123. $$ \cos ^{2} 253+\sin ^{2} 557 $$ $$ \frac{ ext { Simplizy without using a calculatol: }}{\frac{\sin 120 \cdot \cos 210 \cdot an 315 \cdot \cos 27^{\circ}}{\sin 63 \cos 540^{\circ}}} $$

Question

123. $$ \cos ^{2} 253+\sin ^{2} 557 $$ $$ \frac{	ext { Simplizy without using a calculatol: }}{\frac{\sin 120 \cdot \cos 210 \cdot 	an 315 \cdot \cos 27^{\circ}}{\sin 63 \cos 540^{\circ}}} $$

UpStudy AI · Accepted Answer

$$
\textbf{1. Simplify } \cos ^{2}253^\circ+\sin ^{2}557^\circ:
$$

1. $$\cos 253^\circ$$:  
  Notice that 
  $$
  253^\circ = 180^\circ + 73^\circ,
  $$
  so
  $$
  \cos 253^\circ = -\cos 73^\circ.
  $$
  Thus,
  $$
  \cos^2 253^\circ = \cos^2 73^\circ.
  $$

2. $$\sin 557^\circ$$:  
  Reduce the angle modulo $$360^\circ$$: 
  $$
  557^\circ - 360^\circ = 197^\circ.
  $$
  Write
  $$
  197^\circ = 180^\circ + 17^\circ,
  $$
  so
  $$
  \sin 197^\circ = -\sin 17^\circ,
  $$
  and hence,
  $$
  \sin^2 557^\circ = \sin^2 197^\circ = \sin^2 17^\circ.
  $$

3. Now, the expression becomes
  $$
  \cos^2 73^\circ + \sin^2 17^\circ.
  $$
  Notice that by the co-function identity,
  $$
  \cos 73^\circ = \sin (90^\circ-73^\circ)=\sin 17^\circ.
  $$
  Thus,
  $$
  \cos^2 73^\circ = \sin^2 17^\circ.
  $$

4. Therefore,
  $$
  \cos^2 253^\circ+\sin^2 557^\circ = \sin^2 17^\circ+\sin^2 17^\circ = 2\sin^2 17^\circ.
  $$

---

$$
\textbf{2. Simplify } \frac{\sin 120^\circ \cdot \cos 210^\circ \cdot \tan 315^\circ \cdot \cos 27^\circ}{\sin 63^\circ \cdot \cos 540^\circ} \text{ without using a calculator:}
$$

1. Compute each trigonometric value:
  - $$\sin 120^\circ$$:  
   $$
  \sin 120^\circ = \sin (180^\circ-60^\circ)=\sin 60^\circ=\frac{\sqrt{3}}{2}.
  $$

- $$\cos 210^\circ$$:  
   $$
  \cos 210^\circ = \cos(180^\circ+30^\circ) = -\cos 30^\circ = -\frac{\sqrt{3}}{2}.
  $$

- $$\tan 315^\circ$$:  
   $$
  \tan 315^\circ = \tan (360^\circ-45^\circ) = -\tan 45^\circ = -1.
  $$

- $$\cos 27^\circ$$ remains as is.
  - $$\sin 63^\circ$$ remains as is.
  - $$\cos 540^\circ$$:  
   Notice that 
   $$
  540^\circ - 360^\circ = 180^\circ,
  $$
   so
   $$
  \cos 540^\circ = \cos 180^\circ = -1.
  $$

2. Substitute these values into the expression:
  $$
  \frac{\sin 120^\circ \cdot \cos 210^\circ \cdot \tan 315^\circ \cdot \cos 27^\circ}{\sin 63^\circ \cdot \cos 540^\circ} =
  \frac{\left(\frac{\sqrt{3}}{2}\right)\left(-\frac{\sqrt{3}}{2}\right)(-1)\cos 27^\circ}{\sin 63^\circ(-1)}.
  $$

3. Simplify step-by-step:
  - Multiply the first two factors in the numerator:
   $$
  \frac{\sqrt{3}}{2} \cdot \left(-\frac{\sqrt{3}}{2}\right)= -\frac{3}{4}.
  $$
  - Multiply by $$-1$$:
   $$
  -\frac{3}{4} \cdot (-1)= \frac{3}{4}.
  $$
  Thus, the numerator becomes
  $$
  \frac{3\cos 27^\circ}{4}.
  $$

- The denominator is:
   $$
  \sin 63^\circ \cdot (-1) = -\sin 63^\circ.
  $$

4. Therefore, the overall expression is:
  $$
  \frac{\frac{3\cos 27^\circ}{4}}{-\sin 63^\circ} = -\frac{3\cos 27^\circ}{4\sin 63^\circ}.
  $$

5. Use the co-function identity:
  $$
  \cos 27^\circ = \sin (90^\circ-27^\circ)=\sin 63^\circ.
  $$
  Substitute this into the expression:
  $$
  -\frac{3\sin 63^\circ}{4\sin 63^\circ}=-\frac{3}{4}.
  $$

---

$$
\textbf{Final Answers:}
$$

1. $$\cos ^{2}253^\circ+\sin ^{2}557^\circ = 2\sin^2 17^\circ$$.

2. $$\frac{\sin 120^\circ \cdot \cos 210^\circ \cdot \tan 315^\circ \cdot \cos 27^\circ}{\sin 63^\circ \cdot \cos 540^\circ} = -\frac{3}{4}$$.
$$ \cos ^{2}253^\circ+\sin ^{2}557^\circ = 2\sin^2 17^\circ $$ $$ \frac{\sin 120^\circ \cdot \cos 210^\circ \cdot \tan 315^\circ \cdot \cos 27^\circ}{\sin 63^\circ \cdot \cos 540^\circ} = -\frac{3}{4} $$

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