We need to evaluate the expression:
\[ \frac{\sum\limits_{k=1}^{89} \sin k^\circ}{2 \sum\limits_{k=1}^{44} \cos k^\circ + 1}. \]Step 1: Calculate the sum of sines from \(1^\circ\) to \(89^\circ\).
Notice that the sines can be paired symmetrically as:
Using the identity for complementary angles, we have:
\[ \sin x + \sin (90^\circ - x) = \sin x + \cos x = 1, \]where \(x + (90^\circ - x) = 90^\circ\). Each pair sums approximately to 1, and since there are 44 such pairs:
\[ \sum_{k=1}^{89} \sin k^\circ = 44. \]Step 2: Calculate the sum of cosines from \(1^\circ\) to \(44^\circ\).
Similarly, pair cosines as:
Each pair sums to:
\[ 2 \cos 45^\circ = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}. \]With 44 such pairs, the sum is:
\[ \sum_{k=1}^{44} \cos k^\circ = 44 \times \frac{1}{\sqrt{2}} = 22 \sqrt{2}. \]Step 3: Evaluate the given expression.
Substitute the sums into the original expression:
Since 1 is negligible compared to \(44 \sqrt{2}\), approximate as:
\[ \frac{44}{44 \sqrt{2}} = \frac{1}{\sqrt{2}}. \]Final answer:
\[ \boxed{\frac{1}{\sqrt{2}}}. \]| List-I | List-II | ||
|---|---|---|---|
| (A) | $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ | (I) | $[\frac{1}{3} , 1 ]$ |
| (B) | $(x)=|[x]|,x \in [R$ | (II) | Z |
| (C) | $h(x) = |x - [x]| , x \in [R$ | (III) | W |
| (D) | $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ | (IV) | [0, 1) |
| (V) | { -1, 1} | ||
| List I | List II | ||
|---|---|---|---|
| (A) | $\lambda=8, \mu \neq 15$ | 1. | Infinitely many solutions |
| (B) | $\lambda \neq 8, \mu \in R$ | 2. | No solution |
| (C) | $\lambda=8, \mu=15$ | 3. | Unique solution |
If \(\cos \alpha + \cos \beta + \cos \gamma = \sin \alpha + \sin \beta + \sin \gamma = 0,\) then evaluate \((\cos^3 \alpha + \cos^3 \beta + \cos^3 \gamma)^2 + (\sin^3 \alpha + \sin^3 \beta + \sin^3 \gamma)^2 =\)