Question

Find the actual values for sin x by showing all your working if sec x=4/3,tan x<0.

Answer 1

cos² x = 1 - sin² x

cos x = ± √ ( 1 - sin² x )


sec x = 1 / cos x

sec x = 1 / ± √ ( 1 - sin² x )

4 / 3 = 1 / ± √ ( 1 - sin² x ) Take reciprocal value of both sides

3 / 4 = ± √ ( 1 - sin² x ) Raise both sides to on the power of two

9 / 16 = 1 - sin² x

1 - sin² x = 9 / 16 Subtract 1 to both sides

1 - sin² x - 1 = 9 / 16 -1

- sin² x = 9 / 16 -1

- sin² x = 9 / 16 - 16 / 16

- sin² x = - 7 / 16 Multiply both sides by - 1

sin² x = 7 / 16 Take the square root of both sides

sin x = ± √ ( 7 / 16 )

sin x = ± √7 / √16

sin x = ± √7 / 4


tan x < 0 mean tan x is negative

tan x = sin x / cos x

sec x = 4 / 3

cos x = 1 / sec x

cos x = 3 / 4

cos x is positive so:

tan x = sin x / cos x can be negative only if sin x is negative.

This mean:

sin x = - √7 / 4

Answer 2

if secx = 4/3
then cosx = 3/4
also since tanx < 0 we know that x must be in quadrants IV, since only in IV is the cosine positive and the tangent negative

make a sketch of a right-angled triangle in standard position with
base of 3, hypotenuse 4, and height y
x^2 + y^2 = r^2
9 + y^2 = 16
y^2 = 7
y = ± √7, but in IV y = -√7

sinx = -√7/4 , cscx = -4/√7
cosx = 3/4, secx = 4/3 <<<--- our given
tanx = -√7/3 , cotx = -3/√7