Find sin2x, cos2x, and tan2x if tanx=-2 and x terminates in quadrant iv​

easy

Step-by-step explanation:

Find sin2x, cos2x, and tan2x if cos=4/5 and x terminates in quad iv.

cosx=4/5

sinx=-3/5

tanx=-3/4

x in quadrant IV where cos>0, sin<0, and tan<0

..

Identities:

sin2x=2sinxcosx=2*(-3/5)(4/5)=-24/25

cos2x=1-2sin^2x=1-2*(-3/5)^2=1-18/25=7/25

tan2x=(2tanx)/(1-tan^2x)=(2*(-3/4)/(1-(-3/4)^2)=(-6/4)/(1-(9/16))=(-24/16)/(7/16)=-24/7

cosx=4/5 (in Q4)

x≈360-36.87=323.13

2x=646.26

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tan2x=tan(646.26)≈-3.4246..

Exact value=-24/7≈-3.4286

..

cos2x=cos(646.26)≈0.2799..

Exact value=7/25≈0.28

..

sin2x=sin(646.26)≈-0.9600

Exact value=-24/25≈-0.9600..

sin(2x) = 2 sin(x) cos(x) = 2 ( -8/17)(15/16) = -240/289

cos(2x) = 1 - 2* sin^2(x) = 1 - 2 (-8/17)^2 = 1 - 2(64/289) = 1 - 128/289 = 161/289

tan(2x) = 2 tanx / (1-tan^2(x)) =   2 * (-8/15) /   ( 1 -  (-8/15)^2)

                                               = -16/15 / ( (225 - 64)/225

                                                 =16/15 * 225/161 = 240/161-- Answer

one that is bigger than 3 1/2

function 2 has a greater rate of change than function 1

function 2 has a greater y-intercept than function 1

step-by-step explanation: