Finding angles

**Punch** · January 14th 2010, 01:45 AM

Find all the angles between $0^\circ$ and $360^\circ$ which satisfy the equation $sec(2x+30^\circ)=sec(2x-60^\circ)$

**Prove It** · January 14th 2010, 02:08 AM

Originally Posted by Punch

Find all the angles between $0^\circ$ and $360^\circ$ which satisfy the equation $sec(2x+30^\circ)=sec(2x-60^\circ)$

$\sec{(2x + 30^\circ)} = \sec{(2x - 60^\circ)}$

$\cos{(2x + 30^\circ)} = \cos{(2x - 60^\circ)}$ .

Use the identity $\cos{(\alpha \pm \beta)} = \cos{\alpha}\cos{\beta} \mp \sin{\alpha}\sin{\beta}$

So $\cos{2x}\cos{30^\circ} - \sin{2x}\sin{30^\circ} = \cos{2x}\cos{60^\circ} + \sin{2x}\sin{60^\circ}$

$\frac{\sqrt{3}}{2}\cos{2x} - \frac{1}{2}\sin{2x} = \frac{1}{2}\cos{2x} + \frac{\sqrt{3}}{2}\sin{2x}$

$\frac{\sqrt{3} - 1}{2}\cos{2x} = \frac{\sqrt{3} + 1}{2}\sin{2x}$

$\frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \tan{2x}$

$\frac{(\sqrt{3} - 1)^2}{3 - 1} = \tan{2x}$

$\frac{3 - 2\sqrt{3} + 1}{2} = \tan{2x}$

$\frac{4 - 2\sqrt{3}}{2} = \tan{2x}$

$2 - \sqrt{3} = \tan{2x}$

This might seem a bit off track, but if you're good with your half angle identities, you will know that

$\tan{\frac{\theta}{2}} = \frac{1 - \cos{\theta}}{\sin{\theta}}$ .

Using this identity, what does $\tan{15^\circ}$ equal? (It will help you solve your equation...)

**mathaddict** · January 14th 2010, 02:25 AM

Originally Posted by Punch

Find all the angles between $0^\circ$ and $360^\circ$ which satisfy the equation $sec(2x+30^\circ)=sec(2x-60^\circ)$

HI

$\cos (2x-60)=\cos (2x+30)$

$2x-60=2n(180) \pm (2x+30)$

$2x-60=2n(180) -2x-30$

$2n(180)=4x-30$

$180n=2x-15$ , where n=0,1,2,3

**Punch** · January 14th 2010, 03:02 AM

Originally Posted by Prove It

$\sec{(2x + 30^\circ)} = \sec{(2x - 60^\circ)}$

$\cos{(2x + 30^\circ)} = \cos{(2x - 60^\circ)}$ .

Use the identity $\cos{(\alpha \pm \beta)} = \cos{\alpha}\cos{\beta} \mp \sin{\alpha}\sin{\beta}$

So $\cos{2x}\cos{30^\circ} - \sin{2x}\sin{30^\circ} = \cos{2x}\cos{60^\circ} + \sin{2x}\sin{60^\circ}$

$\frac{\sqrt{3}}{2}\cos{2x} - \frac{1}{2}\sin{2x} = \frac{1}{2}\cos{2x} + \frac{\sqrt{3}}{2}\sin{2x}$

$\frac{\sqrt{3} - 1}{2}\cos{2x} = \frac{\sqrt{3} + 1}{2}\sin{2x}$

$\frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \tan{2x}$

$\frac{(\sqrt{3} - 1)^2}{3 - 1} = \tan{2x}$

$\frac{3 - 2\sqrt{3} + 1}{2} = \tan{2x}$

$\frac{4 - 2\sqrt{3}}{2} = \tan{2x}$

$2 - \sqrt{3} = \tan{2x}$

This might seem a bit off track, but if you're good with your half angle identities, you will know that

$\tan{\frac{\theta}{2}} = \frac{1 - \cos{\theta}}{\sin{\theta}}$ .

