Question

Please help me 20 points

Where does the Pythagorean Identity sin2 Θ + cos2 Θ = 1 come from? Given the Cos Θ = 4/5, how would you use it to find the sine and tangent values of the angle? Explain.

Answer 1

`sin\theta =\pm(4)/(5)\\tan\theta =\pm(3)/(4)`

Explanation:

Kindly refer to the attached image of a right angled
`\triangle ABC`.

Let
`\angle C = \theta`

Sides BC = a

AB = c and

AC = b respectively.

To prove:

`sin^2\theta+cos^2\theta =1`

Proof:

Using Sine and Cosine values:

`sin\theta =(Perpendicular)/(Hypotenuse)\\\Rightarrow sin\theta =(AB)/(AC) = (c)/(b )`

`cos\theta =(Base)/(Hypotenuse)\\\Rightarrow cos\theta =(BC)/(AC) = (a)/(b)`

As per Pythagorean theorem:

`\text{Hypotenuse}^(2) = \text{Base}^(2) + \text{Perpendicular}^(2)\\\Rightarrow b^(2) = a^(2) + c^(2) .......... (1)`

Considering the LHS of
`sin^2\theta+cos^2\theta =1`:

`((c)/(b))^2+((a)/(b))^2\\\Rightarrow (c^2+a^2)/(b^2)`

Now, using equation (1):

`\Rightarrow (b^2)/(b^2) = 1 = RHS`

Hence, proved that :
`sin^2\theta+cos^2\theta =1`

Now, we are given that:

`cos\theta =(4)/(5)`

To find, sine and cosine values of the angle i.e.
`sin\theta = ?` and
`tan\theta = ?`

Using
`sin^2\theta+cos^2\theta =1`:

`sin^2\theta+((4)/(5))^2 =1\\\Rightarrow sin^2\theta = (9)/(25)\\\Rightarrow sin\theta = \pm (3)/(5)`

We know that:

`tan\theta =(sin\theta)/(cos\theta)\\\Rightarrow tan\theta =\pm ((3)/(5))/((4)/(5))\\\Rightarrow \bold{tan\theta =\pm (3)/(4)}`