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How can I verify the Pythagorean Trig Identity using approximations for $\sin x$ and $\cos x$ derived from their infinite series representations?

I can see that the infinite series of $\sin x$ and $\cos x$ should mean that for small angles, $\sin x$ is approximately $x$, and $\cos x$ is approximately $1-.5x^2$. I'd like to use these approximations to verify that $\sin^2x + \cos^2x$ is approximately $1$.

I seem to be off on the wrong track with my approach to this task. First, I simply substituted the approximations derived from the first (or first and second) terms of the series into the Pythagorean identity. Perhaps my algebra is off, but two attempts yielded the result $x^4 = 0$ which fails to confirm either the Pythagorean identity or the approximations. Of course, we know intuitively the results should verify truly.

Can you please suggest an appropriate way to approach this?

Added: I see from the answers below, that I should not have set the substitution in the Pythagorean Trig Identity equal to 1, because that very value of 1 is what I am verifying. Instead, the Pythagorean Trig Identity should be calculated using the approximation substitutions, then compared to the value of 1. If they are approximately close, the verification is done, and the Pythagorean Trig Identity is indeed approximately equal to 1.

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up vote 8 down vote accepted

If $\sin x$ is approximately $x$, and $\cos x$ is approximately $1−.5x^2$, then $(\sin x)^2 + (\cos x)^2$ is approximately $$x^2 + (1-.5 x^2)^2 = x^2 + 1 - x^2 + .25 x^4 = 1 + .25 x^4.$$

Note that the $x^2$ term cancels out, so the result is $1$ with a term of order $x^4$.

As you use more terms of the power series for sin and cos, the result will be closer to 1.

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Marty, I see what you're saying. I set mine = 1, in the style of the Pythagorean Trigonometric Identity, which leads to cancellation of the 1, and x^4 = 0. How do I justify not placing the 1 as I have, and as you did not? – FreeTrader Oct 2 '11 at 5:51
I see from Brian M. Scott's added comment that I should NOT set equal to 1 per the Pythagorean Trig Identity, but rather, should see what answer I get from the substitution, then compare my result to the Pythagorean Trig Identity's result of 1. Marty, you have told me my answer is 1 plus a term of order x^4 which will improve in accuracy the further along the infinite series I am willing to go in calculating my approximation. I understand this now. Thank you very much. – FreeTrader Oct 2 '11 at 6:02

Your algebra seems to be off:

$$\sin^2x + \cos^2 x \approx x^2 + \left(1-\dfrac{x^2}2\right)^2 = x^2 + 1 - x^2 + \dfrac{x^4}4 = 1 + \dfrac{x^4}4,$$ which is approximately $1$ for small values of $x$.

Added: Note that you can’t expect to get exactly $1$, since you’re using an inexact approximation. If, as anon suggests, you were setting this equal to $1$ and solving for $x$, you weren’t vitiating the result: you were just showing that the approximation is perfect only at $x=0$.

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I imagine OP's algebra might have been to set that equal to 1 and then derive $x^4=0$. – anon Oct 2 '11 at 5:41
@anon: Now that you mention it, that does sound likely. Failure of interpretation, not failure of algebra. – Brian M. Scott Oct 2 '11 at 5:44
I did the algebra as you did, up until the right side of your final =. anon is correct, I did set original = 1, which derives x^4 = 0. How do you get the positive 1 + x^4/4 on the right hand side without addressing the 1 in the Pyth Trig ID, Brian? – FreeTrader Oct 2 '11 at 5:53
Brian, I see from your added comment that I should NOT set equal to 1 per the Pythagorean Trig Identity, but rather, should see what answer I get from the substitution, then compare my result to the Pythagorean Trig Identity's result of 1. Thank you! – FreeTrader Oct 2 '11 at 6:00
I would like to accept both answers here, as they combine to produce a solid understanding of the right way to approach this type of problem. Thank you very much. – FreeTrader Oct 2 '11 at 6:03

Why not look at the formal series for sine and cosine, $s(x)$ and $c(x)$, exactly the series you’ve written, and notice that $s'=c$, $c'=-s$, and differentiate the formal expression $s^2+c^2$ to get zero. Reflect on that, and realize that you’ve just shown that the formal series $s^2+c^2$ is constant (not for calculus reasons, but because a nonconstant formal series has nonzero derivative). And the constant in question is clearly $1$.

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$x^4=0$ is correct if interpreted to mean "the identity holds up to terms of degree 4 or higher". This is because terms of degree $\geq 4$ were dropped from the power series for $\sin$ and $\cos$ at the start of the calculation. If terms of degree $\geq n$ are dropped from the series then this order $n$ approximation to $\sin^2 + \cos^2$ will be equal to $1 + ax^n + bx^{n+1} + \dots$ for some numerical coefficients $a,b,\dots$ (that depend on the value of $n$ used).

Near $x=0$ the higher degree terms are smaller and an identity being correct up to higher order means greater accuracy.

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Thanks for the helpful comment. I appreciate it. – FreeTrader Oct 2 '11 at 6:11

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