Question

Prove that
cos²x + cos² (x+Π/3) + cos² (x - Π/3) = 3/2​

Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept}}}$

Here the concept of Trignometric Identities have been used. We see we are given the equation. Now we shall seperate the equation and find the value of different terms. Then we shall apply it together to find the answer.

Let's do it !!

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★ Solution :-

$\\\;\bf{\mapsto\;\;\green{\cos^{2}x\;+\;\cos^{2}\bigg(x\;+\;\dfrac{\pi}{3}\bigg)\;+\;\cos^{2}\bigg(x\;-\;\dfrac{\pi}{3}\bigg)\;=\;\dfrac{3}{2}}}$

Here,

$\\\;\tt{\odot\;\;L.H.S.\;=\;\cos^{2}x\;+\;\cos^{2}\bigg(x\;+\;\dfrac{\pi}{3}\bigg)\;+\;\cos^{2}\bigg(x\;-\;\dfrac{\pi}{3}\bigg)}$

$\\\;\tt{\odot\;\;R.H.S.\;=\;\dfrac{3}{2}}$

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~ For the value of different terms ::

We know that,

$\\\;\sf{\leadsto\;\;\cos\:2x\;=\;2\cos^{2}x\;-\;1}$

Then we get,

$\\\;\sf{\rightarrow\;\;\cos^{2}x\;=\;\bf{\red{\dfrac{\cos\:2x\;+\;1}{2}}}}$

Also,

$\\\;\sf{\rightarrow\;\;\cos^{2}\bigg(x\;+\;\dfrac{\pi}{3}\bigg)\;=\;\bf{\pink{\dfrac{\cos\:2\bigg(x\:+\:\dfrac{\pi}{3}\bigg)\;+\;1}{2}}}}$

And,

$\\\;\sf{\rightarrow\;\;\cos^{2}\bigg(x\;-\;\dfrac{\pi}{3}\bigg)\;=\;\bf{\orange{\dfrac{\cos\:2\bigg(x\:-\:\dfrac{\pi}{3}\bigg)\;+\;1}{2}}}}$

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~ For proving the equation ::

We have,

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{\cos^{2}x\;+\;\cos^{2}\bigg(x\;+\;\dfrac{\pi}{3}\bigg)\;+\;\cos^{2}\bigg(x\;-\;\dfrac{\pi}{3}\bigg)}}$

By applying the first value that we got, we get

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{\bigg(\dfrac{\cos\:2x\;+\;1}{2}\bigg)\;+\;\bigg(\cos^{2}\bigg(x\;+\;\dfrac{\pi}{3}\bigg)\bigg)\;+\;\bigg(\cos^{2}\bigg(x\;-\;\dfrac{\pi}{3}\bigg)\bigg)}}$

Now applying the second value that we got, we get

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{\bigg(\dfrac{\cos\:2x\;+\;1}{2}\bigg)\;+\;\bigg(\dfrac{\cos\:2\bigg(x\:+\:\dfrac{\pi}{3}\bigg)\;+\;1}{2}\bigg)\;+\;\bigg(\cos^{2}\bigg(x\;-\;\dfrac{\pi}{3}\bigg)\bigg)}}$

Now applying the third value that we got, we get

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{\bigg(\dfrac{\cos\:2x\;+\;1}{2}\bigg)\;+\;\bigg(\dfrac{\cos\:2\bigg(x\:+\:\dfrac{\pi}{3}\bigg)\;+\;1}{2}\bigg)\;+\;\bigg(\dfrac{\cos\:2\bigg(x\:-\:\dfrac{\pi}{3}\bigg)\;+\;1}{2}\bigg)}}$

Now multiplying two in respective terms, we get

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{\bigg(\dfrac{\cos\:2x\;+\;1}{2}\bigg)\;+\;\bigg(\dfrac{\cos\:\bigg(2x\:+\:\dfrac{2\pi}{3}\bigg)\;+\;1}{2}\bigg)\;+\;\bigg(\dfrac{\cos\bigg(2x\:-\:\dfrac{2\pi}{3}\bigg)\;+\;1}{2}\bigg)}}$

Now taking out ½ as common, we get

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{\dfrac{1}{2}\;\bigg[1\;+\;1\;+\;1\;+\;\cos\:2x\;+\;\cos\bigg(2x\;+\;\dfrac{2\pi}{3}\bigg)\;+\;\cos\bigg(2x\;-\;\dfrac{2\pi}{3}\bigg)\bigg]}}$

