justifying a trig for inequality (1 Viewer)

[ZHANDONG]

why is cos(𝜋−2)=|cos(2)|, when the LHS is (-ve) and the RHS is (+ve)?

 

[liamkk112]

why is cos(𝜋−2)=|cos(2)|, when the LHS is (-ve) and the RHS is (+ve)?

pi-2 is 1.14, cos1.14 is positive, so lhs isn’t negative (maybe ur calculator was in degrees not rads

 

[ZHANDONG]

pi-2 is 1.14, cos1.14 is positive, so lhs isn’t negative (maybe ur calculator was in degrees not rads

but cos(pi-theta)=-cos(theta)

 

[imagineee]

it isn't

 

[ZHANDONG]

it isn't

updated

 

[Drongoski]

it isn't

It is. It is always true that:



whatever the value of theta.

 

[imagineee]

It is.

original post is edited

 

[ZHANDONG]

ig the justification for |cos(2)|=cos(pi-2) is that the RHS angle is in the 1st quadrant, meaning its cosine is +ve, and evaluating it is exactly the same as it were for |cos(2)|. does my expo make sense??

 

[Nav123]

updated

(1)


so its in the first quadrant

or using your logic:



but from 1

 

[Drongoski]

ig the justification for |cos(2)|=cos(pi-2) is that the RHS angle is in the 1st quadrant, meaning its cosine is +ve, and evaluating it is exactly the same as it were for |cos(2)|. does my expo make sense??

Yes!

For someone doing MX2, your grasp of the fundamentals of trigonometry needs to be better.

 

[ZHANDONG]

Yes!

For someone doing MX2, your grasp of the fundamentals of trigonometry needs to be better.

maybe but I figured it out immediately after