How do you find the exact values of cot, csc and sec for 45 degrees?

2 Answers
Jun 26, 2018

cot45^@=1 color(white)(2/2)csc45^@=sqrt2color(white)(2/2)sec45^@=sqrt2

Explanation:

Recall the definitions of the reciprocal trig identities:

cotx=1/color(steelblue)(tanx) color(white)(2/2) cscx=1/color(red)(sinx) color(white)(2/2) secx=1/color(purple)(cosx)

On the unit circle, the coordinates for 45^@ are

(sqrt2/2,sqrt2/2)

Where the x coordinate is the cos value, and the y coordinate is the sin value.

tanx is defined as sinx/cosx. From this information, we know

color(steelblue)(tanx=1) color(white)(2/2) color(red)(sinx=sqrt2/2) color(white)(2/2) color(purple)(cosx=sqrt2/2)

We can plug these values into our expressions for the reciprocal identities to get

cotx=1/color(steelblue)(1)=1 color(white)(2/2) color(red)(cscx=1/(sqrt2/2)=sqrt2) color(white)(2/2) color(purple)(secx=1/(sqrt2/2)=sqrt2)

Remember for csc45^@ and sec45^@, we had to rationalize the denominator.

bar (ul (| color(white)(2/2) cot45^@=1 color(white)(2/2)csc45^@=sqrt2color(white)(2/2)sec45^@=sqrt2 color(white)(2/2) |) )

Hope this helps!

Jun 27, 2018

cot45^@ = 1/1
" "csc45^@ = sqrt(2)/1
" "" "sec45^@ = sqrt(2)/1

Explanation:

Draw a 45^@ right triangle. A 45^@ right triangle has 2 equal sides plus the hypotenuse, right? If you give the 2 equal sides a length of 1 unit, Pythagoras will tell you that the hypotenuse has a length of sqrt(2).

Label the lengths of the sides and the degrees of the angles. Keep this drawing handy.

You should have these memorized:
sine is opposite/hypotenuse
cosine is adjacent/hypotenuse
tangent is opposite/adjacent

This question is asking for cot, csc and sec. Memorize these facts too:
cosecant is 1/sine
secant is 1/cosine
cotangent is 1/tangent

So thinking about the above things in bold,
cot45^@ = "adjacent/opposite"
csc45^@ = "hypotenuse/opposite"
sec45^@ = "hypotenuse/adjacent"

Now look at your drawing of the 45^@ right triangle and plug in the lengths specified in the above fractions for each trig function.

I hope this helps,
Steve