If all the denominator has 10 then how to solve?

Question

Thita as 17pie/10, 5pie/10, 37pie/10 and 3oie/10 then how to solve?

Answer 1

If there is a common denominator then simply add or subtract (as required) the values in the numerators and then divide the result by the denominator. In this case, however, there's also π in all the numerators. So π can be distributed: π(17+5+37+3)=62π. Now we can divide by the common denominator: 62π/10.

This can be written 6.2π or 31π/5.

But note that this addition of denominators is only valid when simple arithmetic (like addition and subtraction) is involved. It doesn't work if we try to add functions together. For example, sinA+sinB is not the same as sin(A+B); or 2^A+2^B is not the same as 2^A+B.

The quantities in the numerators don't need to be numbers; they can be expressions. For example:

(3sinθ)/10+(4sinθ)/10+sinθ/10+2sinθ/10=sinθ(3+4+1+2)/10=10sinθ/10=sinθ. This works because sinθ can be distributed as a common factor. This makes all the numerators "like" terms. Another example:

(x+3)/8+(3x+8)/8+(2x-5)/8-(6x+5)/8=(x+3x+2x-6x+3+8-5-5)/8=⅛. Here is a succession of algebraic expressions over a common denominator.

Comments

If the denominator was changed in the original question, the relationships would no longer be valid so the answer would need to be calculated using a calculator or tables. The point of the original question was to find out if the student could reduce the angles and apply trigonometrical identities successfully.