How to solve?(sin45+cos60)/(sin60+tan30) Step by step

1 Answer
Steve M
May 25, 2018

# (sin45^o + cos60^o)/(sin60^o + tan30^o) = (sqrt(6) + sqrt(3))/5 #

Explanation:

We seek:

# (sin45^o + cos60^o)/(sin60^o + tan30^o) #

We utilise our knowledge of sine and cosine for standard angles:

# {: (theta, 0^o, 30^o, 45^o, 60^o, 90^o), (sin theta, 0, 1/2, sqrt(2)/2, sqrt(3)/2, 1), (cos theta, 1, sqrt(3)/2, sqrt(2)/2, 1/2, 0) :} #

So then:

# (sin45^o + cos60^o)/(sin60^o + tan30^o) = (sqrt(2)/2 + 1/2)/(sqrt(3)/2 + sin30^o/cos30^o) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = (sqrt(2)/2 + 1/2)/(sqrt(3)/2 + (1/2)/(sqrt(3)/2)) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = (sqrt(2)/2 + 1/2)/(sqrt(3)/2 + 1/sqrt(3)) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = (sqrt(2)/2 + 1/2)/(sqrt(3)/2 + sqrt(3)/3) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = (1/2(sqrt(2) + 1))/(1/6(3sqrt(3) + 2sqrt(3))) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 3(sqrt(2) + 1)/(5sqrt(3)) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 3/5 \ (sqrt(2) + 1) \ sqrt(3)/3 #

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = sqrt(3)/5 \ (sqrt(2) + 1) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = (sqrt(6) + sqrt(3))/5 #

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