Troche trygonometrii

[kolodzieju]

wiedząc, że sin x + cos x =1/√2 oblicz:
a) sin x * cos x
b) sin³x +cos³x
c) |sin x - cos x|
d) sin⁴x + cos⁴x

Wiedząc, że tg α + ctgα = 4 oblicz
a) |tg α + ctgα|
b) tg²α+ctg²α
c) tg³α+ctg³α
d) tg⁴α+ctg⁴α

[irena]

\(sinx+cosx=\frac{1}{\sqrt{2}}\)

a)
\((sinx+cosx)^2=sin^2x+2sinx\ cosx+cos^2x=1+2sinx\ cosx\\2sinx\ cosx= (sinx+cosx)^2-1\\2sinx\ cosx=\frac{1}{2}-1\\2sinx\ cosx=-\frac{1}{2}\\sinx\ cosx=-\frac{1}{4}\)

b)
\(sin^3x+cos^3x=(sinx+cosx)(sin^2x-sinx\ cosx+cos^2x)=(sinx+cosx)(1-sinx\ cosx)=\\=\frac{1}{\sqrt{2}}(1+\frac{1}{4})=\frac{5\sqrt{2}}{8}\)

c)
\((sinx-cosx)^2=sin^2x+cos^2x-2sinx\ cosx\\(sinx-cosx)^2=1-2\cdot(-\frac{1}{4})=\frac{3}{2}\\|sinx-cosx|=\frac{3\sqrt{2}}{2}\)

d)
\(sin^4x+cos^4x=(sin^2x+cos^2x)^2-2sin^2x\ cos^2x=1-2\cdot(sinx\ cosx)^2=\\=1-2\cdot(-\frac{1}{4})^2=\frac{7}{8}\)

[irena]

\(tg\alpha+ctg\alpha=4\)

a)
\((tg\alpha+ctg\alpha)^2-(tg\alpha-ctg\alpha)^2=4tg\alpha\ ctg\alpha=4\cdot1=4\\(tg\alpha-ctg\alpha)^2=(tg\alpha+ctg\alpha)^2-4\\(tg\alpha-ctg\alpha)^2=4^2-4=12\\|tg\alpha-ctg\alpha|=2\sqrt{3}\)

b)
\(tg^2\alpha+ctg^2\alpha=(tg\alpha+ctg\alpha)^2-2tg\alpha\ ctg\alpha=(tg\alpha+ctg\alpha)^2-2=\\=4^2-2=14\)

c)
\(tg^3\alpha+ctg^3\alpha=(tg\alpha+ctg\alpha)(tg^2\alpha+ctg^2\alpha-tg\alpha\ ctg\alpha)=(tg\alpha+ctg\alpha)(tg^2\alpha+ctg^2\alpha-1)=\\=4\cdot(14-1)=52\)

d)
\(tg^4\alpha+ctg^4\alpha=(tg^2\alpha+ctg^2\alpha)^2-2tg^2\alpha\ ctg^2\alpha=(tg^2\alpha+ctg^2\alpha)-2=\\=14^2-2=196-2=194\)