Verified answer

refresh: sin(2x) = 2sinxcosx

so 2sin(x/2)cos(x/2) = sin(2(x/2)) = sinx

Pythagorean identity

sin²x + cos²x = 1

-----------------------------------------

LHS:

Expand

cos²(x/2) - 2sin(x/2)cos(x/2) + sin²(x/2)

By Pythagorean identity

1 - 2sin(x/2)cos(x/2)

1 - sin(2(x/2))

1 - sinx

LHS = RHS

Proved

This is an easy one. The Left Hand Side is in (a+b)^2 form so just open it and you get

={cos(x/2)}^2 + {sin(x/2)}^2 - 2sin(x/2)cos(x/2)

=Now, there are 2 formulae, one is (sinx)^2+(cos)^2=1 and 2sinxcosx=sin2x

Using those relations, you'll get the RHS

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RE:

Prove the trig identity (cos(x/2)-sin(x/2))^2= 1-sinx?

Please help!

We can only work on one side

^2 means squared

/ means divided by

Thank you!

(a--b)^2 = a^2 +b^2 --2ab Same way LHS = (cosx/2 - sinx/2)^2 = cos^2(x/2) + sin^2(x/2) --2sin(x/2) cos(x/2) But cos^2(A) + sin^2(A) = 1 and Sin 2A = 2 sinA cosA So cos^2(x/2) + sin^2(x/2) = 1 and 2sin(x/2) cos(x/2) = Sin (2*x/2) = sin x So the final expression becomes as 1 -- sin x = RHS.

You can make life easier for yourself if you state that x/2 = y and therefore x = 2y

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

Change cos^2(y) because there is no cos on the right side: cos^2(y) = 1 - sin^2(y)

(1 - sin^2(y)) - 2cos(y)sin(y) + sin^2(y) = 1 - 2cos(y)sin(y)

But 2cos(y)sin(y) = sin(2y)

= 1 - sin(2y)

And so we have reached the other side, which through substitution is 1 - sin(2y) because we had established that x = 2y

That's not cheating... that's being clever.

I hope this helps...