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By Help on Sunday, December 29, 2002 - 04:03 pm: Edit |

I have a math test coming up and I am really confused... Can someone pleasse answer the following questions.

1) prove that the values of cos t and cos (-t) are always equal to each other (recall that this is the check for an even function

2) Prove that the values of sin t and sin(-t) are always opposities of each other (recall that this is the check for an odd function).

3)Prove that sin(t+pie)=sin(-t)

4)Prove that The sum(cos t)^2 + (sin t)^2 always equals 1...

Any and all help would be very appreciated. We are currently doing the unit circle and this stuff is getting on my nerves!

By Dumbuket (Dumbuket) on Sunday, December 29, 2002 - 04:08 pm: Edit |

Sorry, buddy, this is college discussion. All of the people here have absurdly high grades and scores and thousands of extra curricular activities. They don't have time for things like math or grammer or stuff.

By vp on Sunday, December 29, 2002 - 04:09 pm: Edit |

well.....for the first two...u can just put it on a co-ordinate plane and show... for the last... the way i do it is draw a right angle triangle...take one side as x, one as y, so the calculate the hypoteneuse.....

Now using x and y express cos and sin...then add up cos^2 and sin^2 and ull get the answer...the middle..one...id do it by drawin a co-ordinate plane...but what is this for anyway.....

By Pat57575 (Pat57575) on Sunday, December 29, 2002 - 04:23 pm: Edit |

cos a= x/r

sin a= y/r

(cos a)^2 + (sin a)^2 = (x/r)^2 + (y/r)^2

---------------------------= (x^2+y^2)/r^2

---------------------------= r^2/r^2

---------------------------= 1

By trig on Sunday, December 29, 2002 - 04:32 pm: Edit |

you need to know this first: cosx=sin(pi/2-x). this can be proven using a right triangle.

sin(t+Pi)=sin(t+2Pi-Pi)=sin(t-Pi)=-sin(Pi-t)

-sin(Pi-t)=-sin(Pi/2-(t-Pi/2))=-cos(t-Pi/2)=-cos(Pi/2-t)=-sint=sin(-t)

my proof looks stuffy, can someone make it easier?

the last one is pretty obvious, let a, b, c be the sides of a right triangle, and c is the hypotenuse

cost=a/c, sint/b/c

cos^2+sin^2=(a/c)^2+(b/c)^2=(a^2+b^2)/c^2=c^2/c^2=1

By FuriousBilly on Sunday, December 29, 2002 - 11:30 pm: Edit |

STOP POSTING MATH PROBLEMS HERE - GO TO ANOTHER FORUM FOR THAT

By trig on Sunday, December 29, 2002 - 11:42 pm: Edit |

There are tons of math problems on the PR board too.

By Owen on Monday, December 30, 2002 - 01:08 am: Edit |

damn i forgot all this math •••• a long time ago. i almost forgot how to do a^2 + b^2=c^2

you guys/girls must have really good teachers. in my last math class (junior year, Pre-Cal) our teacher didnt even go into this stuff. another reason why students like me are at a disadvantage.

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