Using this identity, what does $\tan{15^\circ}$ equal? (It will help you solve your equation...)

I can't continue

**Prove It** · January 14th 2010, 03:09 AM

Originally Posted by Punch

I can't continue

I told you how to continue.

Use the final identity I gave you to evaluate $\tan{15^\circ}$ .

$\tan{15^\circ} = \tan{\frac{30^\circ}{2}}$ .

It will help you.

**Punch** · January 14th 2010, 03:13 AM

Originally Posted by Prove It

I told you how to continue.

Use the final identity I gave you to evaluate $\tan{15^\circ}$ .

$\tan{15^\circ} = \tan{\frac{30^\circ}{2}}$ .

It will help you.

may i ask where did u get the 15 from?

**Prove It** · January 14th 2010, 03:14 AM

Experience.

Try it.

**Punch** · January 14th 2010, 03:15 AM

Originally Posted by Prove It

Experience.

Try it.

I know you are an expert lol but i need to know the way in solving the problems and not by luck... i can't have u sitting beside me in the exam right

**Punch** · January 14th 2010, 03:16 AM

or do you tangent inverse to find what 2x is equals to? or using special angle

**Prove It** · January 14th 2010, 03:27 AM

I would advise

1. Stop arguing with the person who is trying to help you, instead do as you've been instructed so that you learn something and hopefully will be able to remember this for an exam.

2. Use the half-angle identities for sine, cosine and tangent to find the exact surd answers when your angle is $15^\circ$ . You can also find exact values when your angle is $22.5^{\circ}$ , or any successive half angle.

**mr fantastic** · January 14th 2010, 03:30 AM

Originally Posted by Punch

or do you tangent inverse to find what 2x is equals to? or using special angle

If you thought more about post #2 your life would be less stressful.

**Punch** · January 14th 2010, 04:19 AM

Sorry if my words offended you but i truly don't understand and am not arguing with u, i know u r trying to help me and i am truly thankful to u however i really do not understand where the 15 came from...

**differentiate** · January 14th 2010, 05:16 AM

If you work out what tan15 is equal to, it will make sense where the 15 comes from and why it is being used.

alternatively
2 - root3 = tan2x
2 - root3 = (2tanx)/(1 - tan^2x)

simplify to find tanx

**Punch** · January 14th 2010, 05:28 AM

Oh so it's similar to using a special angle... $2 - \sqrt{3} = \tan{15}$
then use double angle formula to solve for x

Don't mind me asking but why is this type of question coming up when I don't learn anything about the special angles of tan15 and such? I have only heard of tan 30, tan 60, tan 90 and so on..

**mathaddict** · January 14th 2010, 05:57 AM

Originally Posted by Punch

Oh so it's similar to using a special angle... $2 - \sqrt{3} = \tan{15}$
then use double angle formula to solve for x

Don't mind me asking but why is this type of question coming up when I don't learn anything about the special angles of tan15 and such? I have only heard of tan 30, tan 60, tan 90 and so on..

hi again punch ,

nowdays questions don't confine to only whatever you have learnt so be prepared . It's good that you learnt the extras . However , in this case , tan 15 is related to whatever you have learnt , the special angle and double angle formula .

$\tan 30=\frac{1}{\sqrt{3}}$

$\tan 2(15)=\frac{1}{\sqrt{3}}$

$\frac{2\tan 15}{1-\tan^2 15}=\frac{1}{\sqrt{3}}$

$1-\tan^2 15=2\sqrt{3}\tan 15$

$\tan^2 15+2\sqrt{3}\tan 15 -1=0$

Using the quadratic formula ,

$<br /> \tan 15=\frac{-2\sqrt{3}\pm \sqrt{12-4(-1)}}{2}<br />$

$=\pm 2-\sqrt{3}$

$-2-\sqrt{3}$ is not possible because tan 15 is in the first quadrant , everything here will be positive .

therefore , $\tan 15=2-\sqrt{3}$

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