MHB What is the quadratic trig identity for cosine when simplified?

AI Thread Summary
The discussion centers on simplifying the quadratic trigonometric identity for cosine, specifically for the expression cos(4x). The initial transformation using double-angle identities leads to the equation cos(4x) = 1 - 8sin^4(x) + 8sin^2(x). Participants explore the relationships between sine and cosine, applying identities to manipulate the equation. The final form presented is cos(4x) = 8sin^4(x) - 8sin^2(x) + 1, showcasing the complexity of the simplification process. The conversation highlights the challenges in breaking down the identity effectively.
karush
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$$\cos\left({4x}\right)
=8\sin^4\left({x}\right)
-8\sin^2\left({x}\right)
+1$$

I thought this would break down
nice from the quadratic but it didn't.
 
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First, apply a double-angle identity for cosine:

$$\cos(4x)=1-2\sin^2(2x)$$

Next, apply the double-angle identity for sine:

$$\cos(4x)=1-2\left(2\sin(x)\cos(x)\right)^2=1-8\sin^2(x)\cos^2(x)$$

Can you proceed? :D
 
$$1-8\sin^4\left({x}\right)+8\sin^2\left({x}\right)$$
=1-8 (\sin^2\left({x}\right)(\sin^2\left({x}\right)-1))$$
 
Last edited:
karush said:
$$1-8\sin^2\left({x}\right)
\cos^2\left({x}\right)
=1-8 (\sin^2\left({x}\right)(\sin^2\left({x}\right)-1))$$

Not quite:

$$\sin^2(x)+\cos^2(x)=1\implies\cos^2(x)=1-\sin^2(x)$$
 
$$1-8\left[\sin^2\left({x}\right)\left(1-\sin^2\left({x}\right)\right)\right]$$
$$1-8\sin^2\left({x}\right)+8\sin^4\left({x}\right)$$
 
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