Posted by Hujinson on Tuesday, April 14, 2009 at 9:20pm.
1. Create an algebraic expression for
sin(arccosxarcsin3x)
2. The cosx=4/5, x lies in quadrant 4. Find sin x/2
3.Determine all solutions in (0,2pie) for sin4xsin^2x=3,
cot^2v(31)cotx=v(3), cos^2x=cosx, and
tan^2x6tanx+4=0
4. Solve; sin(∏/2x)=1/sec
5. Solve; cot+4x+2cot^2x+1=csc4x
6. Solve; tan(x+∏/x=(√(3)+3tanx)/(3√(3)tanx

MathTrig  Reiny, Tuesday, April 14, 2009 at 10:42pm
Wow! You actually want somebody to do that assignment for you, without showing any effort on your part?
I will start off with #2, then you let us know some of your steps and work
use cos 2A = 1  2sin^2 A
then cos x = 1  2sin^2 x/2
4/5 = 1  2sin^2 x/2
solve for sin x/2
you will end up with sin x/2 = ±1/√10
but you are told that x is in quadrant IV, so x/2 must be in quadrant II and
sin x/2 = 1/√10

MathTrig  Hujinson, Tuesday, April 14, 2009 at 11:25pm
Ok, so because sin=opp/hyp and sec=hyp/adj it will be √10/

MathTrig  Reiny, Tuesday, April 14, 2009 at 11:56pm
no, because I solved the equation 4/5 = 1  2sin^2 x/2
2sin^2 x/2 = 1  4/5
sin^2 x/2 = 1/10
take √ of both sides
sin x/2 = ± 1/√10
why are you talking about secant?
the problem dealt only with sines and cosines.
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