SOLUTION: For an angle a in the interval [0,π2) it is given that sin(a)=4/5 Determine cos(a) exactly. From my understanding, the angles is in the first quadrant (0 -> 90 deg), but I don

Algebra ->  Trigonometry-basics -> SOLUTION: For an angle a in the interval [0,π2) it is given that sin(a)=4/5 Determine cos(a) exactly. From my understanding, the angles is in the first quadrant (0 -> 90 deg), but I don      Log On


   

Question 1176366: For an angle a in the interval [0,π2) it is given that sin(a)=4/5
Determine cos(a) exactly.
From my understanding, the angles is in the first quadrant (0 -> 90 deg), but I don't understand, it there a formula I can use?
Found 2 solutions by htmentor, Solver92311:
Answer by htmentor(1343) About Me  (Show Source):
You can put this solution on YOUR website!
Consider the unit circle, with radius = 1. A point on the unit circle has
coordinates (x,y). Then sin(a) = y/r = y/1 = y, and cos(a) = x/r = x.
From the Pythagorean theorem, we know that x^2 + y^2 = r^2 = 1
If sin(a) = 4/5, we can find cos(a) which is equal to x:
1 - (4/5)^2 = x^2 -> x = sqrt(9/25) = 3/5
cos(a) = 3/5

Answer by Solver92311(821) About Me  (Show Source):
You can put this solution on YOUR website!




The sine is positive in QI and QII which means that there are two possible arguments for the function that have a positive sine value, but the cosine is positive in QI and negative in QII. So, as posed the question has two answers as indicated by the in the formula.

The formula is a consequence of the Pythagorean Identity:

Let's say that . We square both sides of this equation to obtain:



Applying the Pythagorean Identity:



A little algebra music, Maestro:



Taking the square root (remembering to account for positive and negative roots):



But we know from earlier that we can substitute to get the desired formula.



Further, given the symmetrical nature of the Pythagorean Identity, it should be intuitively obvious that the following formula holds as well:



The following will help you choose which sign to affix based on your knowledge of the Quadrant location of the function argument: 

Quadrant  Sine Cosine Tangent

   I        +     +      +

   II       +     -      -

   III      -     -      +

   IV       -     +      -


Also, secant has the same sign as cosine and cosecant has the same sign as sin.


John

My calculator said it, I believe it, that settles it

From
I > Ø