Tan(t) = x/a => sin(t) = x/sqrt( a^2 + x^2) ?

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Homework Help Overview

The discussion revolves around a trigonometric relationship derived from the tangent function in the context of a right triangle, specifically relating to the sine function. The original poster seeks clarification on the transition from tan(t) to sin(t) using the sides of a triangle.

Discussion Character

  • Exploratory, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the geometric interpretation of tangent in relation to a right triangle, with some attempting to derive the sine function from the given tangent relationship. There are questions about the reasoning behind the transformation from tan(t) to sin(t).

Discussion Status

Some participants have provided insights into the geometric setup of the problem and the application of the Pythagorean theorem. There appears to be a mix of understanding and exploration of the concepts involved, but no explicit consensus has been reached.

Contextual Notes

There is an emphasis on visualizing the problem through a right triangle, and the discussion includes references to the Pythagorean theorem as a critical component in understanding the relationships between the sides of the triangle.

TristanH
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tan(t) = x/a => sin(t) = x/sqrt( a^2 + x^2) ?

Folks,

In the solution guide to Kline's calculus book, he gives the following in working a problem:

tan(t) = x/a => sin(t) = x/sqrt( a^2 + x^2)

Can someone explain why this is the case?

Thanks,

Tristan
 
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tan(t)=Opposite/Adjacent. Draw a right-angled triangle with the side opposite to the angle ,t, as x and the adjacent side to it as a. Then use pythagoras' theorem.
 


I just figured this out on my own. For anyone who is curious:

tan(t) = x/a means the opposite and adjacent legs of a right triangle with angle t, are x and a respectively. By the Pythagorean Therom, this implies that the hypotenuse is sqrt(a^2 + x^2). Since sin t = opp/hyp, sin t = x/sqrt( a^2 + x^2). QED
 


rock.freak667 said:
tan(t)=Opposite/Adjacent. Draw a right-angled triangle with the side opposite to the angle ,t, as x and the adjacent side to it as a. Then use pythagoras' theorem.

TristanH said:
I just figured this out on my own. For anyone who is curious:

tan(t) = x/a means the opposite and adjacent legs of a right triangle with angle t, are x and a respectively. By the Pythagorean Therom, this implies that the hypotenuse is sqrt(a^2 + x^2). Since sin t = opp/hyp, sin t = x/sqrt( a^2 + x^2). QED

lol ?
 

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