Prove without using a calculator: $$ \quad \frac{\cos 315^{\circ}+1}{\sin 315^{\circ}-1}=-1 $$ Evaluate without using a calculator: $$ \sqrt{4^{\sin 150^{\circ}} .2^{3 an 225^{\circ}}} $$ Show that $$ heta=30^{\circ} $$ is a solution to $$ (\sin heta)^{\sin heta}=\frac{1}{\sqrt{2}} $$

Question

Prove without using a calculator: $$ \quad \frac{\cos 315^{\circ}+1}{\sin 315^{\circ}-1}=-1 $$ Evaluate without using a calculator: $$ \sqrt{4^{\sin 150^{\circ}} .2^{3 	an 225^{\circ}}} $$ Show that $$ 	heta=30^{\circ} $$ is a solution to $$ (\sin 	heta)^{\sin 	heta}=\frac{1}{\sqrt{2}} $$

UpStudy AI · Accepted Answer

1.  
We begin by evaluating the trigonometric functions:
$$
\cos 315^\circ = \cos (360^\circ - 45^\circ) = \cos 45^\circ = \frac{\sqrt{2}}{2}
$$
$$
\sin 315^\circ = -\sin 45^\circ = -\frac{\sqrt{2}}{2}
$$
Substitute these values into the given expression:
$$
\frac{\cos 315^\circ + 1}{\sin 315^\circ - 1} = \frac{\frac{\sqrt{2}}{2} + 1}{-\frac{\sqrt{2}}{2} - 1}
$$
Write the sums with a common denominator:
$$
= \frac{\frac{\sqrt{2} + 2}{2}}{\frac{-\sqrt{2} - 2}{2}} = \frac{\sqrt{2} + 2}{-\sqrt{2} - 2} = -1
$$

2.  
Compute each component inside the square root separately.

First, note that:
$$
\sin 150^\circ = \sin (180^\circ - 30^\circ) = \sin 30^\circ = \frac{1}{2}
$$
Thus,
$$
4^{\sin 150^\circ} = 4^{\frac{1}{2}} = \sqrt{4} = 2
$$
Next, evaluate the tangent:
$$
\tan 225^\circ = \tan (180^\circ + 45^\circ) = \tan 45^\circ = 1
$$
So,
$$
2^{3 \tan 225^\circ} = 2^{3 \cdot 1} = 2^3 = 8
$$
Now, compute the product inside the square root:
$$
4^{\sin 150^\circ} \cdot 2^{3 \tan 225^\circ} = 2 \cdot 8 = 16
$$
Taking the square root:
$$
\sqrt{16} = 4
$$

3.  
For $$\theta = 30^\circ$$, we have:
$$
\sin 30^\circ = \frac{1}{2}
$$
Substitute into the equation:
$$
\left(\sin 30^\circ\right)^{\sin 30^\circ} = \left(\frac{1}{2}\right)^{\frac{1}{2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}
$$
Thus, $$\theta = 30^\circ$$ is indeed a solution to
$$
(\sin \theta)^{\sin \theta} = \frac{1}{\sqrt{2}}
$$
1. $$ \frac{\cos 315^\circ + 1}{\sin 315^\circ - 1} = -1 $$ 2. $$ \sqrt{4^{\sin 150^\circ} \cdot 2^{3 \tan 225^\circ}} = 4 $$ 3. $$ \theta = 30^\circ $$ is a solution to $$ (\sin \theta)^{\sin \theta} = \frac{1}{\sqrt{2}} $$

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