Question

cos(a/2)=+/-sqrt1+cosa/2

Answer 1

Question :-

Prove that

$\rm\:{cos}\dfrac{a}{2}\:=\:\pm \:\sqrt{\dfrac{1 + cosa}{2}}$

$\large\underline{\sf{Solution-}}$

We know that,

$\rm \: cos(x + y) = cosx \: cosy \: - \: sinx \: siny$

Now, Replace x = y = a/2, we get

$\rm \: cos\bigg(\dfrac{a}{2}+\dfrac{a}{2}\bigg) = cos\dfrac{a}{2}\times cos\dfrac{a}{2}\: - \: sin\dfrac{a}{2} \times sin\dfrac{a}{2}$

$\rm \: cosa \: = \: {cos}^{2}\dfrac{a}{2} \: - \: {sin}^{2}\dfrac{a}{2}$

We know,

$\boxed{\tt{ \: \: {sin}^{2}x + {cos}^{2}x = 1 \: \: }}$

So, using this, we get

$\rm \: cosa \: = \: {cos}^{2}\dfrac{a}{2} \: - \: \bigg(1 \: - \: {cos}^{2}\dfrac{a}{2}\bigg)$

$\rm \: cosa \: = \: 2{cos}^{2}\dfrac{a}{2} \: - \:1$

$\rm \: cosa \: + \: 1 \: = \: 2{cos}^{2}\dfrac{a}{2}$

$\rm \: {cos}^{2}\dfrac{a}{2} \: = \: \dfrac{1 + cosa}{2}$

$\rm\implies \:\boxed{\tt{ \: \rm \: {cos}\dfrac{a}{2} \: = \: \pm \: \sqrt{\dfrac{1 + cosa}{2}} \: \: }}$

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ADDITIONAL INFORMATION

$\boxed{\tt{ \: sin2x = 2sinxcosy = \frac{2tanx}{1 + {tan}^{2}x} \: }}$

$\boxed{\tt{ cos2x = {cos}^{2}x - {sin}^{2}x = 1 - {2sin}^{2}x = {2cos}^{2}x - 1 \: }}$

$\boxed{\tt{ \: tan2x = \frac{2tanx}{1 - {tan}^{2}x } \: }}$

$\boxed{\tt{ \: sin3x = 3sinx - {4sin}^{3}x \: }}$

$\boxed{\tt{ \: cos3x = {4cos}^{3}x - 3cosx \: }}$

$\boxed{\tt{ \: tan3x = \frac{3tanx - {tan}^{3}x}{1 - {3tan}^{2}x } \: }}$

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Function & \bf Range \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf y = sinx & \sf - 1 \leqslant y \leqslant 1\\ \\ \sf y = cosx & \sf - 1 \leqslant y \leqslant 1 \\ \\ \sf y = tanx & \sf y \: \in \: ( - \infty , \infty )\\ \\ \sf y = cosec & \sf y \leqslant - 1 \: \: or \: \: y \geqslant 1\\ \\ \sf y = secx & \sf y \leqslant - 1 \: \: or \: \: y \geqslant 1\\ \\ \sf y = cotx & \sf y \: \in \: ( - \infty , \infty ) \end{array}} \\ \end{gathered}$

Answer 2

refer the given attachment