Question

Find the value of :-

$\sf\dfrac{cos^3\beta}{cos\alpha}+\dfrac{sin^3\beta}{sin\alpha}$. If :-

$\sf\dfrac{cos\alpha}{cos\beta}+\dfrac{sin\alpha}{sin\beta}= -1$​

Answer 1

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

We have with us to find the value of

$\red{\rm :\longmapsto\:\sf\dfrac{cos^3\beta}{cos\alpha}+\dfrac{sin^3\beta}{sin\alpha}}$

can be rewritten as

$\rm \: = \:\sf\dfrac{4cos^3\beta}{4cos\alpha}+\dfrac{4sin^3\beta}{4sin\alpha}$

We know,

$\red{\rm :\longmapsto\:cos3x = {4cos}^{3}x - 3cosx} \\ \red{\rm :\longmapsto\:it \: means} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \red{\rm :\longmapsto\: {4cos}^{3}x = cos3x + 3cosx}$

Also,

$\red{\rm :\longmapsto\: {4sin}^{3}x = 3sinx - sin3x}$

So, using this, we get

$\rm \: = \:\sf\dfrac{cos3 \beta + 3cos \beta }{4cos\alpha}+\dfrac{3sin \beta - sin3 \beta }{4sin\alpha}$

$\rm \: = \:\dfrac{cos3 \beta sin \alpha + 3cos \beta sin \alpha + 3sin \beta cos \alpha - sin3 \beta cos \alpha }{4sin \alpha cos \alpha }$

$\rm \: = \:\dfrac{cos3 \beta sin \alpha- sin3 \beta cos \alpha + 3(cos \beta sin \alpha + sin \beta cos \alpha)}{2(2sin \alpha cos \alpha) }$

We know,

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

and

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

and

$\boxed{ \bf{ \: sin2x = 2sinx \: cosx}}$

So, using these,

$\rm \: = \:\dfrac{sin( - 3\beta + \alpha ) + 3sin( \alpha + \beta )}{2sin2 \alpha }$

$\rm \: = \:\dfrac{sin( \alpha - 3\beta ) - sin( \alpha + \beta ) + sin( \alpha + \beta )+ 3sin( \alpha + \beta )}{2sin2 \alpha }$

$\rm \: = \:\dfrac{2cos\bigg[\dfrac{ \alpha - 3\beta + \alpha + \beta }{2} \bigg]sin\bigg[\dfrac{\alpha - 3\beta - \alpha - \beta }{2} \bigg] + 4sin( \alpha + \beta )}{2sin2 \alpha }$

$\rm \: = \:\dfrac{2cos( \alpha - \beta )sin( - 2 \beta ) + 4sin( \alpha + \beta )}{2sin2 \alpha }$

$\purple{\bf \: = \:\dfrac{ - \: 2cos( \alpha - \beta )sin 2 \beta + 4sin( \alpha + \beta )}{2sin2 \alpha } - - - (1)}$

Now, Consider

$\rm :\longmapsto\:\sf\dfrac{cos\alpha}{cos\beta}+\dfrac{sin\alpha}{sin\beta}= -1$

$\rm :\longmapsto\:\dfrac{cos \alpha sin \beta + sin \alpha cos \beta }{sin \beta cos \beta } = - 1$

$\rm :\longmapsto\:\dfrac{sin( \alpha + \beta) }{sin \beta cos \beta } = - 1$

$\rm :\longmapsto\:sin( \alpha + \beta ) = - sin \beta cos \beta$

$\rm :\longmapsto\:2sin( \alpha + \beta ) = - 2 sin \beta cos \beta$

$\rm :\longmapsto\:2sin( \alpha + \beta ) = - sin2 \beta$

So, equation (1), can be rewritten as

$\rm \: = \:\dfrac{ 4\: cos( \alpha - \beta )sin( \alpha + \beta ) + 4sin( \alpha + \beta )}{2sin2 \alpha }$

$\rm \: = \:\dfrac{ (\: cos( \alpha - \beta ) + 1)4sin( \alpha + \beta )}{2sin2 \alpha }$

$\purple{\bf \: = \:\dfrac{ 2(\: cos( \alpha - \beta ) + 1)sin( \alpha + \beta )}{sin2 \alpha } - - (2)}$

Now, we know that

$\rm :\longmapsto\:sin2 \alpha + sin2 \beta = 2sin( \alpha + \beta )cos( \alpha - \beta )$

$\rm :\longmapsto\:sin2 \alpha + sin2 \beta = - sin2 \beta cos( \alpha - \beta )$

$\red{\bigg \{ \because \: - sin2 \beta = 2sin( \alpha + \beta ) \bigg \}}$

$\rm :\longmapsto\:sin2 \alpha = - sin2 \beta - sin2 \beta cos( \alpha - \beta )$

$\rm :\longmapsto\:sin2 \alpha = - sin2 \beta (1 + cos( \alpha - \beta ))$

$\bf :\longmapsto\:sin2 \alpha = 2 sin( \alpha + \beta) (1 + cos( \alpha - \beta )) - - - (3)$

$\red{\bigg \{ \because \: - sin2 \beta = 2sin( \alpha + \beta ) \bigg \}}$

On substituting equation (3) in (2), we get

$\rm \: = \:\dfrac{ sin2 \alpha }{sin2 \alpha }$

$\rm \: = \:1$

Hence,

$\red{\bf :\longmapsto\:\bf\dfrac{cos^3\beta}{cos\alpha}+\dfrac{sin^3\beta}{sin\alpha} = 1}$

Answer 2

Step-by-step explanation:

Solution−

We have with us to find the value of

\red{\rm :\longmapsto\:\sf\dfrac{cos^3\beta}{cos\alpha}+\dfrac{sin^3\beta}{sin\alpha}}:⟼

cosα

cos

3

β

+

sinα

sin

3

β

can be rewritten as

\rm \: = \:\sf\dfrac{4cos^3\beta}{4cos\alpha}+\dfrac{4sin^3\beta}{4sin\alpha}=

4cosα

4cos

3

β

+

4sinα

4sin

3

β

We know,

\begin{gathered}\red{\rm :\longmapsto\:cos3x = {4cos}^{3}x - 3cosx} \\ \red{\rm :\longmapsto\:it \: means} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \red{\rm :\longmapsto\: {4cos}^{3}x = cos3x + 3cosx}\end{gathered}

:⟼cos3x=4cos

3

x−3cosx

:⟼itmeans

:⟼4cos

3

x=cos3x+3cosx

Also,

\red{\rm :\longmapsto\: {4sin}^{3}x = 3sinx - sin3x}:⟼4sin

3

x=3sinx−sin3x

So, using this, we get

\rm \: = \:\sf\dfrac{cos3 \beta + 3cos \beta }{4cos\alpha}+\dfrac{3sin \beta - sin3 \beta }{4sin\alpha}=

4cosα

cos3β+3cosβ

+

4sinα

3sinβ−sin3β

\rm \: = \:\dfrac{cos3 \beta sin \alpha + 3cos \beta sin \alpha + 3sin \beta cos \alpha - sin3 \beta cos \alpha }{4sin \alpha cos \alpha }=

4sinαcosα

cos3βsinα+3cosβsinα+3sinβcosα−sin3βcosα

\rm \: = \:\dfrac{cos3 \beta sin \alpha- sin3 \beta cos \alpha + 3(cos \beta sin \alpha + sin \beta cos \alpha)}{2(2sin \alpha cos \alpha) }