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{\dfrac{1}{2}\;\bigg[1\;+\;1\;+\;1\;+\;\cos\:2x\;+\;\cos\bigg(2x\;+\;\dfrac{2\pi}{3}\bigg)\;+\;\cos\bigg(2x\;-\;\dfrac{2\pi}{3}\bigg)\bigg]}}$

$\\\sf{L.H.S.\:=\:\bf{\dfrac{1}{2}\bigg[3+\cos 2x+2\cos\bigg(\dfrac{2x+2x+\dfrac{2\pi}{3}-\dfrac{2\pi}{3}}{2}\bigg).\cos\bigg(\dfrac{2x-2x+\dfrac{2\pi}{3}+\dfrac{2\pi}{3}}{2}\bigg)\bigg]}}$

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{\dfrac{1}{2}\;\bigg[3\;+\;\cos\:2x\;+\;2\cos\:2x.\cos\dfrac{2\pi}{3}\bigg]}}$

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{\dfrac{1}{2}\;\bigg[3\;+\;\cos\:2x\;+\;2\cos\:2x.\bigg(-\:\cos\dfrac{\pi}{3}\bigg)\bigg]}}$

This will give us,

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{\dfrac{1}{2}\;\bigg[3\;+\;\cos\:2x\;+\;\bigg(-\:\cos\:2x\bigg)\bigg]}}$

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{\dfrac{1}{2}\;\bigg[3\;+\;\cos\:2x\;-\;\cos\:2x\bigg]}}$

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{\dfrac{1}{2}\;\bigg[3\;+\;0\bigg]}}$

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{\dfrac{1}{2}\;\bigg[3\bigg]}}$

$\\\;\bf{\Longrightarrow\;\;L.H.S.\;=\;\bf{\blue{\dfrac{3}{2}}}}$

Also we know that,

$\\\;\bf{\Longrightarrow\;\;R.H.S.\;=\;\dfrac{3}{2}}$

Combining both, we get

$\\\;\bf{\Longrightarrow\;\;L.H.S.\;=\;R.H.S.\;=\;\bf{\blue{\dfrac{3}{2}}}}$

This gives us the answer. So its proved.

$\\\;\qquad\qquad\underline{\boxed{\tt{\purple{Hence,\;\;Proved}}}}$

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★ More to know :-

$\\\;\sf{\leadsto\;\;\sin(-\:x)\;=\;-\:\sin\:x}$

$\\\;\sf{\leadsto\;\;\cos(-\:x)\;=\;\cos\:x}$

$\\\;\sf{\leadsto\;\;\cos(x\:+\:y)\;=\;\cos\:x\cos\:y\;-\;\sin\:x\:\sin\:y}$

$\\\;\sf{\leadsto\;\;\cos(x\:-\:y)\;=\;\cos\:x\cos\:y\;+\;\sin\:x\:\sin\:y}$

$\\\;\sf{\leadsto\;\;\sin(x\:+\:y)\;=\;\sin\:x\cos\:y\;+\;\cos\:x\:\sin\:y}$

$\\\;\sf{\leadsto\;\;\sin(x\:-\:y)\;=\;\sin\:x\cos\:y\;-\;\cos\:x\:\sin\:y}$

Answer 2

Question:-

Prove : cos²x+cos²(x+π/3)+cos²(x-π/3)=3/2

Answer:-⤵️

☆We know that, cos2x = 2cos²x - 1

→ 1 + cos2x = 2cos²x

→ cos²x = (1+cos2x)/2

Using this :

→ (1+cos2x)/2 + [1+cos2(x+π/3)]/2 + [1+cos2(x-π/3)]/2

→ ½[3 + cos2x + cos2(x+π/3) + cos2(x-π/3)]

→ ½[3 + cos2x + cos(2x+2π/3) + cos(2x-2π/3)]

☆We know : cos(A+B) + cos(A-B) = 2 cosA cosB

→ ½[3 + cos2x + 2 cos2x cos2π/3]

→ ½[3 + cos2x + 2 cos2x cos(π-π/3)]

→ ½[3 + cos2x + 2 cos2x (-cosπ/3)]

→ ½[3 + cos2x - ½×2 cos2x]

→ ½[3 + cos2x - cos2x]

→ ½ × 3

→ 3/2 (RHS)

Hence Proved ✔