Math Is Fun Forum

lcbuys

Member

Registered: 2007-05-26

Posts: 2

trig functions

Let t be the angle between o and pi/zero such that sin=1/4. Then cos t=_______ sin(-t)_________cos (-t)_______
I understand that it takes place in quadrant one I just don't understand how to even start the problem.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: trig functions

If I'm interpreting this right, you want to find constants a and b such that cos t = asin (-t) * bcos(-t).

The two constants will just multiply together, so you actually only need one: cos t = ksin (-t) * cos(-t).

Some useful rules of trigonometry are that sin t = -sin (-t) and that cos t = cos (-t).
Therefore, sin (-t) * cos (-t) = -sin t * cos t.

Using this in the original equation gives that cos t = -k sin t cos t.
Simplify and rearrange: k = -1/sin t.

We're told that sin t = 1/4, so k = -1/(1/4) = -4.

And if you really want two constants, just pick any two that multiply to give that. 2 and -2 will work.

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Why did the vector cross the road?
It wanted to be normal.

harman23

Member

Registered: 2007-05-26

Posts: 2

Re: trig functions

i think it is right answer

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: trig functions

Just incase that's not what he meant, I'll give an alternate answer.
Inv sin of 0.25 is 14.47 degrees on your calculator.
cos 14.47 degrees = 0.968245836
sin -14.47 degrees = -1/4
cos -14.47 degrees = 0.968245836
Sine is vertical height in a unit circle at that angle where the slant hits the circle.
Cosine is the horizontal run in a unit circle at that angle where the slant hits the circle.
Measured from the origin going up or right until you are level with these points on the circle.
If you are going right then when you are lined up, up-and-down with the point on the circle.

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