Question

Wykaż, że: a) cos^4x-sin^4x=cos2x

b)|sinx/2|=[pod pierwiastkiem] (1-cos2x)/2

c) (sinx/2 + cosx/2)^2 = 1+sinx

d) 1/(sinxcosx) - cosx/sinx = tgx

e) cosx= cos^2x/2 - sin^2x/2

f) sinxcos^2x+sin^3x=sinx

g) 2cos^2x*tgx+2=(sinx+cosx)2

h) tg^2x/(1+tg^2x) = sin^2x

Chodzi o rozwiązanie tożsamości, wykazać, że P=L

Answer 1

a)

cos^4 x - sin ^4 x = ( cos ^2 x - sin ^2 x)*( cos^2 x + sin ^2 x) =

=( cos^2 x - sin ^ 2 x)* 1 = cos^ 2 x - sin ^2 x = cos 2x

L = P

b)

sin(x/2) = p [ 1 - cos ^2 (x/2)]

sin ^2(x/2) + cos ^2 (x/2) = 1

sin^2 (x/2) = 1 - cos ^2 (x/2)

sib (x/2) = p [ 1 - cos^2 (x/2) ]

c)

[ sin(x/2) + cos(x/2)]^2 = sin^2 (x/2) + 2* sin (x/2)*cos (x/2) + cos^2 (x/2)=

= sin^2(x/2) + cos^2 (x/2) + sin x = 1 + sin [ 2*(x/2)] = 1 + sin x

d)

1 / [ sin x cos x ]  - cos x / sin x = 1/[sin x cos x ] - ( cos^2 x)/(sin x cos x] =

= [ 1 - cos ^2 x]/[ sin x  cos x] = sin ^2 x /[ sin x cos x ] = sin x / cos x = tg x

e)

cos x = cos ^2 (x/2) - sin ^2 (x/2)

cos ^2 (x/2) - sin^2 (x/2) = cos [2*(x/2)] = cos x

f)

sin x  cos^2 x  + sin ^3 x = sin x *[ cos ^2 x + sin ^2 x] = sin x *1 = sin x

g)

( sin x + cos x ) ^2 = sin^2 x + 2 sin x cos x + cos^2 x = 1 + sin 2 x

2 cos ^2 x *tg x + 2= 2 cos^2 x* [ sin x / cos x] + 2 = 2 cos x sin x + 2 =

= 2 + sin 2x

Nie

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h)

tg^2 x / [1+ tg^2 x] = [ sin^2 x / cos ^2 x]/[ cos^2x / cos^2 x + +sin^2x/cos^2x ] = [ sin^2 x / cos ^2 x] / [ 1/ cos^2 x] =

= [ sin ^2 x / cos ^2 x]* cos^2 x = sin ^2 x