Solving Cos4x + Sin4x = 1: Can I Sqroot Both Terms?

  • Thread starter uzman1243
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In summary, the conversation discusses whether or not the equation cos^4x + sin^4x = 1 is true. While it can be written as cos^4x + sin^4x = (cos^2x + sin^2x)^2 - 2cos^2xsin^2x, it is not a correct form. Using complex analysis, it can be shown that cos^4x + sin^4x = cos(4x) = (x^4 + y^4)/r^4, which is not equal to 1.
  • #1
uzman1243
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I am a bit confused here.

Cos2x + sin2x = 1

Thus can I say

Cos4x + sin4x = 1

If I just sqroot each term:
sqroot Cos4x + sqroot sin4x = sqroot (1) = 1

?
 
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  • #2
No. Sqrt(a+b) and sqrt(a)+sqrt(b) is different. If you take sqrt on both sides, it has to be sqrt(a+b)=sqrt(c).
But you can still write##cos^4x+sin^4x=(cos^2x+sin^2x)-2cos^2xsin^2x##
 
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  • #3
AdityaDev said:
No. Sqrt(a+b) and sqrt(a)+sqrt(b) is different. If you take sqrt on both sides, it has to be sqrt(a+b)=sqrt(c).
But you can still write##cos^4x+sin^4x=(cos^2x+sin^2x)-2cos^2xsin^2x##

You can write it, but it wouldn't be correct. Instead:

##cos^4x+sin^4x=(cos^2x+sin^2x)^2-2cos^2xsin^2x = 1 - 2cos^2xsin^2x##
 
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  • #4
I think that the simplest form will be:
$$...=1-2\cos^2x\sin^2x=1-\frac{1}{2}\sin^2(2x)$$
because:
$$2\sin x\cos x=\sin(2x)$$
 
  • #5
PeroK said:
You can write it, but it wouldn't be correct. Instead:

##cos^4x+sin^4x=(cos^2x+sin^2x)^2-2cos^2xsin^2x = 1 - 2cos^2xsin^2x##
it was a typo
 
  • #6
Using complex analysis makes it still more elegant. On one hand: [itex]e^{4ix}=cos(4x)+isin(4x) [/itex]. On the other hand: [itex]e^{4ix}=(cos(x)+isin(x))^{4}= cos^{4}(x) + 4icos^{3}(x)sin(x) + 6i^{2}cos^{2}(x)sin^{2}(x)+4i^{3}cos(x)sin^{3}(x)+i^{4}sin^{4}(x) [/itex]. Multiplying out: [itex]e^{4ix}=cos^{4}(x) + 4icos^{3}(x)sin(x) - 6cos^{2}(x)sin^{2}(x)-4icos(x)sin^{3}(x)+sin^{4}(x)[/itex]. Thus: [itex]cos(4x)+isin(4x)=cos^{4}(x)+sin^{4}(x) - 6cos^{2}(x)sin^{2}(x) + i(4cos^{3}(x)sin(x) )-4cos(x)sin^{3}(x)) [/itex]. Taking the real part: [itex]cos(4x)=cos^{4}(x)+sin^{4}(x) - 6cos^{2}(x)sin^{2}(x) [/itex]
 
  • #7
From the definition of trigonometric function
cos=x/r
sin=y/r
From the Pythagorean theorem
r^2=x^2+y^2

cos^4(x) + sin^4(x)
=(x^4+y^4)/r^4
=((x^2+y^2)^2-2x^2y^2)/(r^4)
=(r^4-2x^2y^2)/r^4
!=1
 
Last edited:

1. Can I solve the equation by taking the square root of both terms?

No, taking the square root of both terms would not be a valid method to solve this equation.

2. What is the most efficient way to solve this equation?

The most efficient way to solve this equation would be by using the double angle formula for cosine and sine, which would simplify the equation to a quadratic form.

3. Can I use a calculator to solve this equation?

Yes, you can use a calculator to solve this equation. However, it is important to use the correct setting for trigonometric functions and to double check your solution.

4. Is there more than one solution to this equation?

Yes, there are multiple solutions to this equation. In fact, there are an infinite number of solutions for this equation.

5. Can I use the unit circle to solve this equation?

Yes, you can use the unit circle to solve this equation by finding the angles where cos4x and sin4x equal 1/2, and then using the double angle formula to find the corresponding values of x.

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