SOLUTION: (a) Evaluate the expression under the given conditions. sin(θ/2); tan(θ) = −5/12, θ in Quadrant IV (assume 0 < θ < 2π) (b) Evaluate the expression under the given condi

Algebra ->  Trigonometry-basics -> SOLUTION: (a) Evaluate the expression under the given conditions. sin(θ/2); tan(θ) = −5/12, θ in Quadrant IV (assume 0 < θ < 2π) (b) Evaluate the expression under the given condi      Log On


   

Question 1160204: (a) Evaluate the expression under the given conditions.
sin(θ/2); tan(θ) = −5/12, θ in Quadrant IV (assume 0 < θ < 2π)

(b) Evaluate the expression under the given conditions.
tan(2θ); cos(θ) = 3/5, θ in Quadrant I
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
Evaluate the expression under the given conditions.
sin%28theta%2F2%29;
tan%28theta%29+=+-5%2F12, theta in Quadrant IV (assume 0+%3C+theta+%3C+2pi%29
tan%28theta%29+=opposite%2Fadjacent+=>opposite=-5, adjacent=12
hypothenuse=sqrt%28%28-5%29%5E2%2B12%5E2%29
hypothenuse=sqrt%2825%2B144%29
hypothenuse=sqrt%28169%29
hypothenuse=13

cos%28theta%29=adjacent%2Fhypotenuse, then
cos%28theta%29=12%2F13
and
sin%28theta%2F2%29= ±sqrt%281-cos%28theta%29%29%2F2%29...since theta in Quadrant IV, use negative sign
sin%28theta%2F2%29=+-sqrt%281-12%2F13%29%2F2%29
sin%28theta%2F2%29=+-sqrt%28%2813-12%29%2F13%29%2F2%29
sin%28theta%2F2%29=+-sqrt%28%281%2F13%29%2F2%29
sin%28theta%2F2%29=+-sqrt%282%2F13%29%29
sin%28theta%2F2%29=+-sqrt%2826%29%2F13


Evaluate the expression under the given conditions.
tan%282theta%29;
since+tan%282theta%29=+%282tan%28theta%29%29%2F%281-tan%5E2%28theta%29%29, tan%28theta%29=sin%28theta%29%2Fcos%28theta%29, and cos%28theta%29 is given, we have to find sin%28theta%29
​


cos%28theta%29+=+3%2F5, theta in Quadrant I

cos%28theta%29=adjacent%2Fhypotenuse->adjacent=3 and hypotenuse=5
then
opposite=sqrt%285%5E2-3%5E2%29
opposite=sqrt%2825-9%29
opposite=sqrt%2816%29
opposite=4
so, sin%28theta%29=4%2F5
then
tan%28theta%29=opposite%2Fadjacent
tan%28theta%29=4%2F3
now we can find
+tan%282theta%29=+%282tan%28theta%29%29%2F%281-tan%5E2%28theta%29%29......substitute tan%28theta%29
+tan%282theta%29=+%282%284%2F3%29%29%2F%281-%284%2F3%29%5E2%29
+tan%282theta%29=+%288%2F3%29%2F%281-16%2F9%29
+tan%282theta%29=+%288%2F3%29%2F%289%2F9-16%2F9%29
+tan%282theta%29=+%288%2Fcross%283%291%29%2F%28-7%2Fcross%289%293%29
+tan%282theta%29=+%288%2F1%29%2F%28-7%2F3%29
+tan%282theta%29=-24%2F7

Answer by ikleyn(52864) About Me  (Show Source):
You can put this solution on YOUR website!
.

            Regarding part  (a),  the solution and the answer by @Mathlover1 are both incorrect;

            therefore, I came to bring the correct solution. 


(a)  if  tan%28theta%29 = -5%2F12   and  theta is in QVI, then'


     cos%28theta%29 = 12%2F13;


     theta%2F2  lies in QII, where sine is positive,


     and therefore  sin%28theta%2F2%29 = + sqrt%28%281-cos%28theta%29%29%2F2%29 = sqrt%28%281-12%2F13%29%2F2%29 = sqrt%281%2F26%29 = sqrt%2826%29%2F26,


      totally different formula than solution by @MathLover1.


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