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laila [671]
3 years ago
11

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

Mathematics
1 answer:
o-na [289]3 years ago
5 0

Answer:

\frac{3}{2}

Step-by-step explanation:

Using the addition formulae for cosine

cos(x ± y) = cosxcosy ∓ sinxsiny

---------------------------------------------------------------

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

--------------------------------------------------------------------

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

--------------------------------------------------------------------------

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}

                 

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