<s>[INST] Using the half angle formula Find $\sin\frac { x }{ 2 }, \cos\frac { x }{ 2 }$, and $\tan\frac { x }{ 2 }$ from $\sin x=\frac { 3 }{ 5 } ,\quad 0°<x<90°$
What I did:
$$\sin\frac { x }{ 2 } =\sqrt { \frac { 1-\frac { 4 }{ 5 }  }{ 2 }  } =\sqrt { 10 } $$
$$\cos\frac { x }{ 2 } =\sqrt { \frac { 1+\frac { 4 }{ 5 }  }{ 2 }  } =\sqrt { \frac { 9 }{ 10 }  } =\frac { 3 }{ \sqrt { 10 }  } $$
$$\tan\frac { x }{ 2 } =\frac { \frac { 3 }{ 5 }  }{ 1+\frac { 4 }{ 5 }  } =\frac { \frac { 3 }{ 5 }  }{ \frac { 9 }{ 5 }  } =\frac { 1 }{ 3 } $$
The last one seems to be the only one that corresponds to the correct answer. The first two are wrong and I can't seem to find where I went wrong.
 [/INST] As $0<\dfrac x2<45^\circ\implies$ all the trigonometric ratios $>0$
$$\sin^2\frac x2=\frac{1-\cos x}2=\cdots=\frac1{10}\implies\sin\frac x2=+\frac1{\sqrt{10}}=\frac{\sqrt{10}}{10}$$
Similarly for $\cos\dfrac x2$
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