Question

PLEASE HELP IM DESPERATE precalculus. also 50 points!! please god help

Answer 1

Answer:

For B thru F these options will vary but here how you do it

B. Step 1 Draw the 4 Quadrants.

Then Draw the Triangle in the lower right quadrant which we call quadrant 4. Label the X axis as Adjacent and positive. Label the Y axis as Opposite and negative. Label the Slanted side as the hypotune and AS POSITVE SINCE HYPOTENUSE IS ALWAYS POSITIVE.

FOR C. IN QUADRANT 2, PLOT A POINT AT 0,12 AND AT (-5,0). CONNECT THE DOTS AND IT FORMS A TRIANGLE. Label the X axis as adjacent and negative and the y axis as positve and opposite and label the slanted side hypotunese and positive.

FOR D Draw a straight line along the x axis then draw a slanted line passing through (5,-1). In between them put the theta symbol in there.

The labeling is the same for C.

For E. Since tan must be positve and secant must be positve, our triangle must be in the 1st Quadrant. Draw any right triangle as long it is in the first quadrant

The x axis is adjacent and positve. The y axis is opposite and positve. The hypotenuse is the slanted side and it is positve.

For F. Since sin is negative and cos is positve the triangle is in the 4th quadrant. Draw any triangle in the 4th quadrant and the labeling is the same for Problem B.

2. We can find the sec of cos by flipping cosine.

$\cos( \frac{x}{y} ) = \sec( \frac{y}{x} )$

$\cos( \frac{1}{2} ) = \sec(2 )$

Sec is 2.

To find the cotangent, first let find the sin then tan.

We can use the identity

$\cos( {theta}^{2} ) + \sin( {theta}^{2} ) = 1$

Let plug in the number

$\cos( \frac{ {1}^{2} }{{2}^{2} } ) + \sin(x {}^{2} ) = 1$

$\cos( \frac{1}{4} ) + \sin(x {}^{2} ) = 1$

$\sin(x {}^{2} ) = 1 - \frac{1}{4}$

$\sin(x {}^{2} ) = \frac{3}{4}$

$\sin(x) = \frac{ \sqrt{3} }{ \sqrt{4} }$

$\sin(x) = \frac{ \sqrt{3} }{2}$

Since sin is negative, sin x=

$- \frac{ \sqrt{3} }{2}$

Now let apply the formula

$\frac{ \sin(x) }{ \cos(x) } = \tan(x)$

$\frac{ \frac{ - \sqrt{3} }{2} }{ \frac{1}{2} } = \tan(x)$

$- \sqrt{3}$

Now let find cotangent we can the reciprocal of

tan.

$\tan= - \sqrt{3}$

$\cot = - \frac{1}{ \sqrt{3} }$

Rationalize denominator

$\frac{ - 1}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } = \frac{ \sqrt -{ 3} }{3}$

cotangent equal

$- \frac{ \sqrt{ 3} }{3}$