find the values ( trig)?

Question

if tan x = 8/15 find the values of sin x/2 sec x/2 and tan x/2

Answer 1

p=perpendicular , b= base , h= height

tan x = 8/15 = p/b

by pythagores theorem
p^2 + b^2 = h^2

so
8^2 + 15^2 = h^2
64+225 = h^2
289 = h^2
sqrt(289) = h

so
h = 17

therefore
sin x = p/h = 8/17
cos x = b/h = 15/17

we know that
2 cos^2(A/2) - 1 = cos A
cosA/2 = sqrt [ (cosA +1)/2 ]

so
cos x/2 = sqrt[ (15/17 + 1) /2]
= sqrt[ (32/17 ) /2]
= sqrt[ 16/17]
= 4/sqrt(17)

sec x/2 = sqrt(17) / 8=1.030

sin x/2 = sqrt(1- cos^2 x/2)= 0.2424

tan x/2 = sin x/2 / cos x/2 = 0.2496


sqrt[ (15/17 + 1) /2]

Answer 2

x is an angle of the (8, 15, 17) triangle.

If c = cos(x/2), then:
2c^2 - 1 = 15/17
c = +/- (4 / sqrt(17) )
sec(x/2) = +/- sqrt(17)/4

x/2 is an angle of the ( 1, 4, sqrt(17) ) triangle.

sin(x/2) = +/- 1 / sqrt(17) = +/- sqrt(17) / 17

tan(x/2) = +/- 1 / 4

All values are acceptable with either sign, since x/2 can be in any quadrant.

Answer 3

if tanx=8/15 x = 0.49 +kpirad and x/2 =0.245 +k*pi/2 rad
if we take the interval 0.2pi
sin x/2 =0.2425 or 0.9707 or -0.2425 -0.9701
the same for sec =1/cos
tan x/2=0,25 tan x/2=-4

Answer 4

x = arctan(8/15) which can be found from tables or your calculator; now knowing this, find x/2, so that
sin x/2
sec x/2
tan x/2
are all easy