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meerkat18
so base on your question the possible answer to that kind of question and the solution are the following and i hope you will understand the formula and free to ask some questions if needed.

2cos2x2=1+cosx=1−1517=217,cos2x2=117 cosx2=−117−−√....(1) sinx2=117−−√,90<x2<135 sinx2>0,cosx2<0,tanx2<0 2sin2x2=1−cosx=1−−1517=3217,sinx2=417−−√.....(2) divide (2) by (1) and get the value of tan x/2
ardni313

[tex]\rm sin\dfrac{1}{2}x=\sqrt{ \dfrac{32}{54}}\\\\cos\dfrac{1}{2}x=-\sqrt{ \dfrac{2}{54} }\\\\tan\dfrac{1}{2}x=-4[/tex]

Further explanation

Trigonometry is the science of mathematics that studies the relationship between sides and angles in triangles

In a cartesian plane, the angle α can vary from 0 ° to 360 °, counterclockwise

This coordinate field is divided into 4 quadrants:

  • quadrant 1: 0 ° - 90 °
  • quadrant 2: 90 ° - 180 °
  • quadrant 3: 180 ° - 270 °
  • quadrant 4: 270 ° - 360 °

For half angle we can find it from double angle

Trigonometry Formulas for Double Angles:

sin 2A = 2 sin A cos A

cos 2A = 1 - 2 sin²A = 2cos²A-1

tan 2A = 2 tan A / 1-tan²A

then:

cos 2A = 2cos²A-1

[tex]\rm cos\:A=2cos^2\dfrac{1}{2}A-1\\\\2cos^2\frac{1}{2}A=1+cos\:A\\\\cos^2\frac{1}{2}A=\dfrac{1+cos\:A}{2}\\\\cos\:\frac{1}{2}A=\sqrt{\dfrac{1+cos\:A}{2} }[/tex]

cos 2A = 1-2 sin²A

in the same way

[tex]\rm sin\dfrac{1}{2}A=\sqrt{\dfrac{1-cos\:A}{2} }[/tex]

[tex]\rm tan\dfrac {1} {2}A = \sqrt {\dfrac {1-cos\:A} {1+cos\:A}}[/tex]

input the value : cos x = −15/17

Because x in the quadrant III (180° < x < 270°) , then

  [tex]\rm \dfrac{1}{2}x\Rightarrow quadrant\:II[/tex]

For quadrant I, only sin have positif value, for cos and tan have negatif value

[tex]\rm sin\dfrac{1}{2}x=\sqrt{\dfrac{1-cos\:x}{2} }[/tex]

[tex]\rm sin\dfrac{1}{2}x=\sqrt{\dfrac{1-(-\dfrac{15}{17}) }{2} }\\\\sin\dfrac{1}{2}x=\sqrt{\dfrac{1\dfrac{15}{17} }{2} }=\sqrt{\dfrac{32}{54} }[/tex]

[tex]\rm cos\:\frac{1}{2}x=-\sqrt{\dfrac{1+cos\:x}{2} }\\\\cos\:\frac{1}{2}x=-\sqrt{\dfrac{1-\dfrac{15}{17} }{2} }=-\sqrt{\dfrac{2}{54} }[/tex]

[tex]\rm tan\dfrac {1} {2}x = -\sqrt {\dfrac {1-cos\:x} {1+cos\:x}}\\\\tan\dfrac {1} {2}x = -\sqrt {\dfrac {1+\frac{15}{17} } {1-\frac{15}{17} }}\\\\tan\dfrac{1}{2}x=-\sqrt{\dfrac{\dfrac{32}{17}}{\dfrac{2}{17} } }=-\sqrt{16}=-4[/tex]

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