Homework Help

Prove sinx=cos(90-x) and tanx=1/tan(90-x) in geometry's way,and prove tanx=sinx/cosx in...

user profile pic

levvy | eNoter

Posted October 28, 2013 at 12:09 AM via iOS

dislike 1 like

Prove sinx=cos(90-x) and tanx=1/tan(90-x) in geometry's way,and prove tanx=sinx/cosx in algebra's way, thank you so much!!!^_^

Tagged with math

1 Answer | Add Yours

user profile pic

embizze | High School Teacher | (Level 1) Educator Emeritus

Posted October 28, 2013 at 2:24 AM (Answer #1)

dislike 0 like

Consider right triangle `Delta ABC` with acute angles A and B. Note that A and B are complements; that is the measure of angle A equals 90 minus the measure of angle B.

Now by right triangle trigonometry the sine of an angle is defined to be the ratio of the leg opposite the angle to the hypotenuse, while the definition of the cosine of an angle is the ratio of the leg adjacent to the angle to the hypotenuse.

Thus `sinA=a/c` and `cosB=a/c` . But B=90-A so `sinA=cos(90-A)` as required.

The tangent of an angle is defined as the leg opposite the angle to the leg adjacent to the angle. `tanA=a/b,tanB=b/a` . Then `tanA=1/(tan(B))=1/(tan(90-A))` as required.

------------------------------------------------------------------

Show `tanx=(sinx)/(cosx)` :

Using the set-up above:

`tanA=a/b=((a/c))/((b/c))=(sinA)/(cosA)`

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes