Question #02e7c

Question

Question #02e7c

2 Answers

Impact of this question

1121 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-11T08:35:48

Given

$cos 4 x = 1 - b {sin}^{2} x {cos}^{2} x$ $cos4x = 1 - bsin^2x cos^2x$

$\Rightarrow 1 - 2 {sin}^{2} 2 x = 1 - \frac{b}{4} (2 sin x cos x) 2$ $=>1-2sin^2 2x = 1 - b/4(2sinx cosx)2$

$\Rightarrow 2 {sin}^{2} 2 x = \frac{b}{4} {sin}^{2} 2 x$ $=>2sin^2 2x = b/4sin^2 2x$

$\Rightarrow \frac{b}{4} = 2$ $=>b/4=2$

$\Rightarrow b = 8$ $=>b=8$

Answer 2 · 2016-11-28T14:14:53

b = 8

Explanation:

$cos 4 x = 1 - b {sin}^{2} x . {cos}^{2} x$ $cos 4x =1 - bsin^2 x.cos^2 x$ (1)
Use 2 trig identities:
$cos 2 a = 1 - 2 {sin}^{2} a$ $cos 2a = 1 - 2sin^2 a$ , and
sin 2a = 2sin a.cos a

Replace a by x, we get:
$cos 4 x = 1 - 2 {sin}^{2} (2 x)$ $cos 4x = 1 - 2sin^2 (2x)$
Since sin 2x = (2sin x.cos x), then:
$2 {sin}^{2} (2 x) = 2 {(2 sin x . cos x)}^{2}$ $2sin^2 (2x) = 2(2sinx.cos x)^2$
There for:
$cos 4 x = 1 - 2 {(2 sin x . cos x)}^{2} = 1 - 8 {sin}^{2} x . {cos}^{2} x$ $cos 4x = 1 - 2(2sin x.cos x)^2 = 1 - 8sin^2 x.cos^2 x$ . (2)
Compare (1) and (2), we get b = 8.

Science

Math

Humanities

... and beyond

Question #02e7c

2 Answers

Explanation:

Related questions

Impact of this question