Prove thatcos 2 theta + cos² ( theta +120° ) + cos ² ( theta -120°

1.	Prove thatcos 2 theta + cos² ( theta +120° ) + cos ² ( theta -120°) = ⅔
Answer» Correct Statement is $\sf \: {cos}^{2}\theta + {cos}^{2}(\theta + 120\degree) + {cos}^{2}(\theta - 120\degree) = \dfrac{3}{2}$ $\large\underline{\sf{Solution-}}$ $\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}} \end{gathered}$ $\boxed{ \sf \: {cos}^{2}x = \dfrac{1 + cos2x}{2}}$ $\boxed{ \sf \: cos(x + y) + cos(x - y) = 2cosxcosy}$ $\boxed{ \sf \: cos(\pi \: + x) = - cosx}$ Let's solve the problem now!! Consider, $\sf \: {cos}^{2}\theta + {cos}^{2}(\theta + 120\degree) + {cos}^{2}(\theta - 120\degree)$ $\sf \: = \: \dfrac{1 + cos2\theta}{2} + \dfrac{1 + cos(2\theta + 240\degree)}{2} + \dfrac{1 + cos(2\theta - 240\degree)}{2}$ $\sf \: = \: \dfrac{1}{2}\bigg(3 + cos2\theta + cos(2\theta + 240\degree) + cos(2\theta - 240\degree) \bigg)$ $\sf \: = \: \dfrac{1}{2}\bigg(3 + cos2\theta + 2cos2\theta \: cos240\degree \bigg)$ $\sf \: = \: \dfrac{1}{2}\bigg(3 + cos2\theta + 2cos2\theta \: cos(180\degree + 60\degree) \bigg)$ $\sf \: = \: \dfrac{1}{2}\bigg(3 + cos2\theta - 2cos2\theta \: cos60\degree\bigg)$ $\sf \: = \: \dfrac{1}{2}\bigg(3 + cos2\theta - \cancel2cos2\theta \times \dfrac{1}{ \cancel2} \bigg)$ $\sf \: = \: \dfrac{1}{2}\bigg(3 + \cancel{ cos2\theta} - \cancel{cos2\theta} \bigg)$ $\sf \: = \: \dfrac{3}{2}$ ${\boxed{\boxed{\bf{Hence, Proved}}}}$ ─━─━─━─━─━─━─━─━─━─━─━─━─ Additional Information :- Trigonometry Formulas sin(−θ) = −sin θ cos(−θ) = cos θ tan(−θ) = −tan θ cosec(−θ) = −cosecθ sec(−θ) = sec θ cot(−θ) = −cot θ Product to Sum Formulas sin x sin y = 1/2 [cos(x–y) − cos(x+y)] cos x cos y = 1/2[cos(x–y) + cos(x+y)] sin x cos y = 1/2[sin(x+y) + sin(x−y)] cos x sin y = 1/2[sin(x+y) – sin(x−y)] Sum to Product Formulas sin x + sin y = 2 sin [(x+y)/2] cos [(x-y)/2] sin x – sin y = 2 cos [(x+y)/2] sin [(x-y)/2] cos x + cos y = 2 cos [(x+y)/2] cos [(x-y)/2] cos x – cos y = -2 sin [(x+y)/2] sin [(x-y)/2] Sum or Difference of angles cos (A + B) = cos A cos B – sin A sin B cos (A – B) = cos A cos B + sin A sin B sin (A+B) = sin A cos B + cos A sin B sin (A -B) = sin A cos B – cos A sin B tan(A+B) = [(tan A + tan B)/(1 – tan A tan B)] tan(A-B) = [(tan A – tan B)/(1 + tan A tan B)] cot(A+B) = [(cot A cot B − 1)/(cot B + cot A)] cot(A-B) = [(cot A cot B + 1)/(cot B – cot A)] cos(A+B) cos(A–B)=cos^2A–sin^2B=cos^2B–sin^2A sin(A+B) sin(A–B) = sin^2A–sin^2B=cos^2B–cos^2A Multiple and Submultiple angles sin2A = 2SINA cosA = [2tan A /(1+tan²A)] cos2A = cos²A–sin²A = 1–2sin²A = 2cos²A–1= [(1-tan²A)/(1+tan²A)] tan 2A = (2 tan A)/(1-tan²A)

