Answer:
See the explanation.
Step-by-step explanation:
Sin2α = 2SinαCosα.
Cos2α = (Cosα)^2 - (Sinα)^2.
Sin4α can be written as Sin2(2α).
Similarly Cos4α can be written as Cos2(2α).
[tex]Sin 4\alpha = 2Sin 2\alpha \times Cos 2\alpha = 2Sin 2\alpha \times(Cos^{2} \alpha - Sin^{2} \alpha ) = 4Sin \alpha \times Cos \alpha \times (Cos^{2} \alpha - Sin^{2} \alpha )[/tex]
It is given that tanα=3.
Sinα(Sinα) = [tex]\frac{9}{10}[/tex] [[tex]Sec^{2} \alpha = 1 + 3^{2} = 10.\\Cos^{2} \alpha = \frac{1}{10} \\Sin^{2} \alpha = 1 - \frac{1}{10} = \frac{9}{10}[/tex]], the values either be both positive or both negative, as the value of tanα is positive only on first and third quadrant.
Sin4α = [tex]4\times \sqrt{(\frac{9}{10} )} \times \sqrt{(\frac{1}{10} )} \times (\frac{1 - 9}{10} ) = -\frac{4\times8\times3}{100} = -\frac{96}{100}[/tex].
Cos4α = [tex]Cos^{2} 2\alpha - Sin^{2} 2\alpha = (Cos^{2} \alpha - Sin^{2} \alpha )^{2} - Sin^2 {2\alpha } = (Cos^{2} \alpha - Sin^{2} \alpha )^{2} - (2Sin \alpha Cos \alpha )^{2} = (-\frac{8}{10} )^{2} - \frac{4\times9}{100} = \frac{64 - 36}{100} = \frac{28}{100} = \frac{7}{25}[/tex]
