Question

sin²x+sin²(x+π/3)+sin²(x-π/3)
Giải bằng 2 cách

Answer 1

Given Question

Evaluate the following

$\rm :\longmapsto\: {sin}^{2}x + {sin}^{2}\bigg[x + \dfrac{\pi}{3} \bigg] + {sin}^{2} \bigg[x - \dfrac{\pi}{3} \bigg]$

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

Given Trigonometric function is

$\rm :\longmapsto\: {sin}^{2}x + {sin}^{2}\bigg[x + \dfrac{\pi}{3} \bigg] + {sin}^{2} \bigg[x - \dfrac{\pi}{3} \bigg]$

Let first evaluate

$\rm :\longmapsto\:sin\bigg[x + \dfrac{\pi}{3} \bigg]$

$\rm \:  =  \: sinx \: cos\dfrac{\pi}{3} + cosx \: sin\dfrac{\pi}{3}$

$\rm \:  =  \: \dfrac{1}{2}sinx + \dfrac{ \sqrt{3} }{2}cosx$

Now, Consider

$\rm :\longmapsto\:sin\bigg[x - \dfrac{\pi}{3} \bigg]$

$\rm \:  =  \: sinx \: cos\dfrac{\pi}{3} - cosx \: sin\dfrac{\pi}{3}$

$\rm \:  =  \: \dfrac{1}{2}sinx - \dfrac{ \sqrt{3} }{2}cosx$

Now, Consider

$\rm :\longmapsto\: {sin}^{2}x + {sin}^{2}\bigg[x + \dfrac{\pi}{3} \bigg] + {sin}^{2} \bigg[x - \dfrac{\pi}{3} \bigg]$

$\rm = {sin}^{2}x + {\bigg[\dfrac{1}{2}sinx + \dfrac{ \sqrt{3} }{2} cosx\bigg]}^{2} + {\bigg[\dfrac{1}{2}sinx - \dfrac{ \sqrt{3} }{2}cosx\bigg]}^{2}$

We know,

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

So, using this, we get

$\rm =  {sin}^{2}x + 2\bigg[\dfrac{1}{4} {sin}^{2}x + \dfrac{3}{4} {cos}^{2}x\bigg]$

$\rm =  {sin}^{2}x + \dfrac{1}{2} {sin}^{2}x + \dfrac{3}{2} {cos}^{2}x$

$\rm = \dfrac{3}{2} {sin}^{2}x + \dfrac{3}{2} {cos}^{2}x$

$\rm = \dfrac{3}{2} \bigg[{sin}^{2}x + {cos}^{2}x\bigg]$

$\rm \:  =  \: \dfrac{3}{2}$

Hence,

$\rm :\longmapsto\: \boxed{ \tt{ \: {sin}^{2}x + {sin}^{2}\bigg[x + \dfrac{\pi}{3} \bigg] + {sin}^{2} \bigg[x - \dfrac{\pi}{3} \bigg] = \frac{3}{2} \: }}$

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Learn More :-

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

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

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

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

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

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