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If SiNx + cosx = root 2, find the value of (1) sin ^ 3x + cos ^ 3x and (2) TaNx + Cotx
(1) (SiNx + cosx) ^ 2 = sin ^ 2x + 2sinxcosx + cos ^ 2x = 1 + 2sinxcosx = 2, so sinxcosx = 0.5, sin ^ 3x + cos ^ 3x = (SiNx + cosx) * (sin ^ 2x sinxcosx + cos ^ 2x) = (root 2) / 2
(2) Original formula = SiNx / cosx + cosx / sin = (sin ^ 2x + cos ^ 2x) / (sinxcosx) = 2
Simplification; SiNx / (1-cosx) * root sign (TaNx SiNx / TaNx + SiNx) Help
If the formula to be simplified is
SiNx / (1-cosx) * root sign [(TaNx SiNx) / (TaNx + SiNx)]
The simplification is shown in the figure
TaNx = root 2, (cosx SiNx) / (cosx + SiNx) =?
-1/3
{(cosX)^2-(sinx)^2}/(cosx)^2+(sinx)^2={1-(tanx)^2}/{1+(tanx)^2}=-1/3
If SiNx + cosx = root 2, the value of TaNx + 1 / TaNx is?
SiNx + cosx = root 2 square, sin is obtained ² x+2sinxcosx+cos ² X = 21 + 2sinxcosx = 2sinxcosx = 1 / 2, so TaNx + 1 / TaNx = SiNx / cosx + cosx / SiNx = (SIN) ² x+cos ² x)/(cosxsinx)=1/cosxsinx=1/(1/2)=2
Simplify {sin (π + x)} / 1 + cos (3 π - x) under the root sign (TaNx SiNx) / (TaNx + SiNx)
=-sinx/(1-cosx) *√[(1/cosx-1)/(1/cosx+1)] ]
=-sinx/(1-cosx) *[(1-cosx)/|sinx| ]
sinx>0
=-1
sinx
Cos (x-pai / 4) = root sign 2 / 10, X belongs to (PAI / 2,3pai / 4) find SiN x and sin (2x + Pai / 3)
Cos (x - π / 4) = cosxcos π / 4 + sinxsin π / 4 = √ 2 / 2 * (cosx + SiNx) = √ 2 / 10sinx + cosx = 1 / 5cosx = 1 / 5-sinx square cos ² x=1-sin ² x=1/25-2/5*sinx+sin ² xsin ² X-1 / 5 * sinx-12 / 25 = 0 (sinx-4 / 5) (SiNx + 3 / 5) = 0 from X range SiNx > 0
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