Question

If A + B + C = $\pi$, Prove that

cos 4A + cos 4B + cos 4C = -1 + 4 cos 2A cos 2B cos 2C.

Answer 1

We have

LHS = $\text{cos}4A+\text{cos}4B+\text{cos}4C$

=$2\text{cos}(2A+2B)\text{cos}(2A-2B)+\text{cos}4C$

=$2\text{cos}(2\pi -2C)\text{cos}(2A-2B)+(2co{s}^{2}2C-1)$

$[\because A+B+C=\pi \Rightarrow (2A+2B)=(2\pi -2C\left)\right]$

=$2\text{cos}2C2\text{cos}(2A-2B)+2{\text{cos}}^{2}2C-1$

=$2\text{cos}2C\left[\right(2A-2B)+\text{cos}2C]-1$

=$2\text{cos}2C\left[\right(2A-2B)+\text{cos}[2\pi -(2A+2B\left)\right]]-1$

$[\because 2A+2B+2C=2\pi \Rightarrow 2C=2\pi -(2A+2B\left)\right]=2\text{cos}2C\left[\text{cos}\right(2A-2B)+\text{cos}(2A+2B\left)\right]$

=$2\text{cos}2C\left[2\text{cos}2Acos2B\right]-1$

=$-1+4\text{cos}2A\text{cos}2B\text{cos}2C=\text{RHS}$

$\therefore cos4A+cos4B+cos4C=-1+4cos2Acos2Bcos2C$