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If
g(x)=sin⁡2x+sin⁡2(π3+x)+cosx⋅cos(x+π3) and f(54)=2g(x) = \sin^{2}x + \sin^{2}\left( \frac{\pi}{3} + x \right) + cosx \cdot cos\left( x + \frac{\pi}{3} \right)\ and\ f\left( \frac{5}{4} \right) = 2,
then f(g(5)) is equal to

Show answer and solution
g(x)=1−cos2x2+1−cos(2x+2π3)2+cosx⋅cos(x+π3)g(x) = \frac{1 - cos2x}{2} + \frac{1 - cos\left( 2x + \frac{2\pi}{3} \right)}{2} + cosx \cdot cos\left( x + \frac{\pi}{3} \right)
=12[2−(cos⁡2x+cos⁡(2x+2π3))+2cos⁡x⋅cos⁡(x+π3)]= \frac{1}{2}\left\lbrack 2 - \left( \cos{2x} + \cos\left( 2x + \frac{2\pi}{3} \right) \right) + 2\cos x \cdot \cos\left( x + \frac{\pi}{3} \right) \right\rbrack
=12[2−2cos⁡(2x+π3)cos⁡(π3)+cos⁡(2x+π3)+cos⁡(π3)]= \frac{1}{2}\left\lbrack 2 - 2\cos{\left( 2x + \frac{\pi}{3} \right)\cos\left( \frac{\pi}{3} \right)} + \cos\left( 2x + \frac{\pi}{3} \right) + \cos\left( \frac{\pi}{3} \right) \right\rbrack
=12[2+12]=54,∀ x∈R= \frac{1}{2}\left\lbrack 2 + \frac{1}{2} \right\rbrack = \frac{5}{4},\forall\ x \in R
⇒f(g(x))=f(54)=2\Rightarrow f(g(x)) = f\left( \frac{5}{4} \right) = 2
⇒f(g(5))=2\Rightarrow f(g(5)) = 2

Source: Grass-G / M-18 Mathematics Paper 27 Nov Third.docx

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