cosα·tanα-Sinα

cosα·tanα-Sinα

cosa tana-sina
=cosa(sina/cosa)-sina
=sina-sina
=0

How to distinguish sin, cos and tan?
I don't want to know what it means. I just need to know how to distinguish it

Special angle trigonometric function value sin0 = 0, sin30 = 0.5, sin45 = half root 2, sin60 = half root 3, sin90 = 1, cos0 = 1, cos30 = half root 3, cos45 = half root 2, cos60 = 0.5, cos90 = 0, tan0 = 0

The relationship between sin α ± cos α and Tan

2
（sinα±cosα）=1±2sinαcosα
2 2
=1±( 2sinαcosα / sinα+cosα )
2 2 2 2
=1±（2sinαcosα /cosα）/ （sinα+cosα / cosα ）
two
=1± （2tanα/tanα+1）
=1± （-tan2α）
=1± tan2α

sin 0.＋cos 0.＋tan 0.

Sin0 = 0 cos0 = 1 tan0 = 0 so = 1

Simplification and calculation: given cos α = cos β * cos γ, find Tan [(α + β) / 2], Tan [(α - β) / 2]

tan[(α+β)/2]*tan[(α-β)/2]
=sin[(α+β)/2]*sin[(α-β)/2]/{cos[(α+β)/2]*cos[(α-β)/2]}
Sum difference by product
The original formula = (COS α - cos β) / (COS α + cos β)
And cos α / cos β = cos γ
therefore
The original formula = (COS γ - 1) / (COS γ + 1)
We also need to construct an equation
It's too late. Take a seat and go to sleep

sin3(-α)cos(2π+α)tan(- 2- π）

sin(3π-α)cos(2π+α)tan(-α- π）
=sinαcosα(-tanα)
=-sinαcosα(sinα/cosα)
=-sin²α

If sin ^ 4x + cos ^ 4x = 1, then sin θ + cos θ =?

sin^4x+cos^4x=（sin^2x+cos^2x）²-2sin^2xcos^2x=1-2sin^2xcos^2x=1∴2sin^2xcos^2x = 0sinxcosx =0（sinx+cosx）²= sin^2x+cos^2x+2sinxcosx=1+2sinxcosx=1∴sinx+cosx=1

Sin^4xʮcos^4x=

sin^4x-cos^4x=cos^2x-sin^2x=1-2sin^2x=cos4x

sin（x+π/4)=-3/5 sin2x=?
Sina = √ 3sinc B = π / 6 angle a =?

First of all, cos (2x + π / 2) = - sin2x, cos2x = 1-2 (SiNx) &#
So sin2x = - cos (2x + π / 2) = - Cos2 (x + π / 4) = 2 [sin (x + π / 4)] &# 178; - 1 = - 7 / 25
These formulas you will learn later, now use it first, come on!

What is sin (a-pai) equal to

sin(a－pai)=-sin(pai－a)=-sin(a)