Question

Verify the following trigonometric identities:
((sen^2α + cos^2α)^2+(sen^2α-cos^2α)^2=2

Can someone please help?

Answer 1

what is sen?

Answer 2

sin* sorry

Answer 3

lol.ok u know the formula of sin2x and cos2x

Answer 4

sin2x=2sinxcosx

Answer 5

cos2x=cos^2 x-sin^2 x

Answer 6

Have you tried distributing?

Answer 7

I tried to put everything on one side = 0, but I'm not sure if what I'm doing is right??

Answer 8

Hello? I'm reaally not sure of what to do here

Answer 9

recall the identity:
sin^2A+cos^2A = 1

Answer 10

that should sovle the first bit

Answer 11

using ths same identity, convert the cos^2A (in the second bracket) to sin^2A then solve

Answer 12

This statement is not true:
\[
((\sin^2(α) + \cos^2(α))^2+(\sin^2(α)-\cos^2(α))^2=2
\]
Did you mean to write a plus instead of a minus in the 2nd set of parenthesis?
\[
((\sin^2(α) + \cos^2(α))^2+(\sin^2(α)\color{red}{+}\cos^2(α))^2=2
\]

Answer 13

I don't know what happened to the code... :( I think \(\LaTeX\) does not like copy/pasting symbols like alpha directly into the code and replaces them with strings of ???s.

Answer 14

\[[\sin ^{2}\alpha + \cos ^{2}\alpha ]^{2} + [\sin ^{2}\alpha - \cos ^{2}\alpha]^{2} = 2\]

Answer 15

You should take account that sin^2= 1- cos^2 or cos^2= 1 - sin^2
\[[1-\cos ^{2}\alpha+\cos ^{2}\alpha]^{2} + [1-\cos ^{2}\alpha+\cos ^{2}\alpha]^{2}=2\]

Answer 16

I am sorry. for my first equation, it should have been \[[\sin ^{2}\alpha + \cos ^{2}\alpha]^{2} + [\sin ^{2}\alpha + \cos ^{2}\alpha]^{2}\]

Answer 17

carrying on from where I was ...
\[1^{2} + 1^{2} = 2\]

Answer 18

and the identity is proved.