I want to calculate sin 30-cos 60. Can i calculate sin30-cos60=30-60?

chaobas | Student

Not really.

Well we can apply this formula.

sin x = cos(90-x) 

So sin 30 = cos(90-30) = cos 60



sin 30-cos 60 

=cos 60 - cos 60


sanyo | Student
  • nooo, thats a blunder.  sin30-cos60= sin(90-60)-cos60= cos60-cos60=0  since      sin(90-x)=cosx
neela | Student

No please. It is not correct to calculate sin30 - cos60 as 30 - 60. The concept is totally different.

If ABC is a right angled triangle with right angle  at B, then sine and cosine ratios are defined  as below:

sin A = Opposite angle / hypotenuse =  BC/AC

cosC = Adjascent angle/hypotenuse = BC/AC.

Therefore in a right angled triangle if A = x, then C = (90-x) .

Therefore sin A = sinx

sinx = BC/AC...(1).

cosC = cos(90-x).

cos(90-x)  = BC/AC....(2)

Therefore  from (1) and (2) we get:

sinx = cos(90-x)....(3).

If  we put x= 30 in eq (3), we get:

sin30 = cos(90-30) = cos60.

Therefore,  sin30 - cos60 = 0..

Therefore sin30 -cos60 = 0 and   it is not 30 -60.

giorgiana1976 | Student

Definitely not!

You want to calculate the difference of the results of functions sine and cosine and not the difference between their arguments.

The functions sine and cosine don't have the behavior of the identically function, where f(x) = x.

Let's see how it works!

In a right angle triangle, where one cathetus is b and the other one is c and the hypothenuse is a, a cathetus opposite to the angle 30 (meaning pi/6 radians) is half from hypothenuse.

If b is the opposite cathetus to pi/6 angle, that means that b=a/2. In this way, we can find the other cathetus length, using Pythagorean theorem.

a^2=b^2 + c^2

a^2 = a^2/4 + c^2

a^2 - a^2/4 = c^2

3a^2/4 = c^2


sin pi/6=opposite cathetus/hypotenuse

sin pi/6= (a/2)/a

sin pi/6=1/2

cos pi/3=cos 60=adjacent cathetus/hypotenuse

cos pi/3= (a/2)/a

cos pi/3=1/2

sin pi/6 - cos pi/3=1/2 -1/2=0

Note the difference!

30 - 60 = -30 = 330 degrees and sin 30 - cos 60 = 0.

Access hundreds of thousands of answers with a free trial.

Start Free Trial
Ask a Question