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Trigonometry

Hi, I'm not sure how to do this question, so any help would be greatly appreciated! :smile:
Prove that (cosx + sinx)2 + (cosx - sinx)2 = 2
(edited 8 years ago)
Reply 1
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B_9710
18
Expand each set of brackets out and use the identity sin⁡2x+cos⁡2x≡1 \displaystyle \sin^2 x +\cos^2 x \equiv 1 .
Original post by B_9710
Expand each set of brackets out and use the identity sin⁡2x+cos⁡2x≡1 \displaystyle \sin^2 x +\cos^2 x \equiv 1 .


what's with the 3 lines with the = sign? does it mean something significant?
Original post by thefatone
what's with the 3 lines with the = sign? does it mean something significant?


Identical to. Equal for all x.
Reply 4
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KaylaB
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Original post by thefatone
what's with the 3 lines with the = sign? does it mean something significant?


Always equal, no matter the value of x
Original post by KaylaB
Always equal, no matter the value of x


oh ok
Reply 6
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B_9710
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Original post by thefatone
oh ok


x2≡∣x∣ \sqrt{x^2} \equiv |x| .
Original post by B_9710
x2≡∣x∣ \sqrt{x^2} \equiv |x| .


does the lines mean the magnitude of x?
Reply 8
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B_9710
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Original post by thefatone
does the lines mean the magnitude of x?

Yes the absolute value. |-2|=2
Original post by B_9710
Yes the absolute value. |-2|=2


wait so you put the root of x² is exactly the same as the size of x? wat isn't that obvious??
Reply 10
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B_9710
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Original post by thefatone
wait so you put the root of x² is exactly the same as the size of x? wat isn't that obvious??

People often make the mistake that x2=x \sqrt{x^2}=x .
But it doesn't. Try it with a negative value of x and you'll see why what I have put here cannot be true.
(edited 8 years ago)
Original post by B_9710
People often make the mistake that sqrtx2=x sqrt{x^2}=x .
But it doesn't. Try it with a negative value of x and you'll see why what I have put here cannot be true.


so does the formula x2≡±x\sqrt {x^2} \equiv \pm x ?
(edited 8 years ago)
Reply 12
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B_9710
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Original post by thefatone
so does the formula x2≡±x\sqrt x^2 \equiv \pm x ?

No. There is no value that returns a negative value from the square root. You can't square root something and get a negative answer.
Original post by B_9710
No. There is no value that returns a negative value from the square root. You can't square root something and get a negative answer.


oops fixed it
x2≡±x\sqrt {x^2} \equiv \pm x

but surely you square the number first? since that's inside the square root?
Reply 14
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B_9710
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Original post by thefatone
oops fixed it
x2≡±x\sqrt {x^2} \equiv \pm x

but surely you square the number first? since that's inside the square root?


What you have put above is incorrect. You cannot square root something and get a negative answer. Let's take -1 as an example.
(−1)2=1=1 \sqrt{(-1)^2}=\sqrt{1} = 1 .
Original post by B_9710
What you have put above is incorrect. You cannot square root something and get a negative answer. Let's take -1 as an example.
(−1)2=1=1 \sqrt{(-1)^2}=\sqrt{1} = 1 .


but doesn't 1=1? \sqrt 1 =1 ? my calculator tells me this is true....

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