angles help. sin0.5x

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• June 22nd 2009, 06:28 AM
helloying
angles help. sin0.5x
Given that x is an obtuse angle and sin x =3/5, find, without the use of calculator, cos0.5x

please help me thanks. Also, i have something to clarify,in cases like this: $2\sin x\cos x = 3\cos^2 x$solve for x between 0 to 180 deg
can i just cancel out cos x ? then i will have tan x =3/2 x=56.3

but since i cancel out the cos x so it is suppose to be zero so i just put cos x =0 and x=90

the zero part doesnt sound very logical to me.is it right to do that?
• June 22nd 2009, 06:58 AM
mathaddict
Quote:

Originally Posted by helloying
Given that x is an obtuse angle and sin x =3/5, find, with the use of calculator, cos0.5x

please help me thanks. Also, i have something to clarify,in cases like this: $2\sin x\cos x = 3\cos^2 x$solve for x between 0 to 180 deg
can i just cancel out cos x ? then i will have tan x =3/2 x=56.3

but since i cancel out the cos x so it is suppose to be zero so i just put cos x =0 and x=90

the zero part doesnt sound very logical to me.is it right to do that?

(1)sin x =3/5 (x is in quadrant 2)

x=143.13 degree

Then use the calculator to find cos 0.5(143.13)

(2) $2\sin x\cos x = 3\cos^2 x$

No u cannot cancel like that .

3cos^2x-2sinxcosx=0

cosx(3cosx-2sinx)=0

cosx=0 , x = ..

3cosx=2sinx

3/2=tanx , x=..
• June 22nd 2009, 07:28 AM
helloying
Quote:

Originally Posted by mathaddict
(1)sin x =3/5 (x is in quadrant 2)

x=143.13 degree

Then use the calculator to find cos 0.5(143.13)

(2) $2\sin x\cos x = 3\cos^2 x$

No u cannot cancel like that .

3cos^2x-2sinxcosx=0

cosx(3cosx-2sinx)=0

cosx=0 , x = ..

3cosx=2sinx

3/2=tanx , x=..

thank you very much for ur reply but i am sorry, for the 1st part i have asked for how to solve it without the use of calculator.i have typed wrongly . i corrected my mistake now.
• June 22nd 2009, 07:40 AM
great_math
Quote:

Originally Posted by helloying
Given that x is an obtuse angle and sin x =3/5, find, without the use of calculator, cos0.5x

please help me thanks. Also, i have something to clarify,in cases like this: $2\sin x\cos x = 3\cos^2 x$solve for x between 0 to 180 deg
can i just cancel out cos x ? then i will have tan x =3/2 x=56.3

but since i cancel out the cos x so it is suppose to be zero so i just put cos x =0 and x=90

the zero part doesnt sound very logical to me.is it right to do that?

sin x=3/5 so $\cos x=\frac{-4}{5}$

Now, $\cos2\theta=2\cos^2\theta-1$

$\cos(2\times0.5x)=2\cos^20.5x-1$

$\Rightarrow \cos x+1=2\cos^20.5x$

$\Rightarrow \frac1{5}=2\cos^20.5x$

$\Rightarrow \cos^20.5x=\frac1{10}$

$\Rightarrow \cos 0.5x=\frac{1}{\sqrt{10}}$

$\Rightarrow \cos 0.5x\approx\frac1{3.16}\approx 0.31$
• June 22nd 2009, 08:04 AM
mathaddict
Quote:

Originally Posted by helloying
thank you very much for ur reply but i am sorry, for the 1st part i have asked for how to solve it without the use of calculator.i have typed wrongly . i corrected my mistake now.

i felt really awkward when u said with the calculator .. btw without the calculator , you will need to use the half angle formula .

From the double angle formula :

cos 2a = 2cos^2a-1

Let a =x/2

cos x= 2cos^2(x/2)-1

(cosx+1)/2=cos^2(x/2)

$
cos(\frac{x}{2})=\pm\sqrt{\frac{cosx+1}{2}}
$

Now , find $cos\frac{1}{2}x$

substitute x=143.13 and take the positive value .