Name: Trigonometry: Sine of Difference of Two Angles
Uploaded: 2019-07-19
Duration: 2 min 50 s
Description: This video describes the formula of the sine of the difference between two angles. Check out this video to know the formula and to solve problems based on it.

Question

Use the expansions of $\cos (2 x + x)$ and $\cos (2 x - x),$ to prove that: $\cos 3 x + \cos x \equiv 2 \cos 2 x \cos x$

Solve $\cos 3 x + \cos 2 x + \cos x > 0$ for $0 ° < x < 360 ° .$

Accepted Answer

Take $\cos 3 x = \cos (2 x + x) = \cos 2 x \cos x - \sin 2 x \sin x$ ...(i)

$\cos x = \cos (2 x - x) = \cos 2 x \cos x + \sin 2 x \sin x$ ..............(ii)

(i) $+$ (ii)

$0 ° < x < 360 ° .$

Hence, proved.

Using $\cos 3 x + \cos x = 2 \cos 2 x \cos x$

$\Rightarrow \cos 3 x + \cos 2 x + \cos x > 0$

$\Rightarrow (\cos 3 x + \cos x) + \cos 2 x > 0$

$\Rightarrow 2 \cos 2 x \cos x + \cos 2 x > 0$

$\Rightarrow \cos 2 x (2 \cos x + 1) > 0$

$\Rightarrow \cos 2 x > 0 and cosx > - \frac{1}{2}$ ....(iii)

or

$\cos 2 x < 0 and cosx < - \frac{1}{2}$ ....(iv)

Now, $\cos x > - \frac{1}{2}$ when $x < 120 ° or x > 240 °$

$\cos x < - \frac{1}{2}$ when $120 ° < x < 240 °$

$\cos 2 x > 0$ when $0 ° < 2 x < 90 ° or 270 ° < 2 x < 360 °$

i.e., when $0 ° < x < 45 ° or 135 ° < x < 180 °$

$\cos 2 x < 0$ when $45 ° < x < 135 °$

Here, (iii) requires $x < 120 ° and 0 ° < x < 45 °$ so $0 ° < x < 45 °$

Here, (iv) requires $120 ° < x < 240 ° and 45 ° < x < 135 °$ , so $120 ° < x < 135 °$

Use the expansions of $\cos (2 x + x)$ and $\cos (2 x - x),$ to prove that: $\cos 3 x + \cos x \equiv 2 \cos 2 x \cos x$

Solve $\cos 3 x + \cos 2 x + \cos x > 0$ for $0 ° < x < 360 ° .$

Important Questions on Trigonometry

Learn and Prepare for any exam you want !

Use the expansions of cos(2x+x) and cos(2x-x), to prove that: cos 3x+cos x≡2 cos 2x cos x Solve cos 3x+cos 2x+cosx>0 for 0°<x<360°.

Important Questions on Trigonometry

Learn and Prepare for any exam you want !

Use the expansions of $\cos (2 x + x)$ and $\cos (2 x - x),$ to prove that: $\cos 3 x + \cos x \equiv 2 \cos 2 x \cos x$

Solve $\cos 3 x + \cos 2 x + \cos x > 0$ for $0 ° < x < 360 ° .$