Question

Need help on these 2 questions and my last 2 uploads !! Help! please

Answer 1

Answer:

19. $\frac{3\pi}{2}$

20. The work is the answer

Step-by-step explanation:

19. We begin with the equation (for my ease, I will swap theta with x)

$sin(x)+1=cos^{2} (x)$

First, we can use the Pythagorean Identity to substitute in $1-sin^{2} (x)$ for $cos^{2} (x)$

This gives us

$sin(x)+1=1-sin^{2} (x)$

Next, we can subtract 1 from each side

$sin(x)=-sin^{2} (x)$

Now, we can divide each side by $sin(x)$

$1=-sin(x)$

Now we move the negative to the other side

$sin(x)=-1$

Now we can do the inverse of sin to find our value

$x=sin^{-1} (-1)$

With this inverse, we are looking for an angle that has the sin (or the y value) of -1. As this is an inverse sin function, it has the domain restriction of

$[-\frac{\pi }{2} ,\frac{\pi }{2}]$.

From this information, we find that $x=-\frac{\pi}{2}$

But this answer does not fit the initial interval that we were given. Luckily, $\frac{3\pi}{2}$ is the same location as $\frac{\pi}{2}$, so that is our answer.

20. We begin with the expression $sin(360-x)$ and we have to prove that it is equal to $-sin(x)$

First, we need to use the Difference of 2 Angles Trig Identity. This says:

$sin(\alpha -\beta )=sin(\alpha) cos(\beta) -cos(\alpha) sin(\beta)$

When we apply this to our expression, we get

$sin(360) cos(x) -cos(360) sin(x)$

Next, we can simplify our the two finite aspects

$sin(360)=0\\\\cos(360)=1$

When we plug in these two values, we get

$(0) cos(x) -(1) sin(x)$

This simplifies to $-sin(x)$ which is what we were trying to verify.