If Sina + 2cosa = 0, then sin2a + cos2a=

If Sina + 2cosa = 0, then sin2a + cos2a=

sina+2cosa=0
sin^2a+cos^2a=1
→cos^2a=1/5,sin^2a=4/5,sinacosa=-2/5
Sin2a= 2sinacosa=-4/5
And cos2a = 2cos ^ 2a-1 = → cos2a = - 23 / 25
sin2a+cos2a=-4/5+-23/25=-43/25

Given 2cosa = Sina, find the value of (cos2a) / (1 + sin2a)
Given 2cosa = Sina, find the value of (cos2a) / (1 + sin2a)

（cos2a）/(1+sin2a)
=(cosa+sina)(cosa-sina)/(cosa+sina)²
=(cosa-sina)/(cosa+sina)
=(cosa-2cosa)/(cosa+2cosa)
=-1/3

Given Sina = - 2cosa, then sin ^ 2 + 5sinacosa + 2cos ^ 2A=

Sin a = - 2cosa, the square of both sides leads to sin & sup2; a = 4cos & sup2; a, because Sin & sup2; a + cos & sup2; a = 1, then cos & sup2; a = 1 / 5
sin²A+5sinAcosA+2cos²A= 1-cos²A+5*（-2cosA）cosA+2cos²A=1-9cos²A=1-9/5=-4/5

Given Sina = - 2cosa, find the value of sin ^ 2A + 5sina * cosa + 2cos ^ 2A

sina=-2cosasina+2cosa=0sin^2a+4sina*cosa+4cos^2a=01.25sin^2a+5sina*cosa+5cos^2a=0sin^2a+5sina*cosa+2cos^2a=(1.25sin^2a+5sina*cosa+5cos^2a)-0.25sin^2a-3cos^2a=-0.25sin^2a-3cos^2asina=-2cosasin^2a=4cos^...

It is known that Tana = 3
Find sin (pie-a) * cos (a-pie) + 2

From Tana = 3 we get Sina = 3cosa, so Sin & sup2; a = 3sinacosa = 3, so sinacosa = 1,
sin(π－a)*cos(a-π）＋2=-sinacosa+2=-1+2=1

Given Tana = 1 / 2
The square of sina + the square of 11cosa=

sina/cosa=tana=1/2
cosa=2sina
cos²a=4sin²a
Substituting the identity Sin & sup2; a + cos & sup2; a = 1
So Sin & sup2; a = 1 / 5, cos & sup2; = 4 / 5
So the original formula is 9

It is known that Tana = 1 / 3, tanb = - 1 / 7, a, B ∈ (0, π)
(1) Finding the value of tan2a (2) finding the value of 2a-b

（1）
tan2a
=2tana/(1-tana^2)
=2*1/3/(1-1/3^2)
=2/3/(8/9)
=2/3*9/8
=3/4
(2)
tan(2a-b)
=(tan2a-tanb)/(1+tan2atanb)
=(3/4+1/7)/(1-3/4*1/7)
=(25/28)/(25/28)
=1
∵tana>0
0

sin2B＋sin2C=2sin(B＋C)cos(B－C)
How to prove it? Help me push it out. Thank you

Left = sin [(B + C) + (B-C)] + sin [(B + C) - (B-C)]
=sin(B+C)cos(B-C)+cos(B+C)sin(B-C)+sin(B+C)cos(B-C)-cos(B+C)sin(B-C)
=2sin(B+C)cos(B-C)

Why sin2a + cos2a = 1? (2 is square)

Trigonometric function definition
Sina = opposite / hypotenuse
Cosa = adjacent / hypotenuse
So sin2a + cos2a
=(opposite / hypotenuse) ^ 2 + (adjacent / hypotenuse) ^ 2
=(opposite edge ^ 2 + adjacent edge ^ 2) / hypotenuse ^ 2
From Pythagorean theorem
Opposite side ^ 2 + adjacent side ^ 2 = hypotenuse ^ 2
So sin2a + cos2a = 1

sin2a+sin2b-sin2a*sin2b+cos2a*cos2b=1
Verification

Sin2a + sin2b-sin2a · sin2b + cos2a · cos2b = sin2a-sin2a · sin2b + cos2a · cos2b + sin2b = sin2a (1-sin2b) + cos2a · cos2b + 1-cos2b = sin2acos2b + (cos2a-1) · cos2b + 1 = sin2acos2b-sin2a · cos2b + 1 = 1 (2 is square --)