0

When finding trigonometric ratios for 90+theta. Why don’t we make the diagram like this ? I see we are not getting an angle of 90 degree. So , can’t we say there is no values of trigonometric ratio possible for 90+theta.enter image description here

https://www.math-only-math.com/trigonometrical-ratios-of-90-degree-plus-theta.html. So , on this site I checked the proof for this. Below is the passage needed to understand this proof. What I didn’t get is how is sin theta = FE/OE

Take a point C on OA and draw CD perpendicular to OX or OX’.

Again, take a point E on OB such that OE = OC and draw EF perpendicular to OX or OX’.

From the right-angled ∆ OCD and ∆ OEF we get,

∠COD = ∠OEF [since OB ⊥ OA]

and OC = OE.

Therefore, ∆ OCD ≅ ∆ OEF (congruent).

Therefore according to the definition of trigonometric sign, OF = - DC, FE = OD and OE = OC

We observe that in diagram 1 and 4 OF and DC are opposite signs and FE, OD are either both positive.

Again we observe that in diagram 2 and 3 OF and DC are opposite signs and FE, OD are both negative.

According to the definition of trigonometric ratio we get, sin (90° + θ) = FE/ OE

Blue
  • 75,673
Rider
  • 195
  • Doesn't the reference also say sin (90° + θ) =FE/OE? – Star Bright Apr 13 '21 at 17:31
  • "Why don’t we make the diagram like this?" Um ... You got the image from a site that does make the diagram like that, so what's your point? :) ... Instructors are allowed to use different approaches to trig (or any topic). I choose to start w/right-triangle trig for acute angles, deriving as much lore (relations and identities) as possible. I then let the lore drive exploration beyond the bounds of the First Quadrant (see, eg, this answer) before ever introducing the Unit Circle. This serves a narrative of math as a journey of discovery. – Blue Apr 13 '21 at 17:31
  • @Blue I drew a blue colour trifle using edit I,e triangle EOD – Rider Apr 13 '21 at 17:32
  • @BrightStar The perpendicular for triangle EOD is ED. But yes , there is no 90 degree angle. FE /OE means of triangle EFO. That is not where the 90 is theta angle is present – Rider Apr 13 '21 at 17:35
  • 1
    @Blue The link and your thinking is great. – Rider Apr 13 '21 at 17:36
  • 1
    @Blue Thanks a lot. – Rider Apr 13 '21 at 17:37
  • @Rider Sorry that I do not understand your comment "The perpendicular for triangle EOD is ED". In $\triangle EFO, \sin \angle EOF=\sin (90^\circ +\theta)$. All you said in your question is correct, except that "What I didn’t get is how is sin theta = FE/OE" - where does this "sin theta = FE/OE" come from? If you can clarify this, that would be helpful. – Star Bright Apr 13 '21 at 19:15
  • @BrightStar It was in the online site. I also didn’t get how it is FE/OE – Rider Apr 13 '21 at 19:40
  • The online site said this: sin (90° + θ) = FE/OE. – Star Bright Apr 13 '21 at 19:48
  • @BrightStar oh yes. How is sin (90+theta) = FE/OE. . Very sorry for such a typo. Must have disturbed you for no reason. I extremely apologise – Rider Apr 13 '21 at 19:49
  • OK, I understand now:-). Let me write up an answer. – Star Bright Apr 13 '21 at 19:51
  • Please read page 9 of the linked document for general definition of sine function. – Star Bright Apr 13 '21 at 20:46

1 Answers1

0

enter image description here

We need to extend the definitions of trig function from special case in a right triangle to general case in a circle.

In the above graph, \begin{align}\angle COD=\theta, \angle EOC=90^\circ,\angle EOD=90^\circ+\theta \end{align}

By definition, \begin{align} \sin \angle EOD=\sin (90^\circ +\theta)=\frac{FE}{OE} \end{align}

Star Bright
  • 2,338