Question

(cosx-sinx)^2+(cosx+sinx)^2=2
Please help
will give medal

Answer 1

Prove the identity

Answer 2

Will give medal

Answer 3

Distribute (cosx-sinx)^2 and (cosx+sinx)^2 then use the property cos(x)^2+sin(x)^2=1

Answer 4

Have you considered squaring the indicated values?

Answer 5

How would I square it

Answer 6

(cosx - sinx)^2 = cos^2x -2cosxsinx +sin^2x
(cosx+sinx)^2 = cos^2x+2cosxsinx+sin^2x

then add them together

Answer 7

OR - Save some headaches with a little more thought.

Substitute \(\cos(x) + \sin(x) = \sqrt{2}\sin(x+\pi/4)\)
Substitute \(\cos(x) - \sin(x) = \sqrt{2}\cos(x+\pi/4)\)

Rather pops right out after that.

Answer 8

You should know how to square a binomial. You WILL be required to have this skill on examination.

\((a+b)^{2} = a^{2} + 2ab + b^{2}\)

You've done it lots of time, haven't you?

Answer 9

2cos pi/4

Answer 10

No I suck at trig

Answer 11

What is that? It bears no resemblance to anything else we've mentioned.

Answer 12

yeah, i'll say

Answer 13

Ah. Well, it is time to stop sucking. Besides, squaring a binomial is from Algebra I. Come one. Make a better effort. Slowly. Patiently. You'll get it.

Answer 14

I'll do my best

Answer 15

can some show me how to do this

Answer 16

i just gave you half the answer. just add the two expressions

Answer 17

cosxsinx

Answer 18

is that right