Question

cos 3x over cos x=1-4sin^2 x

Answer 1

use
,\[\cos 3x=4\cos ^{3 }x-3\cos x \]

Answer 2

^???????

Answer 3

\[\frac{ \cos 3x }{ \cos x }=\frac{ 4\cos ^{3}x-3\cos x }{ \cos x }\]

Answer 4

take cosx common from the numerator and cancel with denominator.

Answer 5

@surjithayer which rule did you use to get 4cos^3x-3cosx ?

Answer 6

\[then use \sin ^{2}x+\cos ^{2}x=1\]

Answer 7

it is also a formula

Answer 8

I'm not understanding anything

Answer 9

Sketch of my solution:

Let cos 3x=cos(x+2x) use addition rule, simplify.
You will have terms with cos 2x and sin 2x; use the double angle rule, continue to simplify. Eventually, get

\[\cos^2x+\sin^2x=1\]
Done.

Answer 10

\[\cos 3x=\cos \left( 2x+x \right)=\cos 2x \cos x-\sin 2x \sin x\]
\[=\left( \cos ^{2}x-\sin ^{2}x \right)\cos x-2\sin x \cos x \sin x\]
\[=\left\{ \cos ^{2}x-\left( 1-\cos ^{2}x \right) \right\}\cos x-2\cos x \sin ^{2}x\]
\[=(\cos ^{2}x-1+\cos ^{2}x)\cos x-2\cos x \left( 1-\cos ^{2}x \right)\]
\[=2\cos ^{3}x-\cos x-2\cos x+2\cos ^{3}x=4\cos ^{3}x-3\cos x\]

Answer 11

i think now you know the formula for cos 3x

Answer 12

What happens to 1-4sin^2 x

Answer 13

\[\frac{ \cos 3x }{ \cos x }=\frac{ 4\cos ^{3}x-3\cos x }{ \cos x }=\frac{ \cos x \left( 4\cos ^{2}x-3 \right) }{\cos x }\]
\[=4\cos ^{2}x-3=4\left( 1-\sin ^{2}x \right)-3=4-4\sin ^{2}x-3=1-4\sin ^{2}x\]

Answer 14

\[\frac{\cos3x}{cosx}=1-4\sin^2x \implies \frac{\cos(2x+x)}{cosx}=1-4\sin^2x\]\[=\frac{\cos2xcosx-\sin2xsinx}{cosx}=\cos2x-\sin2xtanx=1-4\sin^2x\]

Answer 15

\[\cos^2x-\sin^2x-2sinxcosxtanx=\cos^2x-\sin^2x-2\sin^2x=1-4\sin^2x\]\[\implies \cos^2x+\sin^2x=1~~~QED\]

Answer 16

Work only on the LHS until the last step (combining like terms). It's a little tedious, but not horrible if you use your rules and pay attention to the algebra.

Answer 17

Thanks

Answer 18

No sweat. Do math every day.