How Sin(90° + θ) =sin θ of triangle P'OM'

[Yogesh Tripathi]

Hello everyone. I'm learning Trigonometry right now with myself and at current about how to find the trigonometric ratio of the angle (90° + θ) in terms of θ. I'm quite confused in the Figure. How sin(90° + θ) become equal to sin (90° - θ) of triangle P'OM'. I know that triangle P'OM' and POM are congruent.

 

[blue_leaf77]

$\sin(90^o+\theta) = \cos \theta = \cos(-\theta) = \sin(90^o-\theta)$.

 

[UncertaintyAjay]

Have you heard of a unit circle? Look it up.

 

[Yogesh Tripathi]

Have you heard of a unit circle? Look it up.

yes I know about it. But I didn't able to understand how to find the trigonometric ratio of the angle (90° + θ) in terms of θ from it.

 

[Yogesh Tripathi]

Okay, so for any point on the unit circle: Draw a line from the point to the centre of the circle (0,0) then call the angle that line makes with the x-axis θ. So sin θ given by the y co-ordinate of that point. An angle of 90+θ is what you get when you reflect the above scenario across the y axis. So the y co-ordinate of your point is still the same. If the point was originally (x,y) it is now (-x,y). Since the y co-ordinate represents the sine of the angle and its equal in both cases, sin(90+θ) = sinθ.

Thankyou for the answer. Can you please show me this with figure please. It will help me more.

 

[UncertaintyAjay]

Sorry. Just a minute. I do believe I am wrong. I mistook 90 +θ for 180-θ and said it's equal to θ it's not. Just realized. Really, Really sorry. I have deleted that post. Read what follows.

But, basically you can still do it using a unit circle. For your θ, sin 90+θ is if move over θ from the y axis. So, sin 90+θ becomes (-y,x) [ Remember that (x,y) is your original point.] So if you reflect that across the y-axis you get (y,x) which is sin 90-θ. The following image shows the relationship between θ and 90+θ

And this shows the relationship between θ and 90-θ. So you can see from both these images what I'm talking about.

 

[Yogesh Tripathi]

Sorry. Just a minute. I do believe I am wrong. I mistook 90 +θ for 180-θ and said it's equal to θ it's not. Just realized. Really, Really sorry. I have deleted that post. Read what follows.

But, basically you can still do it using a unit circle. For your θ, sin 90+θ is if move over θ from the y axis. So, sin 90+θ becomes (-y,x) [ Remember that (x,y) is your original point.] So if you reflect that across the y-axis you get (y,x) which is sin 90-θ. The following image shows the relationship between θ and 90+θ

View attachment 94755
And this shows the relationship between θ and 90-θ. So you can see from both these images what I'm talking about.
View attachment 94758

Sorry. Just a minute. I do believe I am wrong. I mistook 90 +θ for 180-θ and said it's equal to θ it's not. Just realized. Really, Really sorry. I have deleted that post. Read what follows.

But, basically you can still do it using a unit circle. For your θ, sin 90+θ is if move over θ from the y axis. So, sin 90+θ becomes (-y,x) [ Remember that (x,y) is your original point.] So if you reflect that across the y-axis you get (y,x) which is sin 90-θ. The following image shows the relationship between θ and 90+θ

View attachment 94755
And this shows the relationship between θ and 90-θ. So you can see from both these images what I'm talking about.
View attachment 94758

How does it proof that angle Sin(90+θ) = sin(90-θ) of triangle P'OM'?

 

[UncertaintyAjay]

your stuff is quite similar to the unit circle. Look at the first diagram on the left in the image you posted .Angle AOP is θ. So angle POP' is 90+θ. So if you reflect triangle P'OM across the y axis, then angle AOP' will be 90-θ . It's easier to see that the two are equal if OP, OP' etc are the radii of a unit circle. Hence the unit circle.