HOW DO YOU PROVE SIN^2X + COS^2X = 1? + EXAMPLE

     
to lớn start off, I understand the proof behind this identity, & I can visualize it in my head with the unit circle.

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But I read this quote:

They only need to remember three facts – that $sin 30^circ = ½$ , that $ an 45^circ =1$, và that $sin^2x + cos^2x =1$ . Just about everything else they need khổng lồ know about trigonometry can be derived from these.

and I realized I don"t have a complete understanding on practical use of the identity. Therefore I am looking for an explanation and some practical examples on why it is so important.

Thanks for your help!


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edited Jul 8, 2013 at 7:57
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Tobias Kienzler
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asked Jul 8, 2013 at 7:39
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The "special angles" in trig are $0^circ$, $30^circ$, $45^circ$, $60^circ$, and $90^circ$ (and their counterparts in quadrants II, III, và IV).

Presumably, the sines and cosines at $0^circ$ và $90^circ$ are obvious to lớn the quoted author, so the quotation boils-down what"s needed to lớn remember values at the remaining angles to lớn three facts.

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"$sin^2 heta+cos^2 heta=1$" allows for conversion between sines and cosines. Always handy.

"$sin 30^circ = 1/2$" immediately gives $cos 60^circ =1/2$. (Complementary angles are cool lượt thích that, since "cosine" means "complementary sine".) With the help of the Pythagorean identity above, we also get $cos 30^circ = sin 60^circ = sqrt3/2$. (If you"re going khổng lồ commit any of these khổng lồ memory, it makes sense to lớn choose the simple fraction, $1/2$. Personally, I think remembering "$cos 60^circ = 1/2$" is easier, because I can "see" it more easily in dropping a perpendicular from the vị trí cao nhất vertex of an equilateral triangle.)

"$ an 45^circ=1$" encodes information about the remaining special angle, saying that (in Quadrant I) $sin 45^circ = cos 45^circ$. Again with the help of the Pythagorean identity, we have $1 = sin^2 45^circ + cos^2 45^circ = 2 sin^2 45^circ$, so that $sin 45^circ = cos 45^circ = 1/sqrt2$.

So, the three facts help lớn recover a total of seven facts. & then getting the remaining secants, tangents, cosecants, & cotangents is a simple matter of using various ratios. The sine-cosine Pythagorean identity also gives rise khổng lồ the others, by dividing-through by $cos^2 heta$ or $sin^2 heta$ ...

$$eginalignfracsin^2 hetacos^2 heta + fraccos^2 hetacos^2 heta = frac1cos^2 heta quad & o quad an^2 heta + 1 = sec^2 heta \<6pt>fracsin^2 hetasin^2 heta + fraccos^2 hetasin^2 heta = frac1sin^2 heta quad & o quad 1 + cot^2 heta = csc^2 hetaendalign$$

... Reducing pressure lớn memorize all three relations.

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There are other (and better?) ways to lớn remember this stuff, but the basic point is that there"s a good deal of informational redundancy in trig. While it may seem lượt thích there are a zillion different things khổng lồ memorize, it doesn"t take long lớn realize that one can focus on a (very) few key facts và re-derive the rest on demand.