1. Who knows Trigonometric Functions?

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.

2. 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.

3. Here is one of them:

Prove:

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

Prove:

4. Here is my full assignment:

5. Originally Posted by Anonymous14
Here is one of them:

Prove:

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

Prove:

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

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

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

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

$\displaystyle =sec x= LHS$

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

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

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

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

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

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

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

From (a) and (b),

LHS=RHS

6. Originally Posted by Anonymous14
Here is one of them:

Prove:

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

Prove:

You know that

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

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

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

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

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

$\displaystyle =\sec x$ = Left side

7. Originally Posted by Anonymous14
Here is one of them:

Prove:

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

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

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

LS=RS

8. Thank you both so much. I will +rep you. I might have a few more questions that I don't understand.

9. Originally Posted by Anonymous14
Here is my full assignment:
Please see these two solution in attachment file, and more coming..........

10. 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.

11. Originally Posted by Anonymous14
See attachment,page2.

12. 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) ?

13. 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

$\displaystyle \sin 29^\circ = 0.48480962$

change $\displaystyle 29^\circ$ into radians,

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

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

Did you get it???

14. Originally Posted by Shyam
Any more questions?

x = 29

$\displaystyle \sin 29^\circ = 0.48480962$

change $\displaystyle 29^\circ$ into radians,

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

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

Did you get it???
Yup, Thanks again for all of your help.

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