Question

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

Answer 1

`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!

Answer 2

`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