Question

Solve using the Pythagorean identity​

Answer 1

$\bold{Sin\theta_{1}=-\frac{4}{5}}$

Answer:

Solution given

Cos$\displaystyle \theta_{1}=\frac{3}{5}$

consider Pythagorean theorem

$\bold{Sin²\theta+Cos²\theta=1}$

Subtracting $Cos²\theta$both side

$\displaystyle Sin²\theta=1-Cos²\theta$

doing square root on both side we get

$Sin\theta=\sqrt{1-Cos²\theta}$

Similarly

$Sin\theta_{1}=\sqrt{1-Cos²\theta_{1}}$

Substituting value of $Cos\theta_{1}$

we get

$Sin\theta_{1}=\sqrt{1-(\frac{3}{5})²}$

Solving numerical

$Sin\theta_{1}=\sqrt{1-(\frac{9}{25})}$

$Sin\theta_{1}=\sqrt{\frac{16}{225}}$

$Sin\theta_{1}=\frac{\sqrt{2*2*2*2}}{\sqrt{5*5}}$

$Sin\theta_{1}=\frac{4}{5}$

Since

In IVquadrant sin angle is negative

$\bold{Sin\theta_{1}=-\frac{4}{5}}$

Answer 2

Answer:

$\sin(\theta_1)=-\frac{4}{5}$

Step-by-step explanation:

We'll use the Pythagorean Identity $\cos^2(\theta)+\sin^2(\theta)=1$ to solve this problem.

Subtract $\cos^2(\theta)$ from both sides to isolate $\sin^2(\theta)$:

$\sin^2(\theta)=1-\cos^2(\theta)$

Substitute $\cos(\theta)=\frac{3}{5}$ as given in the problem:

$\sin^2(\theta_1)=1-(\frac{3}{5}^2)$

Simplify:

$\sin^2\theta_1=1-\frac{9}{25}$

Combine like terms:

$\sin^2\theta_1=\frac{16}{25}$

For $a^2=b$, we have two solutions $a=\pm \sqrt{b}$:

$\sin\theta_1=\pm \sqrt{\frac{16}{25}},\\\begin{cases}\sin \theta_1=\frac{4}{5},\\\sin \theta_1=\boxed{-\frac{4}{5}}\end{cases}$

Since the sine of all angles in quadrant four return a negative output, $\frac{4}{5}$ is extraneous and our answer is $\boxed{\sin(\theta_1)=-\frac{4}{5}}$