# ISI2017-DCG-16

0 votes
36 views

If $\cos x = \dfrac{1}{2}$, the value of the expression $\dfrac{\cos 6x+6 \cos 4x+15 \cos 2x +10}{\cos 5x+5 \cos 3x +10 \cos x}$  is

1. $\frac{1}{2}$
2. $1$
3. $\frac{1}{4}$
4. $0$

recategorized

## 1 Answer

0 votes

Answer: $B$

Solution:

Let $$y = \frac{\cos6x + 6\cos4x + 15\cos2x + 10}{\cos5x + 5\cos3x + 10\cos x}\tag{i}$$

$6\cos4x$ and $15\cos2x$ can be written as:

$$6\cos4x = \cos x + 5\cos4x; \tag{ii}$$

$$15\cos2x = 10\cos2x + 5\cos2x\tag{iii}$$

Now, we know that:

$$2\cos a \cos b = \cos\frac{(a+b)}{2} \ cos \frac{a-b}{2}\tag{iv}$$

Using $(i), \;(ii)\;and\;(iii)$, we get:

$$\implies y = \frac{\underbrace{\cos6x +( \cos4x} + \underbrace{5\cos4x)+(5\cos2x} + \underbrace{10\cos2x) +10}}{\cos5x + 5\cos3x + 10\cos x}\tag{v}$$

Using $(iv)$ and $(v)$, we get:

$$\implies y = \frac{2\cos5x\cos x + 10\cos3x\cos x + 10(\cos2x + \underbrace{1}_\text {= \cos 0})}{\cos5x + 5\cos3x + 10\cos x}$$

$$\implies y = \frac{2\cos x(\cos5x + 5\cos3x) + 10(\cos2x + \underbrace{\cos 0})}{\cos5x + 5\cos3x + 10\cos x}$$

$$\implies y = \frac{2\cos x(\cos 5x + 5\cos3x) +10\times2\cos x \cos x}{\cos5x + 5\cos3x + 10\cos x}$$

$$\implies y = \frac{2\cos x [\cos5 + 5\cos3x + 10\cos x]}{\cos5x + 5\cos3x + 10\cos x}$$

$$\implies y = 2\cos x$$

Now, substitute $x = \frac{1}{2}$, we get:

$$y = 2\times \frac{1}{2} = 1$$

Answer: $\therefore B$ is the correct option.

edited by

## Related questions

0 votes
2 answers
1
36 views
If $\cos ^{2}x+ \cos ^{4} x=1$, then $\tan ^{2} x+ \tan ^{4} x$ is equal to $1$ $0$ $2$ none of these
0 votes
1 answer
2
49 views
If $a,b,c$ are the sides of $\Delta ABC$, then $\tan \frac{B-C}{2} \tan \frac{A}{2}$ is equal to $\frac{b+c}{b-c}$ $\frac{b-c}{b+c}$ $\frac{c-b}{c+b}$ none of these
0 votes
1 answer
3
50 views
The angle between the tangents drawn from the point $(-1, 7)$ to the circle $x^2+y^2=25$ is $\tan^{-1} (\frac{1}{2})$ $\tan^{-1} (\frac{2}{3})$ $\frac{\pi}{2}$ $\frac{\pi}{3}$
1 vote
2 answers
4
110 views
The value of $\dfrac{1}{\log_2 n}+ \dfrac{1}{\log_3 n}+\dfrac{1}{\log_4 n}+ \dots + \dfrac{1}{\log_{2017} n}\:\:($ where $n=2017!)$ is $1$ $2$ $2017$ none of these