok,
did you get
\(s^6+ c^6 = (s^4+c^4 -s^2c^2)\)
??

Answer 24

our aim is very simple, there should be no cos term

Answer 25

how is it coming? :(

Answer 26

could you plz solve the whole problem and send it to me like i did?

Answer 27

wouldn't it be better if you solved it on your own, and i just guide you through steps ?
i don't want to deprive you of the joy of solving question on your own :)

Answer 28

a^3 +b^3 = (a+b) (a^2-ab+b^2)

for s^6 +c^6
a = s^2, b = c^2

Answer 29

a+b = s^2 +c^2 = 1

Answer 30

so
s^6 + c^6 = s^4-2s^2 c^2 +c^4

Answer 31

plug in c^2 = 1-s^2

c^4 = (c^2)^2 = (1-s^2)^2

Answer 32

and you will get 3 terms when you simplify
sin^4 term, sin^2 term, and constant

Answer 33

you can do that ?

Answer 34

if you can, simplifying s^4+c^4 part will be piece of cake

Answer 35

try it, tag me if you get stuck at any step,
and post that step