how does sin^2x-cos^2x =1
could someone explain or write down steps on how the left side becomes 1?
I am positive that you meant that to type
sin^2x + cos^2x =1
label a right angled triangle having sides a, b, and c
with c as the hypotenuse and angle Ø opposite side a.
LS = sin^2Ø + cos^2Ø
= (a/c)^2 + (b/c)^2
= (a^2/c^2) + (b^2/c^2)
= (a^2 + b^2)/c^2 , but a^2 + b^2 = c^2 by Pythagoras
so
LS = c^2/c^2
= 1
= RS
(the choice of variable does not matter, you can call your angle Ø or x or whatever)
the original problem was:
(sin x + cos x)^2 + (sin x - cos x)^2 = 2
In that case you expanded it incorrectly, should have been
LS
=(sin x + cos x)^2 + (sin x - cos x)^2
= sin^2x + 2sinxcosx + cos^2x + sin^2x - 2sinxcosx + cos^2x
= sin^2x+cos^2x + sin^2x + cos^2x
= 1+1
= 2
= RS
To understand how the left side of the equation sin^2x - cos^2x = 1 becomes 1, let's start by considering the Pythagorean identity for sine and cosine:
sin^2x + cos^2x = 1
This identity states that the sum of the squares of sine and cosine of an angle 'x' is always equal to 1.
Now, to derive the equation sin^2x - cos^2x = 1, we need to rearrange the terms. We'll take the Pythagorean identity and subtract 2*cos^2x from both sides:
sin^2x + cos^2x - 2*cos^2x = 1
Combining like terms:
sin^2x - (2*cos^2x - cos^2x) = 1
sin^2x - cos^2x = 1
Therefore, sin^2x - cos^2x is equal to 1.