From SOH CAH TOA
`sin = frac(text(opposite))(text(hypotenuse))`
`cos = frac(text(adjacent))(text(hypotenuse))`
Divide Sin by Cos
`= frac(text(opposite))(text(hypotenuse)) ÷ frac(text(adjacent))(text(hypotenuse))`
`= frac(text(opposite))(text(hypotenuse)) xx frac(text(hypotenuse))(text(adjacent))`
`= frac(text(opposite))(text(adjacent))`
which is Tan. Therefore the Tan of an angle `= frac(text(sin of that angle))(text(cos of that angle))`.
Pythagoras` Theorem is given by `a^2 + b^2 = c^2`
Divide throughout by `c^2` and simplify
`(frac(a)(c))^2 + (frac(b)(c))^2 = 1`
And replacing adjacent ÷ hypotenuse with sin and opposite ÷ hypotenuse with cos:
`(sin x)^2 + (cos x)^2 = 1`
Normally this is written without the brackets, but to ensure that you obtain the sine of the angle before squaring, it is written as
`sin^2 x + cos^2 x = 1`
(`sin x^2` would mean square the angle then take the sine).
If `tan theta = frac(sin^2 theta)(cos theta)`, then what is the value of `theta` given 0 < `theta` < 360?
| Given | `tan theta` | `= frac(sin^2 theta)(cos theta)` |
| Using | `frac(sin theta)(cos theta)` | `= tan theta` |
| substituting | `frac(sin theta)(cos theta)` | `= frac(sin^2 theta)(cos theta)` |
| multiply by `cos theta` | `sin theta` | `= sin^2 theta` |
| divide by `sin theta` | `1` | `= sin theta` |
| `sin-1` both sides | `90` | `= theta` |
Answer: 90º
For 0 < `theta` < 90, what is the value of `theta` when `sin^2 theta = cos theta`? Give your answer to the nearest whole degree.
| Given | `sin^2 theta` | `= cos theta` |
| Using | `sin^2 theta + cos^2 theta` | `= 1` |
| substitute for `sin^2theta` | `cos theta + cos^2 theta` | `= 1` |
| rearrange | `cos^2 theta + cos theta - 1` | `= 0` |
| let `cos theta = x` | `x^2 + x - 1` | `= 0` |
| solve the quadratic | `x` | `= 1.618 or 0.618` |
| substitute back | `cos theta` | `= 1.618 or 0.618` |
| not possible: | `cos theta` | `= 1.618` |
| valid answer: | `cos theta` | `= 0.618` |
| cos-1 both sides | `theta` | `= 51.8` |
| nearest degree | `theta` | `= 52` |
Answer: 52º