more identities.(in a hurry)

**Dave19** · July 16th 2008, 07:16 PM

need the work already have answers.

1)what is this equivalent to cos(180degrees + B) answer:-cosB

2)use an identity to find an exact value for sin 105degrees.

answer:root6+root2/4

3)use an identity to find an exact value for 22.5degrees.

answer:root2-1

**Isomorphism** · July 16th 2008, 07:42 PM

Originally Posted by Dave19

need the work already have answers.

1)what is this equivalent to cos(180degrees + B) answer:-cosB

$\cos (180^{\circ} + B) = \cos 180^{\circ} \cos B - \sin 180^{\circ} \sin B$

But $\cos 180^{\circ} = -1$ and $\sin 180^{\circ} = 0$

$\cos (180^{\circ} + B) = \cos 180^{\circ} \cos B - \sin 180^{\circ} \sin B = (-1) \cos B = -\cos B<br />$

2)use an identity to find an exact value for sin 105degrees.

answer:root6+root2/4

$\sin 105^{\circ} = \sin (90^{\circ} + 15^{\circ}) = \cos 15^{\circ}$

Use the double angle formula: $\cos 2x = 2cos^2x - 1$

put $x = 15^{\circ}$ , since you already know $\cos 30^{\circ} = \frac{\sqrt{3}}2$ .

3)use an identity to find an exact value for 22.5degrees.

answer:root2-1

DO you mean sin of 22.5 degrees or cosine of 22.5 degrees? Whoat does the phrase "exact value of 22.5 degrees mean"?

**Dave19** · July 16th 2008, 08:01 PM

sorry its tan and can you try my other questions if you can find them.

**Dave19** · July 16th 2008, 08:03 PM

its under urgent homework help i think

**Isomorphism** · July 16th 2008, 08:06 PM

Originally Posted by Dave19

sorry its tan and can you try my other questions if you can find them.

$\tan 2x = \frac{2\tan x}{1 - \tan^2 x}$ and $\tan 45 = 1$

Now if x = 22.5, 2x = 45. So

$\tan 2x = \tan 45 = 1 = \frac{2\tan x}{1 - \tan^2 x}$

$1 = \frac{2\tan x}{1 - \tan^2 x}$

Now solve the above equation for tan x. It is a quadratic equation which you can easily solve.

**Dave19** · July 16th 2008, 08:13 PM

i dont no how that will equal root2-1

**Isomorphism** · July 16th 2008, 08:27 PM

Originally Posted by Isomorphism

$\tan 2x = \frac{2\tan x}{1 - \tan^2 x}$ and $\tan 45 = 1$

Now if x = 22.5, 2x = 45. So

$\tan 2x = \tan 45 = 1 = \frac{2\tan x}{1 - \tan^2 x}$

$1 = \frac{2\tan x}{1 - \tan^2 x}$

Now solve the above equation for tan x. It is a quadratic equation which you can easily solve.

$1 = \frac{2\tan x}{1 - \tan^2 x} \Rightarrow 1 - \tan^2 x = 2\tan x$

If it is hard for you to see, put $u =\tan x$ ,

$1 - \tan^2 x = 2\tan x \Rightarrow 1 - u^2 = 2u \Rightarrow u^2 +2u -1 = 0$

Now try solving it. Come on show us some effort... You want to learn, dont you?

Ask me if you dont know...

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