Math, asked by sanya55, 1 year ago

heya!!Here is a question from class11 maths trigonometry

➡CONTENT QUALITY NEEDED⬅

Q.
prove \: that \:  \\  \cos {}^{2} x + cos {}^{2} ({x}{}  +  \frac{\pi}{3}) + cos {}^{2}  (  x -  \frac{\pi}{3} ) =  \frac{3}{2}
Pls solve with explanation


VijayaLaxmiMehra1: Now check my answer
VijayaLaxmiMehra1: I've edit it
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Answers

Answered by rohitkumargupta
10
we know,
cos(90 + ø) = sinø
cos²x = (1 + sin2x)/2

we have to \underline{prove\: :-}cos^{2} x + cos {}^{2} ({x}{} + \frac{\pi}{3}) + cos {}^{2} ( x - \frac{\pi}{3} ) = \frac{3}{2}

now,

(1 + cos2x)/2 + [1 + cos(2x + \frac{2\pi}{3})]/2 + [ 1 + cos(2x - \frac{2\pi}{3})]/2
\to\to\to\to\to\to\to\to\to\therefore\boxed{1 + cos2X = 2cos^2x}

1/2 [ 1 + cos2x + 1 + cos(2x + \frac{2\pi}{3}) + 1 + cos(2x - \frac{2\pi}{3})]

1/2[3 + cos2x + cos(2x + \frac{2\pi}{3}) + cos(2x - \frac{2\pi}{3})]
\to\to\to\to\to\to\to\to\to\therefore\boxed{cos(x + y) + cos(x - y) = 2cosx * cosy}

1/2[ 3 + cos2x + (2cos(2x) × cos\frac{2\pi}{3} ]

1/2[3 + cos2x + 2cos2X × cos(120) ]

1/2[3 + cos2X + 2cos2X × cos(90 + 30) ]

1/2[3 + cos2X + 2cos2X × (-sin30)]
\to\to\to\to\to\to\to\to\to\therefore\boxed{cos(\frac{\pi}{2} + \theta) = -sin\theta}

1/2[3 + cos2X + 2cos2X × (-1/2)]
\to\to\to\to\to\to\to\to\to\therefore\boxed{sin30\degree= 1/2}

1/2[ 3 + cos2X - cos2x]

3/2

sanya55: thanks
rohitkumargupta: :-)
FuturePoet: wow great
sanya55: yeah!!
sanya55: really awesome answers
rohitkumargupta: AchA
rohitkumargupta: thanks thanks:-)
sanya55: :-))
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