# 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

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.

$$\sin^4x-\cos^4x=(\sin^2x-\cos^2x)(\sin^2x+\cos^2x)$$

Now use $\displaystyle \cos^2x + \sin^2x=1$ to eliminate $\cos^2x$

We have that $y^2-(1-y)^2=2y-1$ for any value of $y$. Let $y=\sin^2(x)$ to prove the equality.

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.

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.