How do you evaluate #tan 27°#?

1 Answer
George C.
Dec 2, 2017

#tan 27^@ = sqrt(5)-1 - sqrt(5 - 2 sqrt(5))#

Explanation:

I like this question because it highlights the fact that you can find algebraic expressions for trigonometric functions for any multiple of #3^@#. The same is not true of integer number of degrees that are not multiples of #3#.

Some typical values that you are probably familiar with are:

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

#sin 30^@ = cos 60^@ = 1/2#

#sin 60^@ = cos 30^@ = sqrt(3)/2#

Some that you may be less familiar with are:

#sin 18^@ = cos 72^@ = 1/4(sqrt(5)-1)#

#sin 72^@ = cos 18^@ = 1/4 sqrt(10+2sqrt(5))#

We can find trigonometric values for other angles by using half angle formulas or formulas for sine and cosine of sums and differences.

In our particular example, we have:

#sin 27^@ = sin (45^@ - 18^@)#

#color(white)(sin 27^@) = sin 45^@ cos 18^@ - sin 18^@ cos 45^@#

#color(white)(sin 27^@) = sqrt(2)/2 1/4 sqrt(10+2sqrt(5)) - 1/4(sqrt(5)-1) sqrt(2)/2#

#color(white)(sin 27^@) = sqrt(2)/8 (sqrt(10+2sqrt(5)) - sqrt(5)+1)#

#cos 27^@ = cos (45^@ - 18^@)#

#color(white)(cos 27^@) = cos 45^@ cos 18^@ + sin 45^@ sin 18^@#

#color(white)(cos 27^@) = sqrt(2)/2 1/4 sqrt(10+2sqrt(5)) + sqrt(2)/2 1/4 (sqrt(5)-1)#

#color(white)(cos 27^@) = sqrt(2)/8 (sqrt(10+2sqrt(5)) + sqrt(5)-1)#

So:

#tan 27^@ = sin 27^@/cos 27^@#

#color(white)(tan 27^@) = (sqrt(2)/8 (sqrt(10+2sqrt(5)) - sqrt(5)+1))/(sqrt(2)/8 (sqrt(10+2sqrt(5)) + sqrt(5)-1))#

#color(white)(tan 27^@) = (sqrt(10+2sqrt(5)) - sqrt(5)+1)/(sqrt(10+2sqrt(5)) + sqrt(5)-1)#

#color(white)(tan 27^@) = ((sqrt(10+2sqrt(5)) - sqrt(5)+1)^2)/((sqrt(10+2sqrt(5)) + sqrt(5)-1)(sqrt(10+2sqrt(5)) - sqrt(5)+1))#

#color(white)(tan 27^@) = ((10+2sqrt(5))-2(sqrt(5)-1)sqrt(10+2sqrt(5)) + (sqrt(5)-1)^2)/((10+2sqrt(5)) - (sqrt(5)-1)^2)#

#color(white)(tan 27^@) = ((10+2sqrt(5))-2(sqrt(5)-1)sqrt(10+2sqrt(5)) + (6-2sqrt(5)))/((10+2sqrt(5)) - (6-2sqrt(5)))#

#color(white)(tan 27^@) = (16-2(sqrt(5)-1)sqrt(10+2sqrt(5)))/(4(sqrt(5)+1))#

#color(white)(tan 27^@) = ((16-2(sqrt(5)-1)sqrt(10+2sqrt(5)))(sqrt(5)-1))/(4(sqrt(5)+1)(sqrt(5)-1))#

#color(white)(tan 27^@) = 1/16(16-2(sqrt(5)-1)sqrt(10+2sqrt(5)))(sqrt(5)-1)#

#color(white)(tan 27^@) = sqrt(5)-1 - sqrt(5 - 2 sqrt(5))#

I have skipped a few steps at the last hurdle, due to the length of the answer, but the last two expressions are equivalent.

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