SOLUTION: hey, i am doing a exam review packet for my math class. i am having a ton of problems with all this trig stuff, i get the unit circle and everything like that but for some reason i
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Question 308803: hey, i am doing a exam review packet for my math class. i am having a ton of problems with all this trig stuff, i get the unit circle and everything like that but for some reason i am having a tough time getting it.
here are a few of the problems that i am having trouble with.
prove
cos (270-X)= -sinX
sin(X+(pi/2))=cosX
find exact solutions
2cosX- square root of 2=0
2cos^2 X-cosX-1=0
cos^2 -sin^2+1=0
4sin^2 X+4cosX -1=0
sin((pi/3)-(pi/2))
cos((3pi/2)+(pi/3))
simplify into a single trig func
(sec^2 X)(cotX)(sin^2 X)
(sinX)(cot^2 X+1)
(tan^2 X)/(sec X+1)
(cos 2X+sin^2 X)/ (sin 2X)
while i would love the answers =P please also give me a quick run down of why the answer is what it is. i have gone to my teacher but i am still having trouble so i am hope the internet can help me out thanks so much!!!
Answer by user_dude2008(1862) (Show Source): You can put this solution on YOUR website!
cos (270-X)= -sinX
cos(270)cos(x)+sin(270)sin(x)= -sin(x)
0*cos(x)-1*sin(x)= -sin(x)
-sin(x)= -sin(x)
-----------------------------------------
sin(x+(pi/2))=cos(x)
sin(x)cos(pi/2)+cos(x)sin(pi/2)=cos(x)
sin(x)*0+cos(x)*1=cos(x)
cos(x)=cos(x)
----------------------------------------
2cos(x)-sqrt(2)=0
2cos(x)=sqrt(2)
cos(x)=sqrt(2)/2
x = -pi/4+2pi*n or x = pi/4+2pi*n
----------------------------------------
2cos^2(x)-cos(x)-1=0
2z^2-z-1=0
(2z+1)(z-1)=0 .... z = cos(x)
(2cos(x)+1)(cos(x)-1)=0
2cos(x)+1=0 or cos(x)-1=0
cos(x)=-1/2 or cos(x)=1
cos(x)=-1/2 ----> x = 2pi/3+2pi*n or x = 4pi/3+2pi*n
or
cos(x)=1 ----> x = 2pi*n
-------------------------------------------------
cos^2(x)-sin^2(x)+1=0
cos^2(x)+(1-sin^2(x))=0
cos^2(x)+cos^2(x)=0
2cos^2(x)=0
cos^2(x)=0
cos(x)=0
x = pi/2+2pi*n or x = 3pi/2+2pi*n
----> x = pi/2+pi*n
---------------------------------------------
4sin^2(x)+4cos(x) -1=0
4(1-cos^2(x))+4cos(x) -1=0
4-4cos^2(x)+4cos(x) -1=0
-4cos^2(x)+4cos(x) +3=0
4cos^2(x)-4cos(x)-3=0
(2cos(x)+1)(2cos(x)-3)=0
2cos(x)+1=0 or 2cos(x)-3=0
cos(x)=-1/2 or cos(x)=3/2
cos(x)=-1/2 ----> x = 2pi/3+2pi*n or x = 4pi/3+2pi*n
or
cos(x)=3/2 ---> no solutions
-----------------------------------------------
sin((pi/3)-(pi/2))
sin((2pi/6)-(3pi/6))
sin(-pi/6)
-sin(pi/6)
-1/2
------------------------------
cos((3pi/2)+(pi/3))
cos((9pi/6)+(2pi/6))
cos(11pi/6)
sqrt(3)/2
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