sec B=(-17/8), pi<B<(3pi/2) find sin B, cos B, and tan B....help please!

Expert Answers
gsenviro eNotes educator| Certified Educator

sec B = -17/8 and pi < B < 3pi/2

In the given range of B, both sin and cos have negative values and therefore, tan has a positive value.

now, sec is inverse of cos.

Thus, sec B = 1/cosB = -17/8

means, cos B = -8/17

sin B = `sqrt (1-cos^2B) = sqrt(1-(-8/17)^2) = - 15/17`  

and tan B = sinB/cosB = `(-15/17)/(-8/17) = 15/8`

Hope this helps.

kspcr111 | Student

From the below tables givne in the attachments we can easily find the values of sin B, cos B, tan B

Given ,

`sec B=(-17/8), pi<B<(3pi/2) `  ie the angle B is in the Third quadrant .

From the table I , we get the values and from the table II , we assign the signs that is the final answer 

Now , Finding 

1)

`sin B = (+- (sqrt(sec^2 B -1 )/(sec B)))`

        = `(+- (sqrt((-17/8)^2 -1)/(-17/8)))`

        = `(+- (sqrt((17^2 - 8^2)/8^2)/(-17/8)))`

        =`(+- (sqrt((17^2 - 8^2))/8)/(-17/8))`

         = `(+- (sqrt((17^2 - 8^2))/(-17)))`

         = `(+- (sqrt((289 - 64))/(-17)))`

        = `(+- (sqrt((225))/(-17)))`

        = `(+-(-15)/17)`

As the angle is in the third quadrent  sin is negative (see table II )

so ,

`sin B =((-15)/17)`

2)

Cos B = `(1/ sec B)`

         = `(1/((-17)/8))`

         = `((-8)/17)`

3)

Tan B = `(+- sqrt(sec^2 B -1 ))`

          = `(+- sqrt(((-17)/8)^2 -1 ))`

          = `(+- sqrt(((289)/64) -1 ))`

          = `(+- sqrt(((289-64)/64) ))`

           = `(+- sqrt(((225)/64) ))`

            = `(+- (15/8) )`

As the angle is in the Third quadrent tan is positive (see table II )

so,

`Tan B =(15/8)`

simple :)

         

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