It is mainly a process to find the root of Quarter 2 × sin (π / 4 - x) + 6 × cos (π / 4 - x) by using the formula of transformation one

It is mainly a process to find the root of Quarter 2 × sin (π / 4 - x) + 6 × cos (π / 4 - x) by using the formula of transformation one

It is suggested that the second part of the root sign is as follows: the second part of the root sign is two × (one half of sin (π / 4 - x) + the second part of the root sign 3 × cos (π / 4 - x)) = the second part of the root sign two × (COS π / 6 × sin (π / 4 - x) + sin π / 6 × cos (π / 4 - x)) = two part root two × sin (π / 4 - x + π / 6) = half root

It is known that π in 4 is less than 3 in 4, 0 is less than π in 4, cos α = 5 / 13, sin β = 5 / 13 Help me

π/4＜α＜3π/4
cosα=5/13
Sin α = radical (1-cos ^ 2 α) = 12 / 13
0＜β＜π/4
sinβ=5/13
Cos β = radical (1-sin ^ 2 β) = 12 / 13
sin(α+β)=sinαcosβ+cosαcosβ
= 12/13 * 12/13 + 5/13 * 5/13
= 1

The known quarter Math homework help users 2017-10-25 report Use this app to check the operation efficiently and accurately!

sin2a
=sin<(a-b)+(a+b)>
=sin(a-b)cos(a+b)+sin(a+b)cos(a-b)
=5 / 12 times - 4 / 5-3 / 5 times 12 / 13
=-173/195
Should Right, at least the idea must be right

Given cos (π / 4-A) = 3 / 5, sin (5 π / 4 + b) = - 12 / 13, a ∈ (π / 4,3 / 4 π), B ∈ (0, π / 4), find the value of sin (a-b)

If cos (π / 4-A) = 3 / 5, sin (5 π / 4 + b) = - 12 / 13, a ∈ (π / 4,3 / 4 π), B ∈ (0, π / 4), then
π/4-a∈(-π/2,0),
sin(π/4-a)=-4/5
5π/4+B∈(5π/4,3π/2),
cos(5π/4+B)=-5/13
sin(a-B)=cos(π/2-a+B)=-cos(3π/2-a+B)=-cos[（π/4-a）+(5π/4+B)]=-[cos(π/4-a)cos(5π/4+B)-sin(π/4-a)sin(5π/4+B)]=-[（3/5）*（-5/13）-（-4/5）*（-12/13）]=63/65

α, β∈ (3 π) is known 4，π)，sin(α+β)＝−3 5，sin(β−π 4)＝12 13, then cos (α + π) 4)=______ .

α, β∈ (3 π) is known
4，π)，sin(α+β)＝−3
5，
sin(β−π
4)＝12
13，α+β∈(3π
2，2π)，β−π
4∈(π
2，3π
4)，
∴cos(α+β)＝4
5，cos(β−π
4)＝−5
13，
∴cos(α+π
4)=cos[(α+β)−(β−π
4)]
=cos(α+β)cos(β−π
4)+sin(α+β)sin(β−π
4)
=4
5•(−5
13)+(−3
5)•12
13＝−56
Sixty-five
So the answer is: - 56
Sixty-five

① It is known that cos (π / 4-A) = 3 / 5, sin (3 π / 4 + b) = 5 / 13

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

It is known that cos (π / 4 + α) = - 3 / 5, sin (3 π / 4 + β) = 5 / 13 and PI / 4

Pie / 4

Given cos (π / 4-A) = 3 / 5, sin (5 π / 4 + b) = - 12 / 13, a ∈ (π / 4,3 π / 4) B ∈ (0, π / 4), find the value of sin (a + b)

a∈(π/4,3π/4)
π/4-a∈(-π,0)
sin(π/4-a)0
sin²(π/4+b)+cos²(π/4+b)=1
cos(π/4+b)=5/13
sin(a+b)=sin[(π/4+b)-(π/4-a)]
=sin(π/4+b)cos(π/4-a)-cos(π/4+b)sin(π/4-a)
=16/65

Given that 0 ＜ B ＜ π / 4 ＜ a ＜ 3 / 4 π, cos (π / 4-A) = 4 / 5, sin (3 π / 4 + b) = 5 / 13, the value of sin (a-b) is calculated

﹤ 0 ﹤ B ﹤ π / 4 ﹤ a ﹤ 3 ﹤ 4 ﹤ 0 ﹤ B \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\) - [B + (3 π / 4)] + π} = - sin {[a - (π

Is sin α + cos α = √ 2Sin (α + π / 4) a formula?

It's the formula, the auxiliary angle formula
acosA+bsinA=√(a^2+b^2)sin(A+M) (tanM=a/b)