The value of cos

Question: The value of $\cos \frac{\pi}{65} \cos \frac{2 \pi}{65} \cos \frac{4 \pi}{65} \cos \frac{8 \pi}{65} \cos \frac{16 \pi}{65} \cos \frac{32 \pi}{65}$ is (a) $\frac{1}{8}$ (b) $\frac{1}{16}$ (c) $\frac{1}{32}$ (d) none of these Solution: (d) none of these We have, $\cos \frac{\pi}{65} \cos \frac{2 \pi}{65} \cos \frac{4 \pi}{65} \cos \frac{8 \pi}{65} \cos \frac{16 \pi}{65} \cos \frac{32 \pi}{65}$ $=\frac{2 \sin \frac{\pi}{65}}{2 \sin \frac{\pi}{65}} \cos \frac{\pi}{65} \cos \frac{2 \pi}{65}...

Evaluate each of the following

Question: Evaluate each of the following $\cos ^{2} 30^{\circ}+\cos ^{2} 45^{\circ}+\cos ^{2} 60^{\circ}+\cos ^{2} 90^{\circ}$ Solution: We have to find the following expression $\cos ^{2} 30^{\circ}+\cos ^{2} 45^{\circ}+\cos ^{2} 60^{\circ}+\cos ^{2} 90^{\circ} \ldots \ldots$ (1) Now, $\cos 45^{\circ}=\frac{1}{\sqrt{2}}, \cos 30^{\circ}=\frac{\sqrt{3}}{2}, \cos 60^{\circ}=\frac{1}{2}, \cos 90^{\circ}=0$ So by substituting above values in equation (1) We get, $\cos ^{2} 30^{\circ}+\cos ^{2} 45^{...

The weights (in kg) of 15 students are:

Question: The weights (in kg) of 15 students are: 31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36, 44, 45, 42, 30. Find the median. If the weight 44 kg is replaced by 46 kg and 27 kg by 25 kg, find the new median. Solution: Given the numbers are31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36, 44, 45, 42, 30 Arranging the numbers in ascending order 27, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 44, 45. n = 15 (odd) $\therefore$ New Median $=\left(\frac{\mathrm{n}+1}{2}\right)^{\text {th }}$ value $=\l...

Question: $\frac{\sec 8 A-1}{\sec 4 A-1}=$ (a) $\frac{\tan 2 A}{\tan 8 A}$ (b) $\frac{\tan 8 A}{\tan 2 A}$ (c) $\frac{\cot 8 A}{\cot 2 A}$ (d) none of these. Solution: (b) $\frac{\tan 8 A}{\tan 2 A}$ We have, $\frac{\sec 8 A-1}{\sec 4 A-1}=\frac{\frac{1}{\cos 8 A}-1}{\frac{1}{\cos 4 A}-1}$ $=\frac{\cos 4 A}{\cos 8 A} \times \frac{1-\cos 8 A}{1-\cos 4 A}$ $=\frac{\cos 4 A}{\cos 8 A} \times \frac{2 \sin ^{2} 4 A}{2 \sin ^{2} 2 A} \quad\left(2 \sin ^{2} \theta=1-\cos 2 \theta\right)$ $=\frac{(2 \co...

Find the median of the following data:

Question: Find the median of the following data: 41, 43, 127, 99, 61, 92, 71, 58, 57. If 58 is replaced by 85, what will be the new median? Solution: Given the numbers are41, 43, 127, 99, 61, 92, 71, 58, 57 Arranging the numbers in ascending order 41, 43, 57, 58, 61, 71, 92, 99, 127 n = 9 (odd) $\therefore$ New Median $=\left(\frac{\mathrm{n}+1}{2}\right)^{\text {th }}$ value $=\left(\frac{9+1}{2}\right)^{\text {th }}$ value If 58 is replaced by 85 Then the new values be in order are: 41, 43, 57...

8 sin

Question: $8 \sin \frac{x}{8} \cos \frac{x}{2} \cos \frac{x}{4} \cos \frac{x}{8}$ is equal to (a) 8 cosx (b) cosx (c) 8 sinx (d) sinx Solution: (d) sinx We have, $8 \sin \frac{x}{8} \cos \frac{x}{2} \cos \frac{x}{4} \cos \frac{x}{8}$ $=4 \times\left(2 \sin \frac{x}{8} \cos \frac{x}{8}\right) \cos \frac{x}{2} \cos \frac{x}{4}$ $=4 \times \sin \frac{x}{4} \cos \frac{x}{2} \cos \frac{x}{4}$ $=2 \times\left(2 \sin \frac{x}{4} \cos \frac{x}{4}\right) \cos \frac{x}{2}$ $=2 \times \sin \frac{x}{2} \cos...

