Given cos α = - 3 / 5, find sin α, Tan α

Given cos α = - 3 / 5, find sin α, Tan α

sin²α+cos²α=1
So sin α = ± 4 / 5
tanα=sinα/cosα
therefore
sinα=-4/5,cosα=4/3
sinα=4/5,cosα=-4/3

It is known that sin α = 3 5, find cos α, Tan α

∵sin2α+cos2α=1，sinα=3
5，
∴cos2α=16
25，
When α is the first quadrant angle, cos α = 4
5, Tan α = sin α
cosα=3
4；
When α is the second quadrant angle, cos α = - 4
5, Tan α = sin α
cosα=-3
4．

Sin (a + b) = - 3 / 5. A belongs to (90180) B belongs to (0,90) Sina / 2 + cosa / 2 = 2 times root 3 / 3. Sin (a + b) = - 3 / 5. A belongs to (90180) B belongs to (0,90). We need to explain cosa and SINB in detail. Thank you very much!

Using trigonometric function formula: ① sin2 α = 2Sin α· cos α; ② sin (α + β) = sin α· cos β + cos α · sin β; ∵ sin (A / 2) + cos (A / 2) = 2 √ 3 / 3, and a ∈ (90 ° 180 °), then a / 2 ∈ (45 °, 90 °)

If cosa = - 1 / 2, a ∈ (0, π), SINB = - √ 3 / 2, B ∈ (3 π / 2,2 π), then sin (a + b) =?

According to cos? A + sin? A = 1, Sina = ±√ 3 / 2, a ∈ (0, π), Sina = √ 3 / 2
cosb=-1/2
sin(a+b)=sinacosb+cosasinb=√3/2(-1/2)+(-1/2)*(-√3/2)=0

It is proved that cos a + cos (120 + b) + cos (120-b) / (SINB + sin (120 + a) - sin (120-a) = Tan (a + b) / 2

First, remove the brackets from the left numerator and denominator, and then write the right side in the form of the ratio of sin and COS. Then we can use the method from result to cause to see the necessary conditions for the equation to be established, and then we can solve the problem

Given cosa = - 3 / 5, a belongs to (90 degrees, 180 degrees), find sin a / 2, cos A / 2, Tan A / 2

cosa=1-2sin^2a/2=-3/5
2sin^2a/2=8/5
sin^2a/2=4/5
sina/2=+-2*5^1/2/5
Ninety

Tan θ = (Sina COSA) / (Sina + COSA) find Sina cosa / sin θ

Tan θ = (Sina COSA) / (Sina + COSA) = (tana-1) / (Tana + 1) = Tan (a-45 °) θ = α - 45 ° or θ = 180 ° + α - 45 ° = α + 135 ° when θ = α - 45 °, Sina cosa / sin θ = Sina cosa / 1 / root 2 (sin α - cos α) = radical 2 when θ = α + 135 °, Sina cosa / sin θ = si

Given Sina + CoSb = 3 / 4, cosa + SINB = - 5 / 4, find sin (a + b)

2 + 2sinacosb + 2C is obtained from the following formula: Sina + CoSb = 3 / 4, ν (Sina + CoSb) Ω = 9 / 16, sin? 2A + 2sinacosb + cos? B = 9 / 16 (1)

Simplify sin (a + b) cosa-1 / 2 [sin (2a + b) - SINB]. Thank you

sin（a+B）cosa-1/2[ sin（2a+B）-sinB]
=sin（a+B）cosa-1/2[ sin（a+B+a）-sinB]
=sin(a+B)cosa-(1/2)[sin(a+B)cosa+cos(a+B)sina-sinB]
=(1/2)[sin(a+B)cosa-cos(a+B)sina+sinB]
=(1/2)[sin(a+B-B)+sinB]=sinB

Cosa * SINB = 1 / 2 [sin (a + b) - sin (a-b)]

You can prove it from right to left
Right = (1 / 2) [sin (a + b) - sin (a-b)]
=(1/2)[sinacosb+cosasinb-(sinacosb-cosasinb)]
=(1/2)*2cosasinb
=cosasinb
=Left,
So the equation holds