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1. A) Is there a number x ∈ R, so cos x = 0.8 and sin x = 0.2? b) Same question for cos x = 0.8 and sin x = 0.6 2. Show that for any x ∈ R: a) sin^{3}x + cos^{3} = (sin x + cos x)(1- sin x * cos x) b) sin^{4} x + cos^{4} x = 1 - 2sin^{2} x * cos^{2}x

Respuesta :

ukshedrack

Answer:

1Ai) [tex]x = 36.9^{o}[/tex]

1Aii) [tex]x = 11.5^{o}[/tex]

Step-by-step explanation:

Question 1a

[tex]cosx=0.8\\x= cos^{-1}(0.8)\\ x= 36.9^{o}[/tex]

[tex]sinx=0.2\\x=sin^{-1}(0.2)\\ x=11.5^{o}[/tex]

Question 2a

[tex]sin^{3}x+cos^{3}x = sin^{2}x(sinx)+cos^{2}x(cosx)\\ \\= (1-cos^{2}x)sinx+(1-sin^{2}x)cosx\\ \\= sinx-sinxcos^{2}x+cosx-sin^{2}xcosx\\ \\=sinx-sin^{2}xcosx+cosx-sinxcos^{2}x\\[/tex]

Factorize

[tex]sinx(1-sinxcosx)+cosx(1-sinxcosx)\\\\=(sinx+cosx)(1-sinxcosx)\\\\=(sinx+cosx)(1-sinx*cosx)[/tex]

Question 2b

[tex]sin^{4}x+ cos^{4}x=(sin^{2}x)^{2} + (cos^2}x)^{2} \\But , a^{2} +b^{2}=(a+b)^{2}-2ab\\ \\a = sin^{2}x\\ b = cos^{2}x\\\\Therefore, (sin^{2}x)^{2} + (cos^2}x)^{2} = (sin^{2}x+cos^{2}x)^{2}-2sin^{2}xcos^{2}x \\\\But, sin^{2}x+cos^{2}x = 1\\\\Therefore, (sin^{2}x+cos^{2}x)^{2}-2sin^{2}xcos^{2}x = 1 - 2sin^{2}xcos^{2}x\\\\= 1 - 2sin^{2}x*cos^{2}x[/tex]

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