Without using a calculator evaluate #sin45^o/(cos30^o +sin60^o#?

1 Answer
Feb 15, 2018

#(sqrt(6))/6#

Explanation:

Soh Cah Toa

Soh #->" sin = " ("opposite")/("hypotenuse")#

Cah #->"cos = "("adjacent")/("hypotenuses")#

Toa #-> "tan = "("opposite")/("adjacent")#

Tony B

Given: #(sin(45^o))/(cos(30^o)+sin(60^o)) #

Note that #sin(45^o)=1/sqrt(2)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider the #cos(30^o)+sin(60^o)#

Looking at the diagram we have

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

so #1/(cos(30^o)+sin(60^o))=1/sqrt(3)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Putting it all together we have:

#1/sqrt(2)xx1/(sqrt(3))color(white)("d") -> color(white)("d")1/(sqrt(2)sqrt(3))#

#color(white)("dddddddddd")->color(white)("d")1/(sqrt(6))#

It is considered bad mathematical practice to have a root as or in the denominator if avoidable

Multiply by 1 and you do not change the value. However, one comes in many forms.

#color(white)("dddddddddd")->color(white)("d")1/(sqrt(6))xx1#

#color(white)("dddddddddd")->color(white)("d")1/(sqrt(6))xx(sqrt6)/(sqrt6)#

#color(white)("dddddddddd")->color(white)("d")(sqrt(6))/6#