Prove thatcos 2 theta + cos² ( theta +120° ) + cos ² ( theta -120°) = ⅔

Answer»

Correct Statement is

$\sf \: {cos}^{2}\theta + {cos}^{2}(\theta + 120\degree) + {cos}^{2}(\theta - 120\degree) = \dfrac{3}{2}$

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

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}} \end{gathered}$

$\boxed{ \sf \: {cos}^{2}x = \dfrac{1 + cos2x}{2}}$

$\boxed{ \sf \: cos(x + y) + cos(x - y) = 2cosxcosy}$

$\boxed{ \sf \: cos(\pi \: + x) = - cosx}$

Let's solve the problem now!!

Consider,

$\sf \: {cos}^{2}\theta + {cos}^{2}(\theta + 120\degree) + {cos}^{2}(\theta - 120\degree)$

$\sf \: = \: \dfrac{1 + cos2\theta}{2} + \dfrac{1 + cos(2\theta + 240\degree)}{2} + \dfrac{1 + cos(2\theta - 240\degree)}{2}$

$\sf \: = \: \dfrac{1}{2}\bigg(3 + cos2\theta + cos(2\theta + 240\degree) + cos(2\theta - 240\degree) \bigg)$

$\sf \: = \: \dfrac{1}{2}\bigg(3 + cos2\theta + 2cos2\theta \: cos240\degree \bigg)$

$\sf \: = \: \dfrac{1}{2}\bigg(3 + cos2\theta + 2cos2\theta \: cos(180\degree + 60\degree) \bigg)$

$\sf \: = \: \dfrac{1}{2}\bigg(3 + cos2\theta - 2cos2\theta \: cos60\degree\bigg)$

$\sf \: = \: \dfrac{1}{2}\bigg(3 + cos2\theta - \cancel2cos2\theta \times \dfrac{1}{ \cancel2} \bigg)$

$\sf \: = \: \dfrac{1}{2}\bigg(3 + \cancel{ cos2\theta} - \cancel{cos2\theta} \bigg)$

$\sf \: = \: \dfrac{3}{2}$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

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Additional Information :-

Trigonometry Formulas

sin(−θ) = −sin θ

cos(−θ) = cos θ

tan(−θ) = −tan θ

cosec(−θ) = −cosecθ

sec(−θ) = sec θ

cot(−θ) = −cot θ

Product to Sum Formulas

sin x sin y = 1/2 [cos(x–y) − cos(x+y)]

cos x cos y = 1/2[cos(x–y) + cos(x+y)]

sin x cos y = 1/2[sin(x+y) + sin(x−y)]

cos x sin y = 1/2[sin(x+y) – sin(x−y)]

Sum to Product Formulas

sin x + sin y = 2 sin [(x+y)/2] cos [(x-y)/2]

sin x – sin y = 2 cos [(x+y)/2] sin [(x-y)/2]

cos x + cos y = 2 cos [(x+y)/2] cos [(x-y)/2]

cos x – cos y = -2 sin [(x+y)/2] sin [(x-y)/2]

Sum or Difference of angles

cos (A + B) = cos A cos B – sin A sin B

cos (A – B) = cos A cos B + sin A sin B

sin (A+B) = sin A cos B + cos A sin B

sin (A -B) = sin A cos B – cos A sin B

tan(A+B) = [(tan A + tan B)/(1 – tan A tan B)]

tan(A-B) = [(tan A – tan B)/(1 + tan A tan B)]

cot(A+B) = [(cot A cot B − 1)/(cot B + cot A)]

cot(A-B) = [(cot A cot B + 1)/(cot B – cot A)]

cos(A+B) cos(A–B)=cos^2A–sin^2B=cos^2B–sin^2A

sin(A+B) sin(A–B) = sin^2A–sin^2B=cos^2B–cos^2A

Multiple and Submultiple angles

sin2A = 2SINA cosA = [2tan A /(1+tan²A)]

cos2A = cos²A–sin²A = 1–2sin²A = 2cos²A–1= [(1-tan²A)/(1+tan²A)]

tan 2A = (2 tan A)/(1-tan²A)