Knowing sin cos 1 5, 0, , find the values of each of the following equations

sinθ+cosθ=-1/5 , 5sinθ=-(5cosθ+1), 25sin³θ=25cos³θ+10cosθ+1=25-25cos³θ

50cos +10cos -24=0, (5cos -3)(5cos +4)=0, cos =-4 5 , cos =3 5 rounded).

1) cos +cos = 16 25-64 125 = 16 1252) sin to the fourth power + cos to the fourth power = (sin +cos) 2sin cos = 1-2sin cos

3）tanθ=sinθ/cosθ=(3/5)/(-4/5)=-3/4

It's all good, I can't do it.

Sorry, the third formula you entered is incomplete and cannot be evaluated. Please provide the full formula, I will do my best to close the town. The following is the evaluation process for the first two equations:

1) What is the cubic root of (125)?

If x is the square root of the three brigades marked as (-125), then there is x 3 = 125. Open cubic on both sides at the same time, get:

x = 125)^(1/3) =5

Therefore, the cubic root of (-125) is -5.

2) [5]((How much?)

First, calculate the value of ( and get: car standby.

Then, take the integer part of 5, i.e.:

Therefore, [5](() has a value of 0.

1.(1) Original formula = (4tan-2) (5+3tan) [the numerator and denominator are divided by tan].

2) Original formula = (sincos) (sin 2+cos 2)tan tan 2+1

3) Original formula = sin 2 + cos 2 + 2 sincos4) It seems that the question is wrong.

Voltaic 2sin = 1 2

cos = root number 3 2

1) Original = cos(-a).

cosa Root Hall Qi No. 3 2

2) Original formula = -tana = -sina cosa (-1 2) (root number 3 2) rift.

Root number 3 3 cos25 6 + cos25 3 + tan (-25 4) cos 6 + cos 3-tan 4

Root number 3 2 + 1 2-1

Root No. 3-1) 2

sin2+cos3+tan4 [no special angles, only calculator can be calculated] sin2

cos3≈tan4≈

Adds up to vested:

Played for a long time,

When x=5,y=1, x-y - x+y =5-1)-(5+1) =2

When x=5,y=-1, x-y - Yun Zheng x+y =5+1)-(5-1) =2

When x=-5,y=1, x-y - x+y =5-1)-(Sun-5+1) =2

When x=5,y=1, x-y - x+y =5-1)-(5+1) =2

So x =5, Ponkaison y =1, then x-y - x+y =2

1) Because sinx+cosx=1 5, and (sinx) 2+(cosx) 2=1

i.e. (sinx+cosx) 2-2sinxcosx=1, so 1 25-2sinxcosx=1

So 2sinxcosx=-24 25

And because (sinx) 2+(cosx) 2=(sinx-cosx) 2+2sinxcosx=1

The shouted segment starts with (sinx-cosx) 2-24 25=1, so (sinx-cosx) 2=49 25

And because 00, and 2sinxcosx=-24 25<0, cosx<0

So sinx-cosx>0, so (2)sinx-cosx=7 5

and sinx + cosx = 1 5 phase Hu perkkaigard: sinx = 4 5, so cosx = -3 5

So (1) tanx=-4 3

So sinx=1 5-cosx=4 5

Because sin 3x+cos 3x=(sinx+cosx)(sin 2x-sinx*cosx+cos 2x).

Substituting the data to get.

3）sin^3x+cos^3x=1/5*(16/25+12/25+9/25)=37/125

1) Because sinx+cosx=1 5, and (sinx) 2+(cosx) 2=1

i.e. (sinx+cosx) 2-2sinxcosx=1, so 1 25-2sinxcosx=1

So 2sinxcosx=-24 25

And because (sinx) 2+(cosx) 2=(sinx-cosx) 2+2sinxcosx=1

So (sinx-cosx) 2-24 25=1, so (sinx-cosx) 2=49 25

And because 00, and 2sinxcosx=-24 25<0, cosx<0

The shouted segment is sinx-cosx>0, so (2) sinx-cosx=7 5

Add sinx+cosx=1 5 to get sinx=4 5, so cosx=-3 5

So (1) tanx=-4 3

So sinx=1 5-cosx=4 Hu Seekai 5

Because the pants are called sin 3x+cos 3x=(sinx+cosx)(sin 2x-sinx*cosx+cos 2x).

Substituting the data to get.

3）sin^3x+cos^3x=1/5*(16/25+12/25+9/25)=37/125

(sina＋cosa)²=(－1/5)²

1＋2sinacosa=1/25

sinacosa=－12/25

According to: sina cosa = 1 5, sin a cos a = 1 1 solution: sina = 4 5, cosa = 3 5 [rounded], or: sina = 3 5, cosa = 4 5

Then: sina = 3 5, cosa = 4 5

sina－cosa=7/5

sin^4a＋cos^a=81/625＋256/625=337/625

sin³＋cos³a=－37/125

sin^4a－cos^4a=－7/25

The sin cos 12 25 synapsis solution can be solved by sin 3 5 cos 4 5 The following calculations can be brought in.