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Debora [2.8K]
3 years ago
9

Solve using the Pythagorean identity​

Mathematics
2 answers:
Lynna [10]3 years ago
7 0
<h3>\bold{Sin\theta_{1}=-\frac{4}{5}}</h3>

Answer:

Solution given

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

consider Pythagorean theorem

\bold{Sin²\theta+Cos²\theta=1}

Subtracting Cos²\thetaboth 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}

<h3>Solving numerical</h3>

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}}

Ilia_Sergeevich [38]3 years ago
3 0

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}}

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