Pythagoream Identities

Question

Use Pythagorean identities to simplify.'

A. Find sin x if cot x= sqrt3/2 and cos x < 0

B. Find cot x if cos x= 1/7 and sin x <0

Answer 1

cotx=sqr3/2 = adjacent side/opposite side cosx <0 means it's in quadrant III, the only quadrant where cotx>0 and cosx<0. In quadrant III, sinx<0, -1<sinx<0

sinx = opposite side/hypotenuse. cotx = adjacent side/opposite side =sqr3/2 = -sqr3/-2

opposite side squared = hypotenuse squared -minus adjacent side squared,

4=h^2-3

h^2 = 7

h =sqr7

sinx =-2/sqr7 = -2sqr7/7 = (-2/7)sqr7 = about - 0.756

B) cosx=1/7, sinx<0. It's quadrant IV

cosx = adjacent over hypotenuse. cotx = opposite/hypotenuse

sin= opposite/hypotenuse

hypotenuse =7

sinx= -sqr(49-1)/7 = -sqr48/7 = about -0.99

Answer 2

The problem presents clues as to how to determine the answer.

Where is cosine < 0, that is negative?

And

Where is tangent > 0, that is positive?

This tells you in which quadrant tangent it.

The next is to draw and label a diagram of the tangent.

Once you have the two legs of the triangle, use Pythagoras to determine the hypotenuse. Then you can determine the sine

The same kind of sleuthing is done for the second one.

Answer 3

The key here is to manipulate the equations given to you so that you can use the Pythagorean identities to solve them. For the first one, we will use the Identity the contains cot(x)

1 + cot²(x) = csc²(x)

We start with what is given by the problem,

cot(x) = √3/2

Plug this into the Identity to get,

1 + (√3/2)² = csc²(x)

1 + (3/4) = csc²(x)

7/4 = csc²(x)

Now, remember that csc²(x) = 1 / sin²(x)

7/4 = 1 / sin²(x)

4/7 = sin²(x) (reciprocal)

√(4/7) = sin(x) (square root both sides)

[because cos(x) < 0 and cot(x) >, we know sin(x) < 0, so we can discard the positive solution]

-2/√7 = sin(x)

-(2√7)/7 = sin(x) (rationalize)

You can use the same process to solve the second problem, but use the Pythagorean Identity involving cos(x).