Skola2030 paraugā dotais sasniedzamais rezultāts: risina kompleksu matemātisku problēmu, sasaistot trigonometrijas un citu matemātikas apakšnozaru zināšanas.
Piemērs:
Aprēķini ${\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x$, ja $\mathrm{sin}x+\mathrm{cos}x=c$.
Risinājums

1) $\mathrm{sin}x+\mathrm{cos}x=c$ abas puses kāpina kvadrātā, pielieto formulu ${\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x=1$

$\begin{array}{l}{\left(\mathrm{sin}x+\mathrm{cos}x\right)}^{2}={c}^{2}\\ {\mathrm{sin}}^{2}x+2\mathrm{sin}x\cdot \mathrm{cos}x+{\mathrm{cos}}^{2}x={c}^{2}\\ 2\mathrm{sin}x\cdot \mathrm{cos}x+1={c}^{2}\\ \mathrm{sin}x\cdot \mathrm{cos}x=\frac{{c}^{2}-1}{2}\end{array}$
Šo rezultātu izmantosim tālāk.

1. variants tālākam risinājumam
$\begin{array}{l}{\mathrm{sin}}^{4}\mathrm{\alpha }+{\mathrm{cos}}^{4}\mathrm{\alpha }=\\ ={\left({\mathrm{sin}}^{2}\mathrm{\alpha }+{\mathrm{cos}}^{2}\mathrm{\alpha }\right)}^{2}-2{\mathrm{sin}}^{2}\mathrm{\alpha }\cdot {\mathrm{cos}}^{2}\mathrm{\alpha }=\\ =1-2{\mathrm{sin}}^{2}\mathrm{\alpha }\cdot {\mathrm{cos}}^{2}\mathrm{\alpha }=\\ =1-2{\left(\frac{{c}^{2}-1}{2}\right)}^{2}=\\ =1-\frac{{c}^{4}-2{c}^{2}+1}{2}=\\ =\frac{1-{c}^{4}+2{c}^{2}}{2}\end{array}$

2. variants tālākam risinājumam

$\mathrm{sin}x+\mathrm{cos}x=c$ kāpinām 4. pakāpē, izmantojam Ņūtona binoma izvirzījumu, koeficientus nolasām no Paskāla trijstūra 4. rindas (skat. zemāk)

$\begin{array}{l}{\left(\mathrm{sin}x+\mathrm{cos}x\right)}^{4}=\\ ={\mathrm{sin}}^{4}x+4{\mathrm{sin}}^{3}x\cdot \mathrm{cos}x+6{\mathrm{sin}}^{2}x{\cdot \mathrm{cos}}^{2}x+4\mathrm{sin}x{\cdot \mathrm{cos}}^{3}x+{\mathrm{cos}}^{4}x\end{array}$

$\begin{array}{l}{\mathrm{sin}}^{4}x+4{\mathrm{sin}}^{3}x\cdot \mathrm{cos}x+6{\mathrm{sin}}^{2}x{\cdot \mathrm{cos}}^{2}x+4\mathrm{sin}x{\cdot \mathrm{cos}}^{3}x+{\mathrm{cos}}^{4}x={c}^{4}\\ {\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x+\underset{¯}{4{\mathrm{sin}}^{3}x\cdot \mathrm{cos}x+4\mathrm{sin}x{\cdot \mathrm{cos}}^{3}x}+6{\mathrm{sin}}^{2}x{\cdot \mathrm{cos}}^{2}x={c}^{4}\\ {\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x+4\mathrm{sin}x\mathrm{cos}x\left({\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x\right)+6{\mathrm{sin}}^{2}x{\cdot \mathrm{cos}}^{2}x={c}^{4}\\ {\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x+4\mathrm{sin}x\mathrm{cos}x\cdot 1+6{\mathrm{sin}}^{2}x{\cdot \mathrm{cos}}^{2}x={c}^{4}\\ {\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x={c}^{4}-4\underset{¯}{\underset{¯}{\mathrm{sin}x\mathrm{cos}x}}-6{\left(\underset{¯}{\underset{¯}{\mathrm{sin}x\cdot \mathrm{cos}x}}\right)}^{2}\end{array}$

Iepriekš jau noskaidrojām, ka
$\mathrm{sin}x\cdot \mathrm{cos}x=\frac{{c}^{2}-1}{2}$

${\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x={c}^{4}-4\cdot \frac{{c}^{2}-1}{2}-6\cdot {\left(\frac{{c}^{2}-1}{2}\right)}^{2}$
$\begin{array}{l}{c}^{4}-\frac{4\left({c}^{2}-1\right)}{2}-\frac{6{\left({c}^{2}-1\right)}^{2}}{4}=\\ =\frac{2{c}^{4}}{2}-\frac{4\left({c}^{2}-1\right)}{2}-\frac{3\left({c}^{4}-2{c}^{2}+1\right)}{2}=\\ =\frac{2{c}^{4}-4{c}^{2}+4-3{c}^{4}+6{c}^{2}-3}{2}=\\ =\frac{-{c}^{4}+2{c}^{2}+1}{2}\end{array}$
Atbilde: ${\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x=\frac{1-{c}^{4}+2{c}^{2}}{2}$
$\begin{array}{l}{\left(a+b\right)}^{n}=\\ ={C}_{n}^{0}{a}^{n}{b}^{0}+{C}_{n}^{1}{a}^{n-1}{b}^{1}+{C}_{n}^{2}{a}^{n-2}{b}^{2}+...+{C}_{n}^{n-1}{a}^{1}{b}^{n-1}+{C}_{n}^{n}{a}^{0}{b}^{n}\end{array}$