Respuesta :

msm555

[tex]\bold{Sin\theta_{1}=-\frac{4}{5}}[/tex]

Answer:

Solution given

Cos[tex]\displaystyle \theta_{1}=\frac{3}{5}[/tex]

consider Pythagorean theorem

[tex]\bold{Sin²\theta+Cos²\theta=1}[/tex]

Subtracting [tex]Cos²\theta[/tex]both side

[tex]\displaystyle Sin²\theta=1-Cos²\theta[/tex]

doing square root on both side we get

[tex]Sin\theta=\sqrt{1-Cos²\theta}[/tex]

Similarly

[tex]Sin\theta_{1}=\sqrt{1-Cos²\theta_{1}}[/tex]

Substituting value of [tex]Cos\theta_{1}[/tex]

we get

[tex]Sin\theta_{1}=\sqrt{1-(\frac{3}{5})²}[/tex]

Solving numerical

[tex]Sin\theta_{1}=\sqrt{1-(\frac{9}{25})}[/tex]

[tex]Sin\theta_{1}=\sqrt{\frac{16}{225}}[/tex]

[tex]Sin\theta_{1}=\frac{\sqrt{2*2*2*2}}{\sqrt{5*5}}[/tex]

[tex]Sin\theta_{1}=\frac{4}{5}[/tex]

Since

In IVquadrant sin angle is negative

[tex]\bold{Sin\theta_{1}=-\frac{4}{5}}[/tex]

Corsaquix

Answer:

[tex]\sin(\theta_1)=-\frac{4}{5}[/tex]

Step-by-step explanation:

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

Subtract [tex]\cos^2(\theta)[/tex] from both sides to isolate [tex]\sin^2(\theta)[/tex]:

[tex]\sin^2(\theta)=1-\cos^2(\theta)[/tex]

Substitute [tex]\cos(\theta)=\frac{3}{5}[/tex] as given in the problem:

[tex]\sin^2(\theta_1)=1-(\frac{3}{5}^2)[/tex]

Simplify:

[tex]\sin^2\theta_1=1-\frac{9}{25}[/tex]

Combine like terms:

[tex]\sin^2\theta_1=\frac{16}{25}[/tex]

For [tex]a^2=b[/tex], we have two solutions [tex]a=\pm \sqrt{b}[/tex]:

[tex]\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}[/tex]

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

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