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Verified Answer :

LammettHash

Because \(180^\circ, you should expect \(90^\circ<\dfrac x2<135^\circ\), so that \(\dfrac x2\) is an angle in the second quadrant. In this quadrant, you have \(\sin\dfrac x2>0\), \(\cos\dfrac x2<0\), and \(\tan\dfrac x2<0\).

Recall the half-angle identity for sine and cosine:

\(\sin^2\dfrac x2=\dfrac{1-\cos x}2\implies \sin\dfrac x2=\pm\sqrt{\dfrac{1-\cos x}2}\)
\(\cos^2\dfrac x2=\dfrac{1+\cos x}2\implies\cos\dfrac x2=\pm\sqrt{\dfrac{1+\cos x}2}\)

From these, you can derive the tangent half-angle identity, but that won't be needed, since you can just rely on the definition for tangent, \(\tan x=\dfrac{\sin x}{\cos x}\).

You know which sign you expect these expression to have from the given domain, so

\(\sin\dfrac x2=\sqrt{\dfrac{1-\left(-\frac{12}{13}\right)}2}=\dfrac5{\sqrt{26}}\)
\(\cos\dfrac x2=-\sqrt{\dfrac{1+\left(-\frac{12}{13}\right)}2}=-\dfrac1{\sqrt{26}}\)

So,

\(\tan\dfrac x2=\dfrac{\frac5{\sqrt{26}}}{-\frac1{\sqrt{26}}}=-5\)

ardni313

\(\rm sin\dfrac{1}{2}x=\sqrt{\dfrac{25}{26}}\\\\cos\dfrac{1}{2}x=-\sqrt{\dfrac{1}{26}}\\\\tan\dfrac{1}{2}x=-5\)

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  

\(\rm cos\:A=2cos^2\dfrac{1}{2}A-1\\\\2cos^2\frac{1}{2}A=1+cos\:A\\\\cos^2\frac{1}{2}A=\dfrac{1+cos\:A}{2}\\\\cos\:\frac{1}{2}A=\sqrt{\dfrac{1+cos\:A}{2} }\)

cos 2A = 1-2 sin²A  

in the same way

\(\rm sin\dfrac{1}{2}A=\sqrt{\dfrac{1-cos\:A}{2} }\)

\(\rm tan\dfrac {1} {2}A = \sqrt {\dfrac {1-cos\:A} {1+cos\:A}}\)

input the value : cos x = −15/17

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

 \(\rm \dfrac{1}{2}x\Rightarrow quadrant\:II\)

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

\(\rm sin\dfrac{1}{2}x=\sqrt{\dfrac{1-cos\:x}{2} }\)  

\(\rm sin\dfrac{1}{2}x=\sqrt{\dfrac{1-(-\dfrac{12}{13}) }{2} }\\\\sin\dfrac{1}{2}x=\sqrt{\dfrac{1\dfrac{12}{13} }{2} }=\sqrt{\dfrac{25}{26} }\)  

\(\rm cos\:\frac{1}{2}x=-\sqrt{\dfrac{1+cos\:x}{2} }\\\\cos\:\frac{1}{2}x=-\sqrt{\dfrac{1-\dfrac{12}{13} }{2} }=-\sqrt{\dfrac{1}{26} }\)  

\(\rm tan\dfrac {1} {2}x =-\sqrt {\dfrac {1-cos\:x} {1+cos\:x}}\\\\tan\dfrac {1} {2}x =-\sqrt {\dfrac {1+\frac{12}{13} } {1-\frac{12}{13} }}\\\\tan\dfrac{1}{2}x=-\sqrt{\dfrac{\dfrac{25}{13}}{\dfrac{1}{13} } }=-\sqrt{25}=-5\)

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