how to simplify sqrt(8cos^4(t)-16cos^2(t)sin^2(t)+8sin^4(t))

acpodgorski

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Oct 8, 2007
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I figure i need to use some trig identities but I can't figure out which ones...
can i use cos^2+sin^2=1 somehow?

simplify:

sqrt(8cos^4(t) -16cos^2(t)sin^2(t)+8sin^4(t))
 
First factor out 8:

\(\displaystyle \L\\\sqrt{8(cos^{4}t-2cos^{2}tsin^{2}t+sin^{4}t}\)

Now factor the inside:

\(\displaystyle \L\\\sqrt{8(cost+sint)^{2}(cost-sint)^{2}}\)

Now, finish?.
 
and also is that the correct factoring? because i think that would mean that the sin^4t would have to be negative
 
Yes, that's the correct factoring.

It's the same as factoring \(\displaystyle \L\\x^{4}-2x^{2}y^{2}+y^{4}=(x-y)^{2}(x+y)^{2}\)

\(\displaystyle \L\\\sqrt{8(cost-sint)^{2}(cost+sint)^{2})}\)

\(\displaystyle \L\\\sqrt{8}(cost-sint)(cost+sint)\)

Notice the difference of two squares?.

\(\displaystyle \L\\\sqrt{8}(cos^{2}t-sin^{2}t)\)

Now, remeber your identites. Can you simplify further?.
 
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