Find the median of the following observations:

Question: Find the median of the following observations: 46, 64, 87, 41, 58, 77, 35, 90, 55, 92, 33. If 92 is replaced by 99 and 41 by 43 in the above data, find the new median? Solution: Given the numbers are 46, 64, 87, 41, 58, 77, 35, 90, 55, 92, 33 Arranging the numbers in ascending order 33, 35, 41, 46, 55, 58, 64, 77, 87, 90, 92 n = 11 (odd) $\therefore$ Median $=\left(\frac{\mathrm{n}+1}{2}\right)^{\text {th }}$ value $=\left(\frac{11+1}{2}\right)^{\mathrm{th}}$ value $=6^{\text {th }}$ v...

Prove that:

Question: Prove that: $\cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{3 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{5 \pi}{15} \cos \frac{6 \pi}{15} \cos \frac{7 \pi}{15}=\frac{1}{128}$ Solution: $\mathrm{LHS}=\cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{3 \pi}{15} \cos \frac{5 \pi}{15} \cos \frac{6 \pi}{15} \cos \frac{7 \pi}{15}$ $=\cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15}\left(\cos \frac{3 \pi}{15} \cos \frac{6 \pi}{15}\right) \times\left...

Numbers 50, 42, 35, 2x + 10, 2x – 8, 12, 11, 8 are written in descending order and their median is 25, find x.

Question: Numbers 50, 42, 35, 2x + 10, 2x 8, 12, 11, 8 are written in descending order and their median is 25, find x. Solution: Given the number of observation, n = 8 $\therefore$ Median $=\frac{\overline{\overline{2}}^{\text {th }} \text { value }+\left(\frac{\mathrm{n}}{2}+1\right)^{\text {th }} \text { value }}{2}$ $=\frac{\frac{8^{\text {th }}}{2} \text { value }+\left(\frac{8}{2}+1\right)^{\text {th }} \text { value }}{2}$ $=\frac{4^{\text {th }} \text { value }+5^{\text {th }} \text { val...

Evaluate each of the following

Question: Evaluate each of the following $\sin ^{2} 30^{\circ}+\sin ^{2} 45^{\circ}+\sin ^{2} 60^{\circ}+\sin ^{2} 60^{\circ}+\sin ^{2} 90^{\circ}$ Solution: We have to find $\sin ^{2} 30^{\circ}+\sin ^{2} 45^{\circ}+\sin ^{2} 60^{\circ}+\sin ^{2} 90^{\circ} \ldots \ldots$ (1) Now, $\sin 45^{\circ}=\frac{1}{\sqrt{2}}, \sin 30^{\circ}=\frac{1}{2}, \sin 60^{\circ}=\frac{\sqrt{3}}{2}, \sin 90^{\circ}=1$ So by substituting above values in equation (1) We get, $\sin ^{2} 30^{\circ}+\sin ^{2} 45^{\cir...

Find the median of the data:

Question: Find the median of the data: 92, 35, 67, 85, 72, 81, 56, 51, 42, 69 Solution: Numbers are 92, 35, 67, 85, 72, 81, 56, 51, 42, 69 Arranging the numbers in ascending order 35, 42, 51, 56, 67, 69, 72, 81, 85, 92 n = 10(even) $\therefore$ Median $=\frac{\frac{\mathrm{n}^{\text {th }}}{2} \text { value }+\left(\frac{\mathrm{n}}{2}+1\right)^{\text {th }} \text { value }}{2}$ $=\frac{\frac{10^{\text {th }}}{2} \text { value }+\left(\frac{10}{2}+1\right)^{\text {th }} \text { value }}{2}$ $=\f...

Find the vector equation of the plane passing through the intersection of the planes

Question: Find the vector equation of the plane passing through the intersection of the planes $\vec{r} \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})=7, \vec{r} \cdot(2 \hat{i}+5 \hat{j}+3 \hat{k})=9$ and through the point $(2,1,3)$ Solution: The equations of the planes are $\vec{r} \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})=7$ and $\vec{r} \cdot(2 \hat{i}+5 \hat{j}+3 \hat{k})=9$ $\Rightarrow \vec{r} \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})-7=0$ $\vec{r} \cdot(2 \hat{i}+5 \hat{j}+3 \hat{k})-9=0$ The equation of an...

