prove that (cosx+sinx)^2 +(cosx-sinx)^2=2?

Answer 1

All you have to do is expand the brackets

c^2 + 2cs + s^2 + c^2 - 2cs + s^2
= 2*(c^2 + s^2) = 2

Once you see the cos, the sin and the square, you should be looking for sin^2 + cos^2 = 1

Answer 2

Expand it out, and collect terms. I suspect you wind up with 2 cos^2 x + 2 sin^2 x. By Phtag. thm, this is identically 2. So there.
BTW, anywhere you see something like this, it's a good change there is a cos^2 x+sin^2 x=1 waiting to be found.

Answer 3

If you expand those squared binomials, you will have the following:

cos²x + 2cosxsinx + sin²x + cos²x -2cosxsinx + sin²x

Remember the main Pythagorean identity: cos²x + sin²x = 1.....
= 1 + 2cosxsinx + 1 - 2cosxsinx = 2

Answer 4

(cos x + sin x) ^ 2 + (cos x - sin x) ^ 2 = 2 (*)

Put : y = cos x
z = sin x

You substitute y and z in (*)

(y + z )^2 + (y - z)^2 = 2

y^2 + 2yz + z^2 + y^2 -2yz +z^2 = 2

y^2 + z^2 + y^2 + z^2 = 2

2y^2 + 2z^2 = 2 [Again, replaced y and
z values]

2(cos x)^2 + 2(sin x)^2 = 2 [factorized 2]


2[(cos x)^2 + (sin x)^2] = 2 [(cos x)^2 + (sin x)^2 = 1]
( trigo. identity)

2[1] = 2

2 = 2

qed

Answer 5

(cosx+sinx)^2 +(cosx-sinx)^2=
(cos^2 +2sincos +sin^2)+(cos^2-2sincos+sin^2)=
(cos^2+sin^2)+(cos^2+sin^2) = 1+1=2

Answer 6

(cosx+sinx)^2 +(cosx-sinx)^2=2
cos^2x +2sinxcosx +sin^2x+cos^2x -2sinxcosx +sin^2x=2
cos^2x +sin^2x+cos^2x +sin^2x = 2
1+1 =2
2=2

Answer 7

Is this a question. hmmm
expanding both the terms we get
cos^2x+sin^2x+2cosxsinx+cos^2x+sin^2x-2cosxsinx
=1+2cosxsinx+1-2cosxsinx
=2

Answer 8

(cos(x) + sin(x))^2 + (cos(x) - sin(x))^2

(cos(x)^2 + 2cos(x)sin(x) + sin(x)^2) + (cos(x)^2 - 2cos(x)sin(x) + sin(x)^2)

cos(x)^2 + 2cos(x)sin(x) + sin(x)^2 + cos(x)^2 - 2cos(x)sin(x) + sin(x)^2

2cos(x)^2 + 2sin(x)^2
2(cos(x)^2 + sin(x)^2)
2(1)
2

so

(cos(x) + sin(x))^2 + (cos(x) - sin(x))^2 = 2