If anyone is able to do any of these proofs before 11am, that would be super! I've tried and tried and cannot get anything to work out... You guys are my last hope, thanks!

sin^6x + cos^6x = 1 - 3sin^2x * cos^2x

sin^2x * tan^2x + cos^2x * cot^2x = tan^2x + cot^2a - 1

(sin10x + sin6x)/(sin12x + sin4x) - (sin5x - sinx)/(sin7x + sinx) = (2sin2x)/(sin8x)

 

What is sin^6x? How would I do that on a calculator?

Mike.

 

Oh sorry, my math teacher's wierd, it's [Sin(x)]^6

 

Hi,




phew... take me long time for the latex

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Thanks so much, man! I needed that bonus point!!!

 

Not wierd, that is the normal way to write trig functions to a power, if you leave it after it can be confused with being inside the function, sin^2(x) is alot different from sin(x^2).

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Yeah... but she's weird anyway... and mean. I figured that being right out of college, she'd sympathize with use and be nice, but I was wrong. The average grade on the test we took that day was 53%... Plus it was the only test we have all quarter... D:

 

Ouch that hurts, I really hope you guys dont have tests worth 80% of your final grade like I do.

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Haha. There's a module which the final exam worth 80 % too. Some are 50 %.

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No, tests are 40% in her class.
But exams are only, like 30% of the final grade... they aren't too bad. But that test shot my chances of getting a semester 'A', I dropped from a 98.7% to a 71% from one damn test... >.<

 

magician13134,

Only fair that this does not arrive in time for your homework.

sin^6+cos^6 --> (sin^2)^3 +cos^6 -->((1-cos^2)^3-cos^6 --> 1-3cos^2+3cos^4-cos^6+cos^6 --> 1-3cos^2+3cos^4 --> 1-3cos^2*(1-cos^2) --> 1-3cos^2*sin^2

sin^2*tan^2+cos^2*cot^2 --> sin^2*(sec^2-1)+cos^2*(csc^2-1) --> tan^2-sin^2+cot^2-cos^2 --> tan^2-cot^2-1

I believe the third problem is a misprint because the right side of the equation "(2sin2x)/(sin8x)" is 2/sin^6

Ratchit

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The next time stuff like this comes up, or anything that involves differentiation or integration of sin(x) and cos(x) functions, your best bet is to use Euler's Identity:

Where:

e= Base of natural logs, 2.71828...
j= sqrt(-1)

These are complex numbers where the real part gives answers related to the cos(x) function, and the imaginary part gives sin(x) related answers.

It is much easier to manipulate exponentials than it is to mess with sin's and cos's. That is especially true where differentials, integrals, and powers of the functions are involved.

Attached Images

eulers-identity.jpg (10.3 KB, 7 views)

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