How do you prove that ${sin}^{2} (\frac{x}{2}) = \frac{{sin}^{2} x}{2 (1 + cos x)}$ $sin^2(x/2) = (sin^2x)/(2(1+cosx))$ ?

Question

How do you prove that ${sin}^{2} (\frac{x}{2}) = \frac{{sin}^{2} x}{2 (1 + cos x)}$ $sin^2(x/2) = (sin^2x)/(2(1+cosx))$ ?

1 Answer

Impact of this question

10393 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-18T19:04:09

We know that

$cos 2 x = {cos}^{2} x - {sin}^{2} x = 1 - 2 {sin}^{2} x$ $cos2x=cos^2x-sin^2x=1-2sin^2x$

Hence

${sin}^{2} x = \frac{1}{2} \cdot (1 - cos 2 x)$ $sin^2x=1/2*(1-cos2x)$

Set $x \to \frac{t}{2}$ $x->t/2$ so we have

${sin}^{2} (\frac{t}{2}) = \frac{1}{2} \cdot (1 - cos t)$ $sin^2(t/2)=1/2*(1-cost)$

Multiply and divide the LHS with $1 + cos t$ $1+cost$ hence

${sin}^{2} (\frac{t}{2}) = \frac{1}{2} \cdot [\frac{(1 - cos t) \cdot (1 + cos t)}{1 + cos t}] \Rightarrow {sin}^{2} (\frac{t}{2}) = \frac{1}{2} \cdot [\frac{1 - {cos}^{2} t}{1 + cos t}] \Rightarrow {sin}^{2} (\frac{t}{2}) = \frac{1}{2} \cdot [\frac{{sin}^{2} t}{1 + cos t}]$ $sin^2(t/2)=1/2*[((1-cost)*(1+cost))/(1+cost)]=> sin^2(t/2)=1/2*[(1-cos^2t)/(1+cost)]=> sin^2(t/2)=1/2*[sin^2t/(1+cost)]$

Which proves the requested

Science

Math

Humanities

... and beyond

How do you prove that ${sin}^{2} (\frac{x}{2}) = \frac{{sin}^{2} x}{2 (1 + cos x)}$ $sin^2(x/2) = (sin^2x)/(2(1+cosx))$ ?

1 Answer

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you prove that sin2(x2)=sin2x2(1+cosx) sin^2(x/2) = (sin^2x)/(2(1+cosx)) ?

1 Answer

Related questions

Impact of this question

How do you prove that ${sin}^{2} (\frac{x}{2}) = \frac{{sin}^{2} x}{2 (1 + cos x)}$ $sin^2(x/2) = (sin^2x)/(2(1+cosx))$ ?