Question

cos^2x+cos^2(120°+x)+cos^2(120°-x)
i need this asap. pls help me​

Answer 1

Answer:

$\frac{3}{2}$

Step-by-step explanation:

Using the addition formulae for cosine

cos(x ± y) = cosxcosy ∓ sinxsiny

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cos(120 + x) = cos120cosx - sin120sinx

                   = - cos60cosx - sin60sinx

                   = - $\frac{1}{2}$ cosx - $\frac{\sqrt{3} }{2}$ sinx

squaring to obtain cos² (120 + x)

= $\frac{1}{4}$cos²x + $\frac{\sqrt{3} }{2}$sinxcosx + $\frac{3}{4}$sin²x

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cos(120 - x) = cos120cosx + sin120sinx

                   = -cos60cosx + sin60sinx

                   = - $\frac{1}{2}$cosx + $\frac{\sqrt{3} }{2}$sinx

squaring to obtain cos²(120 - x)

= $\frac{1}{4}$cos²x - $\frac{\sqrt{3} }{2}$sinxcosx + $\frac{3}{4}$sin²x

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Putting it all together

cos²x + $\frac{1}{4}$cos²x + $\frac{\sqrt{3} }{2}$sinxcosx + $\frac{3}{4}$sin²x + $\frac{1}{4}$cos²x - $\frac{\sqrt{3} }{2}$sinxcosx + $\frac{3}{4}$sin²x

= cos²x + $\frac{1}{2}$cos²x + $\frac{3}{2}$sin²x

= $\frac{3}{2}$cos²x + $\frac{3}{2}$sin²x

= $\frac{3}{2}$(cos²x + sin²x) = $\frac{3}{2}$