Simple trigonometric identity proof: $\sin^4x-\cos^4x=2\sin^2x-1$ How would you verify that this trigonometric equation is an identity? 
$$\sin^4x-\cos^4x=2\sin^2x-1? $$
The 4th powers are really throwing me off, and i'm still fairly new to this and there is no clear identity from the start like $\cos^2x + \sin^2x=1$
Thanks
 A: Hint.  You can write
$$\sin^4x-\cos^4x=(\sin^2x)^2-(\cos^2x)^2\ ,$$
then use an important factorisation that you would have learned before you ever started trigonometry.
A: $$\sin^4x-\cos^4x=(\sin^2x-\cos^2x)(\sin^2x+\cos^2x)$$
Now use $\displaystyle \cos^2x + \sin^2x=1$ to eliminate $\cos^2x$ 
A: We have that $y^2-(1-y)^2=2y-1$ for any value of $y$. Let $y=\sin^2(x)$ to prove the equality.
A: Your expression can be written as
$$sin^4x−cos^4x=(sin^2x)^2−(cos^2x)^2$$ 
so there you 'll get $sin^2 x-cos^2 x$ which is equal to $2sin^2 x -1$. Then the other term is equal to one, which is $sin^2 x +cos^2 x =1$. 
So multiply those two terms . Then you will get what you want.
A: Factor the expression. Remember that $a^2 - b^2 = (a+b)(a-b)$.
$$sin^4x + cos^4x$$
$$=(sin^2x)^2 + (cos^2x)^2$$
$$=(sin^2x + cos^2x)(sin^2x - cos^2x)$$
Remember the famous trig. identity $sin^2x + cos^2x = 1$. Using this, you can eliminate that part of the expression from your factoring.
$$(sin^2x+cos^2x)(sin^2x-cos^2x)$$
$$=sin^2x-cos^2x$$
You now can use the same identity we used before, $sin^2x + cos^2x = 1$, to rewrite $cos^2x$. We can say that $cos^2x=1-sin^2x$ from simple algebra. This leads to further simplification like so:
$$sin^2x-cos^2x$$
$$=sin^2x-(1-sin^2x)$$
$$=sin^2x-1+sin^2x$$
$$=2sin^2x-1$$
Therefore:
$$\boxed {sin^4x-cos^4x = 2sin^2x-1}$$
This concludes the proof.
