Relating sin and cos

## Relating sin and cos

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).

## Example 1

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

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`