Given Tan, how to find sin and cos

Given Tan, how to find sin and cos

Formula: sin / cos = Tan, sin square + cos square = 1
Figure: if you know Tan, you will know the ratio of two right angle sides, so you can deduce the third side and find sin or cos

Given Tan θ = 2, find 2 + sin θ cos θ - cos θ ^ 2

2+sinθcosθ-cosθ^2
= (2sinθ^2 + cosθ^2 +sinθcosθ)/1
= (2sinθ^2 + cosθ^2 +sinθcosθ)/(sinθ^2 + cosθ^2)
= (2tanθ^2+1+tanθ)/(tanθ^2+1)
= (2*4+1+2)/(4+1)
=11/5

Given Tan θ = 3, find sin θ,

A:
tanx=3,sinx/cosx=3
sinx=3cosx
Substituting (SiNx) ^ 2 + (cosx) ^ 2 = 1, we get:
9(cosx)^2+(cosx)^2=1
cosx=±√10/10
sinx=±3√10/10
So:
sinx=3√10/10,cosx=√10/10
Or:
sinx=-3√10/10,cosx=-√10/10

Is Tan ^ 2 (x) equal to sin ^ 2 (x) / cos ^ 2 (x)?

Yes
Because TaNx = SiNx / cosx
So Tan & # 178; X = (SiNx / cosx) &# 178; = Sin & # 178; X / cos & # 178; X

If π / 4 is less than X and less than π / 2, then cos, sin and Tan are equal

cosX＜sinX＜tanX

Proof: Tan X / 2 = SiN x / (1 + cos x)

sin(x)=2sin(x/2)cos(x/2),cos(x)=2cos(x/2)^2 -1
On the right side of the equation, if x is not equal to (2k + 1) π, K is an integer, because cos (x / 2) cannot be 0

It is known that a is the third quadrant angle and f (a) = sin (π - x) cos (2 π - x) Tan (- x + π) / sin (π + x)
1. Simplify f (a)
2. If cos (A-3 π / 2) = 1 / 5, find f (a)

(1)
f(x)=sin(π-x)cos(2π-x)tan(-x+π)/sin(π+x)
=(sinx)(cosx)(-tanx)/(-sinx)
= cosxtanx
=sinx
(2)
cos(x-3π/2)=1/5
-sinx =1/5
f(x) = sinx = -1/5

It is known that x is the third quadrant angle and {f (x) = sin (π - x) cos (2 π - x) Tan (- x + 3 π / 2)} / {cotxsin (π + x)}
1. Simplify f (x)
2. If cos (x-3 π / 2) = 1 / 5, find the value of F (x)
3. If x = - 1860 °, find the value of F (x)

f(x)=sin(π-x)cos(2π-x)tan(-x+3π/2) }/{cotxsin(π+x)=f(x)=sinxcosxcotx }/-{cotxsinx}=cos^2x/-cosx=-cosxcos(x-3π/2)=1/5=-sinxsinx=-1/5cosx=2√6／5f(x)=-2√6／5x=-1860°f(x)=-cosx=-cos（-1860°）=-co...

Given f (x) = (sin (π - x) * cos (2 π - x)) / cos (- π - x) * Tan (π - x), then the value of F (- 31 π / 3) is

F (x) = (sin (π - x) * cos (2 π - x)) / cos (- π - x) * Tan (π - x) = (sin (π - x) * cosx) / (COS (- π - x + 2 π) * Tan (π - x) = (sin (π - x) * cosx) / (COS (π - x) * Tan (π - x) = Tan (π - x) * cosx) / Tan (π - x) = cosxf (x) is a periodic function, and the period of even function is 2 π f (- 31 π /

If f (x) = sin (π - x) cos (2 π - x) Tan (- x + 3 / 2 π) divided by cos (- π - x), then
What is the value of F (- 31 / 3 π)?

sin（π-x）=sinx,
cos（2π-x）=cosx,
tan（ -x+3/2π）=tan(π/2-x)=cotx=cosx/sinx,
cos（ - π- x）=cos(π-x)=-cosx,
Then f (x) = cos & # 178; X / - cosx = - cosx,
f（- 31/3 π）=-cos(-31/3π)=-cos(-π/3)=-cosπ/3=-1/2.