Question

# Find the value of ${\text{si}}{{\text{n}}^2}18^\circ - {\text{ co}}{{\text{s}}^2}72^\circ$.

${\text{si}}{{\text{n}}^2}18^\circ - {\text{ co}}{{\text{s}}^2}72^\circ$
⟹${\text{si}}{{\text{n}}^2}\left( {90 - 72} \right)^\circ - {\text{ co}}{{\text{s}}^2}72^\circ$ ---- sin (90 – θ) = cosθ, since sine is a periodic function.
⟹${\text{co}}{{\text{s}}^2}72^\circ - {\text{ co}}{{\text{s}}^2}72^\circ$
Hence, ${\text{si}}{{\text{n}}^2}18^\circ - {\text{ co}}{{\text{s}}^2}72^\circ$ = 0