Question

please answer this question​

Answer 1

TO FIND :–

$\\ \dashrightarrow \: { \bold{ \sin^{6} ( \theta) + \cos^{6} ( \theta) = 1 - 3 { \sin}^{2} ( \theta) { \cos}^{2}( \theta) }}$

SOLUTION :–

• Let's take L.H.S. –

$\\ \: \: = \: \: { \bold{ \sin^{6} ( \theta) + \cos^{6} ( \theta) }}$

• We should write this as –

$\\ \: \: = \: \: { \bold{ \{ \sin^{2} ( \theta) \}^{3} + \{ \cos^{2} ( \theta) \}^{3}}}$

• Using identity –

$\\ \dashrightarrow \:{ \bold{ {(a + b)}^{3} = \: \: a^{3} +b^{3} + 3ab(a + b)}}$

$\\ \dashrightarrow \:{ \bold{ a^{3} +b^{3} = {(a + b)}^{3} - 3ab(a + b)}}$

• So that –

$\\ \: \: { \bold{ = \: \: \{ \sin^{2} ( \theta) + \cos^{2} ( \theta) \}^{3} - 3 \: { \sin}^{2} \theta. { \cos}^{2} \theta ( \sin^{2} \theta + \cos^{2} \theta)}}$

$\\ \: \: \because \: \: { \bold{ \sin^{2} ( \theta) + \cos^{2} ( \theta) = 1}}$

$\\ \: \: \therefore \: \: { \bold{ = \: \: \{1 \}^{3} - 3 \: { \sin}^{2} \theta. { \cos}^{2} \theta (1)}}$

$\\ \: \: { \bold{ = \: \: 1- 3 \: { \sin}^{2} \theta { \cos}^{2} \theta }}$

$\\ \: \: { \bold{ = \: \:R.H.S. }}$

$\\ \: \: \: \: \: \: { \underbrace{ \bold{Hence \: \: proved}}}$

Answer 2

Answer:

SOLUTION :–

• Let's take L.H.S. –

\begin{gathered} \\ \: \: = \: \: { \bold{ \sin^{6} ( \theta) + \cos^{6} ( \theta) }} \\ \end{gathered}

=sin

6

(θ)+cos

6

(θ)

• We should write this as –

\begin{gathered} \\ \: \: = \: \: { \bold{ \{ \sin^{2} ( \theta) \}^{3} + \{ \cos^{2} ( \theta) \}^{3}}} \\ \end{gathered}

={sin

2

(θ)}

3

+{cos

2

(θ)}

3

• Using identity –

\begin{gathered} \\ \dashrightarrow \:{ \bold{ {(a + b)}^{3} = \: \: a^{3} +b^{3} + 3ab(a + b)}} \\ \end{gathered}

⇢(a+b)

3

=a

3

+b

3

+3ab(a+b)

\begin{gathered} \\ \dashrightarrow \:{ \bold{ a^{3} +b^{3} = {(a + b)}^{3} - 3ab(a + b)}} \\ \end{gathered}

⇢a

3

+b

3

=(a+b)

3

−3ab(a+b)

• So that –

\begin{gathered} \\ \: \: { \bold{ = \: \: \{ \sin^{2} ( \theta) + \cos^{2} ( \theta) \}^{3} - 3 \: { \sin}^{2} \theta. { \cos}^{2} \theta ( \sin^{2} \theta + \cos^{2} \theta)}} \\ \end{gathered}

={sin

2

(θ)+cos

2

(θ)}

3

−3sin

2

θ.cos

2

θ(sin

2

θ+cos

2

θ)

\begin{gathered} \\ \: \: \because \: \: { \bold{ \sin^{2} ( \theta) + \cos^{2} ( \theta) = 1}} \\ \end{gathered}

∵sin

2

(θ)+cos

2

(θ)=1

\begin{gathered} \\ \: \: \therefore \: \: { \bold{ = \: \: \{1 \}^{3} - 3 \: { \sin}^{2} \theta. { \cos}^{2} \theta (1)}} \\ \end{gathered}

∴={1}

3

−3sin

2

θ.cos

2

θ(1)

\begin{gathered} \\ \: \: { \bold{ = \: \: 1- 3 \: { \sin}^{2} \theta { \cos}^{2} \theta }} \\ \end{gathered}

=1−3sin

2

θcos

2

θ

\begin{gathered} \\ \: \: { \bold{ = \: \:R.H.S. }} \\ \end{gathered}

=R.H.S.