Cos (pai-a) equals - 1 / 4, find sin (2 pai-a)

Cos (pai-a) equals - 1 / 4, find sin (2 pai-a)

From α - s) = π
Cos α = 1 / 4
Therefore, sin α = ± √ (1-cos 2 α) = ± (√ 15) / 4
sin(2π-α)
=sin(-α)
=-sinα
=±(√15)/4

Is cos (& - 3 / 2) equal to sin & or - Sin &?

Cos (& - 3 / 2) = cos (3 / 2 - &)
The cosine in the third quadrant is negative
So cos (& - 3 / 2) = cos (3 / 2 - &)
=-sin&

If sin (α - β) sin β - cos (α - β) cos β = 4 / 5 and α is the second quadrant angle, what is Tan (4 / 4 + α) equal to

sin(α-β)sinβ-coscos=4/5
-cos[(α-β)+β]=4/5
Cos α = - 4 / 5 α in the second quadrant
sinα=3/5
tanα=-3/4
tan(π/4+α)=(tanα+tanπ/4)/[1-tanαtanπ/4]
=(1-3/4)/(1+3/4)
=1/7

What is cos ^ 2 θ - Sin ^ 2 θ equal to? Why?

Cos2 θ this is the formula!

Why is △ y = sin (x + △ x) - SiNx equal to 2Sin (△ X / 2) cos [x + (△ X / 2)] Please give specific reasons,

This is the formula of sum difference product
sina-sinb=2[sin(a-b)/2]*[cos(a+b)/2]

It is known that sin (π - α) = - 2Sin (π) 2 + α), then sin α· cos α =___ .

∵ sin (π - α) = sin α, sin (π 2 + α) = cos α,  sin (π - α) = - 2Sin (π 2 + α) is transformed into: sin α = - 2cos α ①,

How is 2Sin α * cos α equal to sin (2 α)? As the title

sin(2α)=sin(α+α)=sinαcosα+cosαsinα=2sinα*cosα

The simplification of sin (π / 2 + 2x) = 2Sin (π / 4 + x) cos (π / 4 + x)

This is the use of the double angle formula
sin2A=2sinAcosA
Change a to π / 4 + X
The results show that sin (π / 2 + 2x) = 2Sin (π / 4 + x) cos (π / 4 + x)

How to deform y = (sin ^ 2 (x / 4) + cos ^ 2 (x / 4)) ^ 2-2sin ^ 2 (x / 4) cos ^ 2 (x / 4) to (3 / 4) + (1 / 4) cosx?

y=((1-cosx/2)/2+(1+cosx/2)/2)^2-2((1-cosx/2)/2)((1+cosx/2)/2)=(1)^2-1/2 * (1-cosx/2)(1+cosx/2)=1-1/2 * (1-cos^2(x/2))=1-1/2+1/2 * (1+cosx)/2=1/2+1/4+(1/4)cosx=3/4+(1/4)cosx

If f (x) = sin α - cosx, then f '(α) is equal to () A. sinα B. cosα C. 2sinα D. sinα+cosα

f′（x）=sinx
So f '(α) = sin α
Therefore, a