Answer:
Hence, the value of:
[tex]\sin \dfrac{x}{2}=\sqrt\dfrac{1}{26}[/tex]
Step-by-step explanation:
We are given a trignometric formula for the given angle 'x' as:
[tex]\sin x=\dfrac{5}{13}[/tex]
So, we consider a right triangle such that it's perpendicular side is 5 units and the hypotenuse is 13 units.
Hence, we get the third side of the triangle i.e. CB using the Pythagorean Theorem as:
[tex]AB^2=CB^2+AC^2\\\\13^2=CB^2+5^2\\\\169=CB^2+25\\\\CB^2=169-25\\\\CB^2=144\\\\CB=12\ units[/tex]
Hence,
[tex]\cos x=\dfrac{CB}{AB}\\\\\cos x=\dfrac{12}{13}[/tex]
Now, we know that:
[tex]\cos x=1-2\sin^2 \dfrac{x}{2}\\\\2\sin^2 \dfrac{x}{2}=1-\cos x\\\\sin^2 \dfrac{x}{2}=\dfrac{1-\cos x}{2}\\\\\sin^2 \dfrac{x}{2}=\dfrac{1-\dfrac{12}{13}}{2}\\\\\sin^2 \dfrac{x}{2}=\dfrac{13-12}{2\times 13}\\\\\\\sin^2 \dfrac{x}{2}=\dfrac{1}{26}\\\\\sin \dfrac{x}{2}=\sqrt{\dfrac{1}{26}}[/tex]
