Question

Given: sin θ = `p/q`, find cos θ + sin θ in terms of p and q.

Answer 1

Consider the diagram below :

sin θ = `p/q`

i.e.`"perpendicular"/"hypotenuse" = p/h`

Therefore, if length of perpendicular = px,

Length of hypotenuse = qx

Since, hypotenuse² = base² + perpendicular²    ...[Using Pythagoras Theorem]

(qx)² = base² + (px)²

q²x² = p²x^{2 +} base²

q²x² – p²x² = base²

(q² – p²)x² = base²

∴ base = `sqrt(("q"^2 - "p"^"2")x^2)`

∴ base = `xsqrt("q"^2 - "p"^2) = "base"`

Now, cos θ = `"base"/"hypotenuse"`

= `(xsqrt(q^2 - p^2))/(qx)`

Therefore, cos θ + sin θ

= `(xsqrt(q^2 - p^2))/(qx) + p/q`

= `(sqrt(q^2 - p^2))/(q) + p/q`

= `(p + sqrt(q^2 - p^2))/q`

Answer 2

Given: sin θ = `p/q`, with |p| ≤ |q| so `p/q` is a valid sine.

Step-wise calculation:

1. `sin^2θ = p^2/q^2`.

2. cos²θ = 1 – sin²θ

= `1 - p^2/q^2` 

= `(q^2 - p^2)/q^2`

3. `cos θ = ± sqrt(q^2 - p^2)/|q|`. If q > 0 this is `cos θ = ± sqrt(q^2 - p^2)/q`; the sign ± is chosen according to the quadrant of θ.

4. `cos θ + sin θ = ± sqrt(q^2 - p^2)/|q| + p/q`. If q > 0 this simplifies to `(p ± sqrt(q^2 - p^2))/q`.

`cos θ + sin θ = (p ± sqrt(q^2 - p^2))/q`, with the ± matching the sign of cos θ and valid provided |p| ≤ |q|.