Evaluate each of the following

Question: Evaluate each of the following $\cos 60^{\circ} \cos 45^{\circ}-\sin 60^{\circ} \sin 45^{\circ}$ Solution: We have to find the value of the following expression $\cos 60^{\circ} \cos 45^{\circ}-\sin 60^{\circ} \sin 45^{\circ}$....(1) Now $\sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}, \sin 60^{\circ}=\frac{\sqrt{3}}{2}, \cos 60^{\circ}=\frac{1}{2}$ So by substituting above values in equation (1) We get, $\cos 60^{\circ} \cos 45^{\circ}-\sin 60^{\circ} \sin 45^{\circ}$ $=\frac{1}{2...

Find the median of the data:

Question: Find the median of the data: 12, 17, 3, 14, 5, 8, 7, 15 Solution: Numbers are 12, 17, 3, 14, 5, 8, 7, 15 Arranging the numbers in ascending order 3, 5, 7, 8, 12, 14, 15, 17 n = 8 (even) $\therefore$ Median $=\frac{\frac{\bar{n}^{\text {th }}}{2} \text { value }+\left(\frac{n}{2}+1\right)^{\text {th }} \text { value }}{2}$ $=\frac{\frac{8}{2}^{\text {th }} \text { value }+\left(\frac{8}{2}+1\right)^{\text {th }} \text { value }}{2}$ $=\frac{4^{\text {th }} \text { value }+5^{\text {th }...

Prove that:

Question: Prove that: $\sin \frac{\pi}{5} \sin \frac{2 \pi}{5} \sin \frac{3 \pi}{5} \sin \frac{4 \pi}{5}=\frac{5}{16}$ Solution: LHS $=\sin \frac{\pi}{5} \sin \frac{2 \pi}{5} \sin \frac{3 \pi}{5} \sin \frac{4 \pi}{5}$ $=\frac{1}{2}\left(2 \sin \frac{\pi}{5} \sin \frac{4 \pi}{5}\right) \frac{1}{2}\left(2 \sin \frac{2 \pi}{5} \sin \frac{3 \pi}{5}\right)$ $=\frac{1}{4}\left(\cos \left(\frac{\pi}{5}-\frac{4 \pi}{5}\right)-\cos \left(\frac{\pi}{5}+\frac{4 \pi}{5}\right)\right)\left(\cos \left(\frac{2...

Evaluate each of the following

Question: Evaluate each of the following $\sin 60 \cos 30^{\circ}+\cos 60^{\circ} \sin 30^{\circ}$ Solution: We have to find the value of the expression $\sin 60^{\circ} \cos 30^{\circ}+\cos 60^{\circ} \sin 30^{\circ}$....(1) Now $\sin 60^{\circ}=\cos 30^{\circ}=\frac{\sqrt{3}}{2}, \sin 30^{\circ}=\cos 60^{\circ}=\frac{1}{2}$ So by substituting above values in equation (1) We get, $\sin 60^{\circ} \cos 30^{\circ}+\cos 60^{\circ} \sin 30^{\circ}$ $=\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}+\fr...

Find the median of the data:

Question: Find the median of the data: 25, 34, 31, 23, 22, 26, 35, 29, 20, 32 Solution: Numbers are 25, 34, 31, 23, 22, 26, 35, 29, 20, 32 Arranging the numbers in ascending order 20, 22, 23, 25, 26, 29, 31, 32, 34, 35 n = 10 (even) $\therefore$ Median $=\frac{\frac{\mathrm{n}^{\mathrm{th}}}{2} \text { value }+\left(\frac{\mathrm{n}}{2}+1\right)^{\text {th }} \text { value }}{2}$ $=\frac{\frac{10^{\text {th }}}{2} \text { value }+\left(\frac{10}{2}+1\right)^{\text {th }} \text { value }}{2}$ $=\...

Find the equation of the plane through the intersection of the planes

Question: Find the equation of the plane through the intersection of the planes $3 x-y+2 z-4=0$ and $x+y+z-2=0$ and the point $(2,2,1)$ Solution: The equation of any plane through the intersection of the planes, $3 x-y+2 z-4=0$ and $x+y+z-2=0$, is $(3 x-y+2 z-4)+\alpha(x+y+z-2)=0$, where $\alpha \in \mathrm{R}$ ...(1) The plane passes through the point (2, 2, 1). Therefore, this point will satisfy equation (1). $\therefore(3 \times 2-2+2 \times 1-4)+\alpha(2+2+1-2)=0$ $\Rightarrow 2+3 \alpha=0$ ...

