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July 11, 2014

July 11, 2014

Posted by **Hujinson** on Tuesday, April 14, 2009 at 9:20pm.

sin(arccosx-arcsin3x)

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(3-1)cotx=v(3), cos^2x=cosx, and

tan^2x-6tanx+4=0

4. Solve; sin(∏/2-x)=1/sec

5. Solve; cot+4x+2cot^2x+1=csc4x

6. Solve; tan(x+∏/x=(√(3)+3tanx)/(3-√(3)tanx

- Math-Trig -
**Reiny**, Tuesday, April 14, 2009 at 10:42pmWow! 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

- Math-Trig -
**Hujinson**, Tuesday, April 14, 2009 at 11:25pmOk, so because sin=opp/hyp and sec=hyp/adj it will be √10/

- Math-Trig -
**Reiny**, Tuesday, April 14, 2009 at 11:56pmno, 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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