5.1 If $$ 3 \sin \beta=2 $$. and $$ \cos \beta

Question

5.1 If $$ 3 \sin \beta=2 $$. and $$ \cos \beta<0 $$, determine use of a calculator, the value of: 5.1.1 $$ 3 \cos ^{2} \beta-1 $$

UpStudy AI · Accepted Answer

To solve the problem, we start with the given equation:

$$
3 \sin \beta = 2
$$

From this, we can find $$\sin \beta$$:

$$
\sin \beta = \frac{2}{3}
$$

Next, we need to find $$\cos \beta$$. We can use the Pythagorean identity:

$$
\sin^2 \beta + \cos^2 \beta = 1
$$

Substituting $$\sin \beta$$ into the identity:

$$
\left(\frac{2}{3}\right)^2 + \cos^2 \beta = 1
$$

Calculating $$\left(\frac{2}{3}\right)^2$$:

$$
\frac{4}{9} + \cos^2 \beta = 1
$$

Now, we can isolate $$\cos^2 \beta$$:

$$
\cos^2 \beta = 1 - \frac{4}{9}
$$

To perform the subtraction, we convert 1 into a fraction with a denominator of 9:

$$
1 = \frac{9}{9}
$$

Thus, we have:

$$
\cos^2 \beta = \frac{9}{9} - \frac{4}{9} = \frac{5}{9}
$$

Now, we need to find $$3 \cos^2 \beta - 1$$:

$$
3 \cos^2 \beta - 1 = 3 \left(\frac{5}{9}\right) - 1
$$

Calculating $$3 \left(\frac{5}{9}\right)$$:

$$
3 \left(\frac{5}{9}\right) = \frac{15}{9}
$$

Now, substituting this back into the expression:

$$
3 \cos^2 \beta - 1 = \frac{15}{9} - 1
$$

Again, converting 1 into a fraction with a denominator of 9:

$$
1 = \frac{9}{9}
$$

Thus, we have:

$$
3 \cos^2 \beta - 1 = \frac{15}{9} - \frac{9}{9} = \frac{6}{9} = \frac{2}{3}
$$

Finally, we conclude that:

$$
\boxed{\frac{2}{3}}
$$
The value of $$3 \cos^2 \beta - 1$$ is $$\frac{2}{3}$$.

5.1 If \( 3 \sin \beta=2 \). and \( \cos \beta<0 \), determine use of a calculator, the value of: 5.1.1 \( 3 \cos ^{2} \beta-1 \)