Who knows Trigonometric Functions?

**Anonymous14** · July 11th 2009, 09:33 AM

I have an assignment due today at 5 (around 3 hours) and I know I shouldn't of left it this late. I basically get nothing on it, and if anyone here could explain to me how to do it or go through the questions with me, it would be great.
Thanks.

Btw, it is grade 12 functions level.

**Shyam** · July 11th 2009, 09:42 AM

Originally Posted by Anonymous14

I have an assignment due today at 5 (around 3 hours) and I know I shouldn't of left it this late. I basically get nothing on it, and if anyone here could explain to me how to do it or go through the questions with me, it would be great. Thanks.

Btw, it is grade 12 functions level.

Please write the exact questions that you have.

**Anonymous14** · July 11th 2009, 09:45 AM

Here is one of them:

Prove:

sec(x) = 2(cosxsin2x - sinxcos2x)/sin2x

Prove:

**Anonymous14** · July 11th 2009, 09:50 AM

Here is my full assignment:

**alexmahone** · July 11th 2009, 09:50 AM

Originally Posted by Anonymous14

Here is one of them:

Prove:

sec(x) = 2(cosxsin2x - sinxcos2x)/sin2x

Prove:

1) $RHS=\frac{2(cosxsin2x - sinxcos2x)}{sin2x}$

$= \frac{2sin (2x-x)}{sin 2x}$ {Using the sin (A-B) identity}

$=\frac{2sin x}{2sin x cos x}$ { sin 2x=2sinx cosx }

$=\frac{1}{cos x}$

$=sec x= LHS$

---------------------------------------------------------------------

2) $LHS=\frac{1-sin^2x}{cos x}$

$= \frac{cos^2x}{cos x}$

$= cos x$ ------------------------- (a)

$RHS=\frac{sin (2x)}{2sin (x)}$

= $\frac{2sin x cos x}{2 sin x}$

$=cos x$ ---------------------------(b)

From (a) and (b),

LHS=RHS

**Shyam** · July 11th 2009, 09:51 AM

Originally Posted by Anonymous14

Here is one of them:

Prove:

sec(x) = 2(cosxsin2x - sinxcos2x)/sin2x

Prove:

You know that

$\sin (A-B) = \sin A \cos B-\cos A\sin B$

$\sin 2x = 2\sin x\cos x$

Right Side $=\frac{2(\cos x \sin 2x - \sin x\cos 2x)}{\sin 2x}$

$=\frac{\sin(2x-x)}{\sin 2x}$

$=\frac{\sin x}{2\sin x \cos x}=\frac{1}{\cos x}$

$=\sec x$ = Left side

**Shyam** · July 11th 2009, 10:01 AM

Originally Posted by Anonymous14

Here is one of them:

Prove:

$LS=\frac{1-\sin^2 x}{\cos x}=\frac{\cos^2 x}{\cos x}=\cos x$

$RS= \frac{\sin 2x}{2\sin x}$

$=\frac{2\sin x \cos x}{2\sin x} = \cos x$

LS=RS

**Anonymous14** · July 11th 2009, 10:03 AM

Thank you both so much.

I will +rep you. I might have a few more questions that I don't understand.

**Anonymous14** · July 11th 2009, 10:08 AM

**Shyam** · July 11th 2009, 10:26 AM

Originally Posted by Anonymous14

Here is my full assignment:

Please see these two solution in attachment file, and more coming..........

**Anonymous14** · July 11th 2009, 10:54 AM

Originally Posted by Shyam

Please see these two solution in attachment file, and more coming..........

Thanks so much. Please don't do number 6 because I already understand that question perfectly.

**Shyam** · July 11th 2009, 11:05 AM

Originally Posted by Anonymous14

See attachment,page2.

**Anonymous14** · July 11th 2009, 11:26 AM

Thanks, Also, this one...

sin(x) = ??????? similarly tan(pi - x) = ????????? verify on your calculator for x = 29º, or angle of your choice.

Is the first one cos(pi/2 - x) ?

**Shyam** · July 11th 2009, 11:39 AM

Originally Posted by Anonymous14

Thanks so much. Please don't do number 6 because I already understand that question perfectly.

Any more questions?

x = 29

$\sin 29^\circ = 0.48480962$

change $29^\circ$ into radians,

$29^\circ=29^\circ\times\frac{ \pi}{180}=0.506145483\; rad$

$\tan(\pi-x)=\tan(\pi-0.506145483)=\tan(2.635447171)=-0.5543$

Did you get it???

**Anonymous14** · July 11th 2009, 11:54 AM

Originally Posted by Shyam

Any more questions?

x = 29

$\sin 29^\circ = 0.48480962$

change $29^\circ$ into radians,

$29^\circ=29^\circ\times\frac{ \pi}{180}=0.506145483\; rad$

$\tan(\pi-x)=\tan(\pi-0.506145483)=\tan(2.635447171)=-0.5543$

Did you get it???

Yup, Thanks again for all of your help.

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