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

Answer:

$sin\theta =\pm\dfrac{4}{5}\\tan\theta =\pm\dfrac{3}{4}$

Step-by-step 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:

$sin\theta =\dfrac{Perpendicular}{Hypotenuse}\\\Rightarrow sin\theta =\dfrac{AB}{AC} = \dfrac{c}{b }$

$cos\theta =\dfrac{Base}{Hypotenuse}\\\Rightarrow cos\theta =\dfrac{BC}{AC} = \dfrac{a}{b}$

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 $sin^2\theta+cos^2\theta =1$:

$(\frac{c}{b})^2+(\frac{a}{b})^2\\\Rightarrow \frac{c^2+a^2}{b^2}$

Now, using equation (1):

$\Rightarrow \dfrac{b^2}{b^2} = 1 = RHS$

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

Now, we are given that:

$cos\theta =\frac{4}{5}$

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

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

$sin^2\theta+(\dfrac{4}{5})^2 =1\\\Rightarrow sin^2\theta = \dfrac{9}{25}\\\Rightarrow sin\theta = \pm \dfrac{3}{5}$

We know that:

$tan\theta =\dfrac{sin\theta}{cos\theta}\\\Rightarrow tan\theta =\pm \dfrac{\dfrac{3}{5}}{\dfrac{4}{5}}\\\Rightarrow \bold{tan\theta =\pm \dfrac{3}{4}}$