# Trig

Find the exact value of sinx/2 if cosx = 2/3 and 270 < x < 360.
A)1/3
B)-1/3
C)sqrt 6/6
D)-sqrt 6/6

C, since I KNOW cosx is always positive but I don't know the work involved. I know the half angle formula

1. 👍
2. 👎
3. 👁
1. First of all, x/2 will be in the second quadrant, since x is in the fourth quadrant. The sine of x/2 will therefore be positive.

Use the formula for sin (x/2) in terms of cos x.

sin(x/2) = sqrt([1-cos(x)]/2) = sqrt (1/6) = sqrt6/6
You got the right answer, but you it ssmes to have been a lucky guess.

Cos x is NOT always positive, but it is in this case.

1. 👍
2. 👎
2. Thank you. It wasn't really a guess it was either C or D and then I just knew it was positive so that just leaves C.

1. 👍
2. 👎
3. Jon,
Perhaps it would help if you drew an x-y axis system with a unit radius vector in each of the four quadrants.
sin T = y/1 so +
cos T = x/1 so -
tan T = y/x so +
sin T = y/1 so +
cos T = x/1 so - because x is - in q 2
tan T = y/x so -
sin T = y/1 so -
cos T = x/1 so -
tan T = y/x so + because top and bottom both -
sin T = y/1 so -
cos T = x/1 so +
tan T = y/x so -

sin has same sign as its inverse csc
cos has same sign as its inverse sec
tan has same sign as its inverse ctan

1. 👍
2. 👎
sin T = y/1 so +
cos T = x/1 so +
tan T = y/x so +

1. 👍
2. 👎

## Similar Questions

If cos theta = 0.8 and 270

2. ### Math

Use a double or half angle identity to find the exact value of the following expression: if cos x= 4/5 and 270° < x < 360°, find sin 2x Please help! Thanks

3. ### Trigonometry

4. Find the exact value for sin(x+y) if sinx=-4/5 and cos y = 15/17. Angles x and y are in the fourth quadrant. 5. Find the exact value for cos 165degrees using the half-angle identity. 1. Solve: 2 cos^2x - 3 cosx + 1 = 0 for 0

4. ### trigonometry

how do i simplify (secx - cosx) / sinx? i tried splitting the numerator up so that i had (secx / sinx) - (cosx / sinx) and then i changed sec x to 1/ cosx so that i had ((1/cosx)/ sinx) - (cos x / sinx) after that i get stuck

1. ### math;)

The equation 2sinx+sqrt(3)cotx=sinx is partially solved below. 2sinx+sqrt(3)cotx=sinx sinx(2sinx+sqrt(3)cotx)=sinx(sinx) 2sin^2x+sqrt(3)cosx=sin^2x sin^2x+sqrt(3)cosx=0 Which of the following steps could be included in the

2. ### Math

How do I solve this? tan^2x= 2tanxsinx My work so far: tan^2x - 2tanxsinx=0 tanx(tanx - 2sinx)=0 Then the solutions are: TanX=0 and sinX/cosX = 2 sin X Divide through by sinX: we have to check this later to see if allowed (ie sinX

3. ### Trig Identities

Prove the following identities: 13. tan(x) + sec(x) = (cos(x)) / (1-sin(x)) *Sorry for any confusing parenthesis.* My work: I simplified the left side to a. ((sinx) / (cosx)) + (1 / cosx) , then b. (sinx + 1) / cosx = (cos(x)) /

4. ### Calculus

determine the absolute extreme values of the function f(x)=sinx-cosx+6 on the interval 0

Angles x and y are located in the first quadrant such that sinx=3/5 and cosy=5/13. a) Determine an exact value for cosx. b) Determine an exact value for siny.

2. ### Trig Help

Given that cos2x=7/12 and "270 equal or < 2x equal or < 360", find sinx. Please help and Thank you

3. ### precal

1/tanx-secx+ 1/tanx+secx=-2tanx so this is what I did: =tanx+secx+tanx-secx =(sinx/cosx)+ (1/cosx)+(sinx/cosx)-(1/cosx) =sinx/cosx+ sinx /cosx= -2tanxI but I know this can't be correct because what I did doesn't end as a

4. ### math-trig

Use the double angle identities to find sin2x if sinx= 1/square of 17 and cosx