Question

Please help me to prove Sin^2 -Cos^2 =1 .....

Answer 1

Sin^2(x)-Cos^2(x) does not equal one.

Answer 2

\[ \sin ^{2} - \cos ^{2} =1 \] Please Prove this....

Answer 3

\[\sin ^2(x)-\cos^2(x)\neq1\]

Answer 4

But I ve got this from a web site.... Here's the address.... Please go there and see whats wrong with it... .... http://www.cramster.com/practice-problems-p25-0-0-0-0-0-39167-0-cpi0-0.aspx

Answer 5

Which number?

Answer 6

In the Practice-3 Question set.... number 4

Answer 7

\[\sin ^2(x)+\cos^2(x)=1\]\[\sin ^2(x)-\cos^2(x)=-\cos(2x)\]

Answer 8

Oh.... Friutless can you please explain this farther......

Answer 9

Which one?

Answer 10

Your reply for the Question I 've asked.....

Answer 11

ok well this can be proved for sin(x)^2 + cos(x)^2 = 1 because this identity is true. you want to prove this?

THE IDENTITY sin(x)^2 - cos(x)^2 = 1 IS NOT TRUE. DON'T THINK ABOUT WHAT THE WEBSITE SAYS. IT IS NOT TRUE. PERIOD.

what would you like help proving?

Answer 12

It is asking which one from the choices is correct. It is obviously not D.

Answer 13

(i) is the correct answer.

Answer 14

Ok prove this one then... \[\sin ^{8}A-\cos ^{8}A = (\sin ^{2}A-\cos ^{2}A) (1- 2 \sin ^{2}A \cos ^{2}A)\]

Answer 15

apply the FOIL method to the right side (expand it)
\[-\cos^2(A)+\sin^2(A)+2 \cos^4(A) \sin^2(A)-2 Cos^2(A) Sin^4(A)\]

Answer 16

Replace the first two functions with an identity, and factor the rest.

Answer 17

Hmmmm.... friutless in which grade you're studying... ?

Answer 18

high school

Answer 19

In which country...?

Answer 20

usa

Answer 21

Really.... then I think you have done something similar to the syllbus of London A/L isn't you?

Answer 22

I am asking this from you because I am a student who has just started doing A/L s in Sri Lanka (somewhat similar to London A/L ). so I don't know the FOIL method.....
So I can't understand you're reply.....

Answer 23

Given (x-a)(y-b), applying FOIL yeilds xy-xb-ay+ab

Answer 24

Do the same on the trigonometric expressions

Answer 25

wow... I've got the right teacher here... thank you....

Answer 26

did you get the left side from this method?

Answer 27

\[Sin ^{8}A-\cos ^{8}A=( \sin ^{2}A-\cos ^{2}A)(\sin ^{6}A+\cos ^{6}A)+\sin ^{2}Acos ^{6}A-\cos ^{2}Asin ^{6}A\] Here's what I've done....... but I've stucked in it please try this also....

Answer 28

Please help me......