Prove that:

Question: Prove that: $\cos 36^{\circ} \cos 42^{\circ} \cos 60^{\circ} \cos 78^{\circ}=\frac{1}{16}$ Solution: $\mathrm{LHS}=\cos 36^{\circ} \cos 42^{\circ} \cos 60^{\circ} \cos 78^{\circ}$ $=\frac{1}{2} \cos 36^{\circ} \cos 60^{\circ}\left(2 \cos 42^{\circ} \cos 78^{\circ}\right)$ $[2 \cos A \cos B=\cos (A+B)+\cos (A-B)]$ $=\frac{1}{2}\left(\frac{\sqrt{5}+1}{4}\right) \times \frac{1}{2}\left(\cos 120^{\circ}+\cos 36^{\circ}\right)$ $=\left(\frac{\sqrt{5}+1}{16}\right)\left(-\frac{1}{2}+\frac{\s...

Evaluate each of the following

Question: Evaluate each of the following $\sin 45^{\circ} \sin 30^{\circ}+\cos 45^{\circ} \cos 30^{\circ}$ Solution: We have, $\sin 45^{\circ} \sin 30^{\circ}+\cos 45^{\circ} \cos 30^{\circ}$....(1) Now $\sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}, \sin 30^{\circ}=\frac{1}{2}, \cos 30^{\circ}=\frac{\sqrt{3}}{2}$ So by substituting above values in equation (1) We get, $\sin 45^{\circ} \sin 30^{\circ}+\cos 45^{\circ} \cos 30^{\circ}$ $=\frac{1}{\sqrt{2}} \times \frac{1}{2}+\frac{1}{\sqrt{2}...

Find the median of the data:

Question: Find the median of the data: 41, 43, 127, 99, 71, 92, 71, 58, 57 Solution: Numbers are 41, 43, 127, 99, 71, 92, 71, 58, 57 Arranging the numbers in ascending order 41, 43, 57, 58, 71, 71, 92, 99, 127 n = 9 (odd) $\therefore$ Median $=\left(\frac{\mathrm{n}+1}{2}\right)^{\text {th }}$ value $=\left(\frac{9+1}{2}\right)^{\text {th }}$ value $=5^{\text {th }}$ value $=71$...

Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.

Question: Find the equation of the plane with intercept 3 on they-axis and parallel to ZOX plane. Solution: The equation of the plane ZOX is y= 0 Any plane parallel to it is of the form,y=a Since they-intercept of the plane is 3, a= 3 Thus, the equation of the required plane isy= 3...

State whether the following are true or false. Justify your answer.

Question: State whether the following are true or false. Justify your answer. (i) The value of tan A is always less than 1. (ii) $\sec A=\frac{12}{5}$ for some value of angle $\mathrm{A}$. (iii) cos A is the abbreviation used for the cosecant of angle A.(iv) cot A is the product of cot and A. (v) $\sin \theta=\frac{4}{3}$ for some angle $\theta$. Solution: (i) In $\tan A, \angle A$ is acute an angle Therefore, Minimum value of $\angle A$ is $0^{\circ}$ and Maximum value of $\angle A$ is $90^{\ci...

Prove that:

Question: Prove that: $\sin 6^{\circ} \sin 42^{\circ} \sin 66^{\circ} \sin 78^{\circ}=\frac{1}{16}$ Solution: $\mathrm{LHS}=\frac{1}{4}\left(2 \sin 6^{\circ} \sin 66^{\circ}\right)\left(2 \sin 42^{\circ} \sin 78^{\circ}\right)$ $=\frac{1}{4}\left(\cos 60^{\circ}-\cos 72^{\circ}\right)\left(\cos 36^{\circ}-\cos 120^{\circ}\right)$ $[2 \sin A \sin B=\cos (A-B)-\cos (A+B)]$ $=\frac{1}{4}\left(\frac{1}{2}-\sin 18^{\circ}\right)\left(\frac{\sqrt{5}+1}{4}+\frac{1}{2}\right)$ $=\frac{1}{4}\left(\frac{1...

Find the median of the data:

Question: Find the median of the data: 15, 6, 16, 8, 22, 21, 9, 18, 25 Solution: Numbers are 15, 6, 16, 8, 22, 21, 9, 18, 25 Arranging the numbers in ascending order 6, 8, 9, 15, 16, 21, 22, 25 n = 9 (odd) $\therefore$ Median $=\left(\frac{\mathrm{n}+1}{2}\right)^{\text {th }}$ value $=\left(\frac{9+1}{2}\right)^{\text {th }}$ value $=5^{\text {th }}$ value $=16$...