2sin120+3cos765-2sin240-3cos45÷5sin300+3tan225-6cos60

Question

Solve the equation

Answer 1

(2sin120+3cos765-2sin240-3cos45)/(5sin300+3tan225-6cos60)

So numerator = 2sin120+3cos765-2sin240-3cos45

and denominator = 5sin300+3tan225-6cos60

Let us solve numerator first

2sin120+3cos765-2sin240-3cos45

or  2sin(90+30) + 3cos(2*360 +45) - 2sin(180+60) - 3cos45   

or 2cos(30) + 3cos(45) + 2sin(60) - 3cos(45)    ; using sin(90+x) = cosx, cos(n*360 + x) = cosx and sin(180+x) = -sinx

or 2cos(30) + 2sin(60)

or 2sin(60) + 2sin(60)  ; using cos(90-x) = sinx

or 4sin(60)  ----------------------------(1)

Now we solve denominator

5sin300+3tan225-6cos60

or 5sin(360-60) + 3tan(180+45) - 6cos(60)

or -5sin60 + 3tan45 - 6*1/2  ; since sin(n*360-x) =-sinx, tan(180+x)=tanx and cos60=1/2

or -5sin60 + 3 -3 ; since tan45=1

or -5sin60  ----------------------------------(2)

Now on dividing (1) by (2) we get

4sin60/(-5sin60)

or -4/5

So, on solving (2sin120+3cos765-2sin240-3cos45)/(5sin300+3tan225-6cos60) we get

-4/5 or -0.8