If Tan (5 + a) = - 2 and COS > 0, then sin (- PI + a) =/

If Tan (5 + a) = - 2 and COS > 0, then sin (- PI + a) =/

Tana = Tan (π + a) = - 21 + Tan & sup2; a = 1 + 4 = 51 + Tan & sup2; a = 1 + (Sin & sup2; a / cos & sup2; a) = 1 / cos & sup2; A1 / cos & sup2; a = 5cosa = (radical 5) / 5sina = Tana × cosa = - 2 (radical 5) / 5sin (- π + a) = - Sina = 2 (radical 5) / 5

It is known that sina and cosa are two roots of the equation x ^ 2-bx + B = 0 (a). (1) find the value of COS ^ 3 (PI / 2-A) + sin ^ (PI / 2-A) () find Tan (...)
It is known that sina and cosa are two roots of the equation x ^ 2-bx + B = 0 (a). (1) find the value of COS ^ 3 (PI / 2-A) + sin ^ (PI / 2-A) () find the value of Tan (pi-A) - 1 / Tana

sinacosa=bsina+cosa=b
cos^3(pi/2-a)+sin^(pi/2-a)=sin^3(a)+cos^3(a)=(sina+cosa)(sina^2+sinacosa+cosa^2)=b(1+b）
tan(pi-a)-1/tana=-(tana+1/tana)=-(tana^2+1)/tana=-(sina^2+cosa^2)/(sinacosa)=-1/b

sin(x-pi)cos(x-pi)=0.25,sin2x

sin(x-pi)cos(x-pi)=0.25
(-sinx)(-cosx)=0.25
So sin2x = 2sinxcosx = 0.5

cos(pi/4-a)=1/7,sin(3pi/4+b)=11/14
cos(pi/4-a)=1/7
sin(3pi/4+b)=11/14
pi/4

π/4

The function f (x) = a * sin (w * x) + b * cos (w * x) (W > 0) defined on R has a period of PI and f (x)

Cos (PI / 4-A) = 3 / 5 sin (3pi / 4 + b) = 5 / 13 sin (a + b)

Because (3 π / 4 + b) - (π / 4-A) = B + A + π / 2
So cos [(a + b) + π / 2] = cos (a + b) * cos π / 2-sin (a + b) * sin π / 2 = - sin (a + b)
So sin (a + b) = - cos [(a + b) + π / 2] = - cos [(3 π / 4 + b) - (π / 4-b)]
=-[cos(3π/4+B)*cos(π/4-a)+sin(3π/4+B)*sin(π/4-a)]
From π / 4

Solution equation: 4 (sin ^ 6 x + cos ^ 6 x) = 1, X ∈ [0. Pi / 2]

4 (SiNx ^ 6x + cos ^ 6x) = 14 {[(SiNx) ^ 2] ^ 3 + [cosx) ^ 2] ^ 3} = 14 [(SiNx) ^ 2 + (cosx) ^ 2] [(SiNx) ^ 4 - (SiNx) ^ 2 (cosx) ^ 2 + (cosx) ^ 4] = 14 {1-3 (sinxcosx) ^ 2} = 14-3 (sin2x) ^ 2 = 1sin2x = 1 or sin2x = - 12x = 360K + 90 or 2x = 360k-90x = 180K + 45 or x = 180K

The range of function sin5x / SiNx is
The range of (sin5x) \ \ SiNx is

sin5x=sinxcos4x+cosxsin4x
cos4x=2[1-2*(sinx)^2]^2 -1
sin4x=4sinxcosx[1-2(sinx)^2]
Let t = (SiNx) ^ 2,0 < T ≤ 1
Then y = (sin5x) / (SiNx) = 16 (T - 5 / 8) ^ 2 - 5 / 4
So - 5 / 4 ≤ y < 5

Finding the limit: when x → 0 (sin3x SiNx) / X

Sum difference product
=lim(x→0)2cos2xsinx/x
=lim(x→0)2cos2x*lim(x→0)sinx/x
=2×1
=2

The limit of sin3x-sin5x / X when x tends to zero

The original formula = Lim {X - > 0} 2cos4x * sin (- x) / x = - 2lim {X - > 0} cos4x * Lim {X - > 0} SiNx / x = - 2