Question

prove that, cos nA.cos(n+2)A-cos^(2)(n+1)A+sin^(2)A=0​

Answer 1

$\large\underline{\bold{Given \:Question - }}$

Prove that

$\rm :\longmapsto\:cosnA \: cos(n + 2)A - {cos}^{2}(n + 1)A + {sin}^{2}A = 0$

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

Consider LHS,

$\rm :\longmapsto\:cosnA \: cos(n + 2)A - {cos}^{2}(n + 1)A + {sin}^{2}A$

can be rewritten as

$\rm = cosnA \: cos(n + 2)A - \bigg( {cos}^{2}(n + 1)A - {sin}^{2}A \bigg) - - (1)$

Let we simplify first,

$\red{\rm :\longmapsto\:\: {cos}^{2}(n + 1)A - {sin}^{2}A }$

$\rm \: = \: {cos}^{2}(nA+ A)- {sin}^{2}A$

We know that,

$\boxed{ \bf{ \: {cos}^{2}x - {sin}^{2}y = cos(x + y)cos(x - y)}}$

So, using this identity, we get

$\rm \: = \: \:cos(nA + A + A) \: cos(nA + A - A)$

$\rm \: = \: \:cos(nA + 2A) \: cos(nA)$

$\red{\rm \: = \: \:cos(n+ 2)A \: cos(nA)}$

So, equation (1) we have

$\rm = cosnA \: cos(n + 2)A - \bigg( {cos}^{2}(n + 1)A - {sin}^{2}A \bigg)$

On substituting the value, evaluated above, we get

$\rm = cosnA \: cos(n + 2)A - \red{cos(n + 2)AcosnA}$

$\rm = cosnA \: cos(n + 2)A - \red{cosnA \: cos(n + 2)A}$

$\rm \: = \: \:0$

Hence, Proved

Additional Information :-

$\boxed{ \bf{ \: cos(x + y) = cosxcosy - sinxsiny}}$

$\boxed{ \bf{ \: cos(x - y) = cosxcosy + sinxsiny}}$

$\boxed{ \bf{ \: sin(x + y) = sinxcosy + sinycosx}}$

$\boxed{ \bf{ \: sin(x - y) = sinxcosy - sinycosx}}$

$\boxed{ \bf{ \: tan(x + y) = \frac{tanx + tany}{1 - tanxtany}}}$

$\boxed{ \bf{ \: tan(x - y) = \frac{tanx - tany}{1 + tanxtany}}}$

$\boxed{ \bf{ \: {sin}^{2}x - {sin}^{2}y = sin(x + y)sin(x - y)}}$