The values of trigonometric functions sin(x/2) is [tex]\sqrt{\frac{7+4\sqrt{3}}{14}}[/tex], cos(x/2) is [tex]\frac{1}{\sqrt{14(7+4\sqrt{3})}}[/tex] and tan(x/2) is 7+4√3.
Given that the value of trigonometric function csc(x)=7 for 90°<x<180°.
We are given:
csc(x)=7 Where: 90° < x < 180°
This interval indicates that the angle x in the second quadrant and we know that at that quadrant sin(x) and CSC(x) functions are positive and all other trigonometric functions sign are negative.
Now:
sin(x)=1/csc(x)
sin(x)=1/7
Using the trigonometric identity sin²x+cos²x=1, we get
cosx=±√(1-sin²x)
cosx=±√(1-(1/7)²)
cosx=±√((49-1)/49)
cosx=±(4√3)/7
As x is in second quadrant.
Therefore, cosx=-(4√3)/7
consider the inequality, 90°<x<180°
Divide by 2, we get
45°<x/2<90°
This is the angle x/2 in the first quadrant and in that quadrant all functions sign are positive.
Substitute the value of cosx in half angle formula for sine function,
[tex]\begin{aligned}\cos x&=1-2\sin^2\left(\frac{x}{2}\right)\\ -\frac{4\sqrt{3}}{7}&=1-2\sin^2\left(\frac{x}{2}\right)\\ 2\sin^2 \frac{x}{2}&=1+\frac{4\sqrt{3}}{7}\\\sin \frac{x}{2}&=\pm\sqrt{\frac{7+4\sqrt{3}}{14}}\end[/tex]
As x/2 lies in first quadrant then [tex]\sin \frac{x}{2}=\sqrt{\frac{7+4\sqrt{3}}{14}}[/tex]
Again, Substitute the value of cosx in half angle formula for sine function,[tex]\begin{aligned}\sin x&=2\sin\left(\frac{x}{2}\right)\cos\frac{x}{2}\\ \frac{1}{7}&=2\sqrt{\frac{7+4\sqrt{3}}{14}}\cos\left(\frac{x}{2}\right)\\ \cos \frac{x}{2}&=\frac{1}{14\times\frac{\sqrt{7+4\sqrt{3}}}{\sqrt{14}}}\\\cos \frac{x}{2}&=\frac{1}{\sqrt{14(7+4\sqrt{3})}}\end[/tex]
Now find tan(x/2) as shown below, we get
[tex]\begin{aligned}\tan\frac{x}{2}&=\frac{\sin\frac{x}{2}}{\cos \frac{x}{2}}\\&=\frac{\sqrt{7+4\sqrt{3}}}{\sqrt{14}}\times \frac{\sqrt{14}\sqrt{7+4\sqrt{3}}}{1}\\&=7+4\sqrt{3}\end[/tex]
Hence, when trigonometric function csc(x)=7 for 90°<x<180° then sin(x/2) is [tex]\sqrt{\frac{7+4\sqrt{3}}{14}}[/tex], cos(x/2) is [tex]\frac{1}{\sqrt{14(7+4\sqrt{3})}}[/tex] and tan(x/2) is 7+4√3.
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