Question

$\mathbb{PROVE QUESTION} \\ \\ \longrightarrow \rm \cos^{2} x + \cos ^{2} \bigg\{ x + \dfrac{\pi}{3} \bigg\} + \cos^{2} \bigg\{ x - \dfrac{\pi}{3} = 3/2.\ \textless \ br /\ \textgreater \ \ \textless \ br /\ \textgreater \ \mathbb{NOTE IRRELEVANT ANSWER WILL BE REPORTED}$
​

Answer 1

Answer

• ³/₂

Formula Used

$\rm cos2\theta+1=2cos^2\theta\\\\\bullet \;\; \rm cos^2\theta=\dfrac{cos2\theta+1}{2}\\\\\bullet \;\; \rm cosA+cosB=\tiny{2cos\left(\dfrac{A+B}{2}\right).cos\left(\dfrac{A-B}{2}\right)}\\\\\bullet \;\; \rm cos(\pi-\theta)=-sin\theta$

Solution

LHS

$\rm \to cos^2x+cos^2\left(x+\dfrac{\pi}{3}\right)+cos^2\left(x-\dfrac{\pi}{3}\right)$

$\rm \to \dfrac{cos2x+1}{2}+\dfrac{\tiny{cos2\left(x+\dfrac{\pi}{3}\right)+1}}{2}+\dfrac{\tiny{cos2\left(x-\dfrac{\pi}{3}\right)+1}}{2}\\\\\to \rm \dfrac{1}{2}\tiny{\left( 3+cos2x+cos\left(2x+\dfrac{2\pi}{3}\right)+cos\left(2x-\dfrac{2\pi}{3}\right)\right)}\\\\\to \rm \dfrac{1}{2} \tiny{\left(3+cos2x+2cos\left(\dfrac{2x+\dfrac{2\pi}{3}+2x-\dfrac{2\pi}{3}}{2}\right).cos\left(\dfrac{2x+\dfrac{2\pi}{3}-2x+\dfrac{2\pi}{3}}{2}\right)\right)}$

$\rm \to \dfrac{1}{2}\left(3+cos2x+2cos2x.(-sin30^0)\right)\\\\\rm \to \dfrac{1}{2}\left(3+cos2x+2cos2x.\dfrac{-1}{2}\right)\\\\\rm \to \dfrac{1}{2}\left(3+cos2x-cos2x\right)\\\\\rm \to \dfrac{3}{2}\\\\\rm \to RHS$

Since , LHS = RHS ,

Hence proved

Answer 2

Answer:

Answer

³/₂

Formula Used

\begin{gathered}\rm cos2\theta+1=2cos^2\theta\\\\\bullet \;\; \rm cos^2\theta=\dfrac{cos2\theta+1}{2}\\\\\bullet \;\; \rm cosA+cosB=\tiny{2cos(\dfrac{A+B}{2}).cos(\dfrac{A-B}{2})}\\\\\bullet \;\; \rm cos(\pi-\theta)=-sin\theta\end{gathered}

cos2θ+1=2cos

2

θ

∙cos

2

θ=

2

cos2θ+1

∙cosA+cosB=2cos(

2

A+B

).cos(

2

A−B

)

∙cos(π−θ)=−sinθ

Solution

LHS

\begin{gathered}\rm \to cos^2x+cos^2(x+\dfrac{\pi}{3})+cos^2(x-\dfrac{\pi}{3})\\\\\end{gathered}

→cos

2

x+cos

2

(x+

3

π

)+cos

2

(x−

3

π

)

\begin{gathered}\rm \to \dfrac{cos2x+1}{2}+\dfrac{\tiny{cos2(x+\dfrac{\pi}{3})+1}}{2}+\dfrac{\tiny{cos2(x-\dfrac{\pi}{3})+1}}{2}\\\\\to \rm \dfrac{1}{2}\tiny{( 3+cos2x+cos(2x+\dfrac{2\pi}{3})+cos(2x-\dfrac{2\pi}{3}))}\\\\\to \rm \dfrac{1}{2} \tiny{(3+cos2x+2cos(\dfrac{2x+\dfrac{2\pi}{3}+2x-\dfrac{2\pi}{3}}{2}).cos(\dfrac{2x+\dfrac{2\pi}{3}-2x+\dfrac{2\pi}{3}}{2}))}\\\\\end{gathered}

→

2

cos2x+1

+

2

cos2(x+

3

π

)+1

+

2

cos2(x−

3

π

)+1

→

2

1

(3+cos2x+cos(2x+

3

2π

)+cos(2x−

3

2π

))

→

2

1

(3+cos2x+2cos(

2

2x+

3

2π

+2x−

3

2π

).cos(

2

2x+

3

2π

−2x+

3

2π

))

\begin{gathered}\rm \to \dfrac{1}{2}(3+cos2x+2cos2x.(-sin30^0))\\\\\rm \to \dfrac{1}{2}(3+cos2x+2cos2x.\dfrac{-1}{2})\\\\\rm \to \dfrac{1}{2}(3+cos2x-cos2x)\\\\\rm \to \dfrac{3}{2}\\\\\rm \to RHS\end{gathered}

→

2

1

(3+cos2x+2cos2x.(−sin30

0

))

→

2

1

(3+cos2x+2cos2x.

2

−1

)

→

2

1

(3+cos2x−cos2x)

→

2

3

→RHS

Since , LHS = RHS ,

Hence proved

Step-by-step explanation:

mark as verfied answer please