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Calculus, EXT1 C2 2024 HSC 13b

  1. Show that  \(\cos ^4 x+\sin ^4 x=\dfrac{1+\cos ^2 2 x}{2}\).   (2 marks)

  2. Hence, or otherwise, evaluate  \(\displaystyle{\int}_0^{\frac{\pi}{4}}\left(\cos ^4 x+\sin ^4 x\right) d x\).  (3 marks)

Show Answers Only

i.     \(\text{LHS}\) \(=\left[\dfrac{1}{2}(1+\cos (2 x)\right]^2+\left[\dfrac{1}{2}(1-\cos (2 x)\right]^2\)
    \(=\dfrac{1}{4}\left(1+2 \cos (2 x)+\cos ^2(2 x)+1-2 \cos (2 x)+\cos ^2(2 x)\right)\)
    \(=\dfrac{1}{4}\left(2+2 \cos ^{2}(2 x)\right)\)
    \(=\dfrac{1+\cos ^2(2 x)}{2}\)

  
ii.   \(\dfrac{3 \pi}{16}\)

Show Worked Solution

i.     \(\text{LHS}\) \(=\left[\dfrac{1}{2}(1+\cos (2 x)\right]^2+\left[\dfrac{1}{2}(1-\cos (2 x)\right]^2\)
    \(=\dfrac{1}{4}\left(1+2 \cos (2 x)+\cos ^2(2 x)+1-2 \cos (2 x)+\cos ^2(2 x)\right)\)
    \(=\dfrac{1}{4}\left(2+2 \cos ^{2}(2 x)\right)\)
    \(=\dfrac{1+\cos ^2(2 x)}{2}\)

  

ii.     \(\displaystyle{\int}_0^{\frac{\pi}{4}}\left(\cos ^4 x+\sin ^4 x\right) d x\)
    \(=\dfrac{1}{2} \displaystyle{\int}_0^{\frac{\pi}{4}} 1+\cos ^2(2 x) d x\)
    \(=\dfrac{1}{2} \displaystyle{\int}_0^{\frac{\pi}{4}} 1+\dfrac{1}{2}(1+\cos (4 x)) d x\)
    \(=\dfrac{1}{2}\left[\dfrac{3}{2}x +\dfrac{1}{8} \sin (4 x)\right]_0^{\frac{\pi}{4}}\)
    \(=\dfrac{1}{2}\left[\dfrac{3}{2} \times \dfrac{\pi}{4}+\dfrac{1}{8} \sin \pi-0\right]\)
    \(=\dfrac{3 \pi}{16}\)