Question

see the attachment nd plz solve it....​

nd don't spam otherwise your answer will be reported..​

Answer 1

=> cos²x + cos²(x + π/3) + cos²(x - π/3)

=> cos²x + [cosx.cos(π/3) - sinx.sin(π/3)]² + [cosx.cos(π/3) + sinx.sin(π/3)]²

Let cosx.cos(π/3) = a & sinx.sin(π/3) = b

=> cos²x + (a - b)² + (a + b)²

=> cos²x + (a² + b² - 2ab) + (a² + b² + 2ab)

=> cos²x + 2(a² + b²)

=> cos²x + 2[(cosx.cos(π/3))² + (sinx.sin(π/3))²]

=> cos²x + 2[( cosx. 1/2 )² + ( sinx. √3/2 )²]

=> cos² + 2[cos²x/4 + 3sin²x/4]

=> cos²x + cos²x/2 + 3sin²x/2

=> [2cos²x + cos²x + 3sin²x]/2

=> [ 3(cos²x + sin²x) ]/2

=> 3(1)/2

=> 3/2

Solved using:

• cos(x + y) = cosx.cosy - sinx.siny
• cos(x - y) = cosx.cosy - sinx.siny
• Assumption (a & b) is done just to decrease length(calculation)

*better to be seen through browser/desktop-mode

Answer 2

please mark as brainlist