Find the value of
$$\sqrt{\frac{9\sin^2 26^\circ + 9\sin^2 64^\circ + 36}{32 - 7\cos^2 32^\circ - 7\cos^2 58^\circ}}$$

Question

Find the value of 
$$\sqrt{\frac{9\sin^2 26^\circ + 9\sin^2 64^\circ + 36}{32 - 7\cos^2 32^\circ - 7\cos^2 58^\circ}}$$

Accepted Answer

**Solution:**

1. Given expression:
   $$\sqrt{\frac{9\sin^2 26^\circ + 9\sin^2 64^\circ + 36}{32 - 7\cos^2 32^\circ - 7\cos^2 58^\circ}}$$

2. Use the complementary angle identity:
   - Since $\sin 64^\circ = \sin (90^\circ - 26^\circ) = \cos 26^\circ$
   - Similarly, $\cos 58^\circ = \cos (90^\circ - 32^\circ) = \sin 32^\circ$

3. Rewrite numerator:
   $$9\sin^2 26^\circ + 9\sin^2 64^\circ + 36 = 9\sin^2 26^\circ + 9\cos^2 26^\circ + 36$$

4. Since $\sin^2 	heta + \cos^2 	heta = 1$, numerator becomes:
   $$9(\sin^2 26^\circ + \cos^2 26^\circ) + 36 = 9 	imes 1 + 36 = 45$$

5. Rewrite denominator:
   $$32 - 7\cos^2 32^\circ - 7\cos^2 58^\circ = 32 - 7\cos^2 32^\circ - 7\sin^2 32^\circ$$

6. Use $\sin^2 	heta + \cos^2 	heta = 1$ again:
   $$32 - 7(\cos^2 32^\circ + \sin^2 32^\circ) = 32 - 7 	imes 1 = 25$$

7. So the expression simplifies to:
   $$\sqrt{\frac{45}{25}} = \sqrt{\frac{9 	imes 5}{5 	imes 5}} = \sqrt{\frac{9}{5}} = \frac{3}{\sqrt{5}}$$

8. Rationalize the denominator:
   $$\frac{3}{\sqrt{5}} = \frac{3\sqrt{5}}{5}$$

**Answer:** $\frac{3\sqrt{5}}{5}$

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