How would you find the exact value of the six trigonometric function of -45?

1 Answer
George C.
Sep 29, 2016

sin(-45^@) = -sqrt(2)/2

cos(-45^@) = sqrt(2)/2

tan(-45^@) = -1

sec(-45^@) = sqrt(2)

csc(-45^@) = -sqrt(2)

cot(-45^@) = -1

Explanation:

First let's look at sin 45^@ and cos 45^@

Consider the right angled triangle formed by cutting a 1xx1 square in half diagonally...

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Recall that:

sin theta = "opposite"/"hypotenuse"

cos theta = "adjacent"/"hypotenuse"

Hence we find:

sin 45^@ = cos 45^@ = 1/sqrt(2) = sqrt(2)/2

Next note that sin is an odd function and cos an even function.

Hence:

sin(-45^@) = -sin 45^@ = -sqrt(2)/2

cos(-45^@) = cos 45^@ = sqrt(2)/2

Hence we can find the other 4 trigonometric functions:

tan(-45^@) = sin(-45^@)/cos(-45^@) = (-sqrt(2)/2) / (sqrt(2)/2) = -1

sec(-45^@) = 1/cos(-45^@) = 1/(sqrt(2)/2) = 2/sqrt(2) = sqrt(2)

csc(-45^@) = 1/sin(-45^@) = 1/(-sqrt(2)/2) = -2/sqrt(2) = -sqrt(2)

cot(-45^@) = cos(-45^@)/sin(-45^@) = (sqrt(2)/2)/(-sqrt(2)/2) = -1

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