1. ## Trig identites question

Hi
I need help on the following question:

1)Express the function y=cos\theta + sin\theta in the form
$\displaystyle y = Rcos(\theta - \alpha)$

2)Express the following as algebraic expressions in terms of x: $\displaystyle cos(2arccos(x))$

This is what i have done, but i don't understand how this is incorrect.

$\displaystyle make 2arccos(x) = \theta$

$\displaystyle 2cos(arccos(x)) = cos\theta$

$\displaystyle cos\theta = 2x$

$\displaystyle cos\theta = \frac{adj}{hyp} = \frac{2x}{1}$

so that means

$\displaystyle cos\theta = \frac{adj}{hyp} = 2x$

P.S

2. 1)Express the function y=cos\theta + sin\theta in the form
y = sqrt(2)[cos(theta)/sqrt(20 - sin(theta)sqrt(2)
= sqrt(2)[cos(theta-pi/4)

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

3. Originally Posted by sa-ri-ga-ma
1)Express the function $\displaystyle y=cos\theta + sin\theta$ in the form
$\displaystyle y = \sqrt(2)[cos(\theta)\sqrt(20 - sin(\theta)\sqrt(2)$
$\displaystyle + \sqrt(2)[cos(\frac{\theta-\pi}{4})$

2)$\displaystyle 2(arccos(x) = \theta$
$\displaystyle arccos(x) = \frac{\theta}{2}$
$\displaystyle (x) = cos(\frac{\theta}{2})$
$\displaystyle cos(\theta) = 2cos^2(\frac{\theta}{2}) - 1 = 2x^2 -1$
For the first question i don't quite understand what you have done. Where did the square roots come from?

For the second question how did you get
$\displaystyle cos(\theta) = 2cos^2(\frac{\theta}{2}) - 1$

4. Hello Paymemoney
Originally Posted by Paymemoney
Hi
I need help on the following question:

1)Express the function y=cos\theta + sin\theta in the form
$\displaystyle y = Rcos(\theta - \alpha)$
Let
$\displaystyle \cos\theta+\sin\theta=R\cos(\theta-\alpha)$
Expand the RHS using the formula $\displaystyle \cos(A-B)=\cos A \cos B + \sin A \sin B$:
$\displaystyle \Rightarrow\cos\theta+\sin\theta=R\cos\theta\cos\a lpha + R\sin\theta\sin\alpha$
Now compare the coefficients of $\displaystyle \cos\theta$ and $\displaystyle \sin\theta$ on both sides of the equation:
$\displaystyle \left\{\begin{array}{l l}1 = R\cos\alpha&\text{... (1)}\\1 = R\sin\alpha&\text{... (2)}\end{array}\right.$
Square (1) and (2) and add, noting that $\displaystyle \cos^2\alpha+\sin^2\alpha = 1$:
$\displaystyle 1^2+1^2=R^2$

$\displaystyle \Rightarrow R = \sqrt2$

Divide (2) by (1):
$\displaystyle \frac{R\sin\alpha}{R\cos\alpha}=\tan\alpha = 1$

$\displaystyle \Rightarrow \alpha = \pi/4$

Substituting these values of $\displaystyle R$ and $\displaystyle \alpha$ back into the original equation gives:
$\displaystyle \cos\theta+\sin\theta= \sqrt2\cos(\theta-\pi/4)$
2)Express the following as algebraic expressions in terms of x: $\displaystyle cos(2arccos(x))$
Let
$\displaystyle 2\arccos(x) = \theta$
Then:
$\displaystyle \arccos(x) = \tfrac12\theta$

$\displaystyle \Rightarrow x = \cos(\tfrac12\theta)$ ...(3)
Also:
$\displaystyle \cos(2\arccos(x)) = \cos(\theta)$
$\displaystyle =2\cos^2(\tfrac12\theta)-1$, using the formula $\displaystyle \cos2A = 2\cos^2A-1$

$\displaystyle =2x^2-1$, from (3)

5. so from the formula: cos2A are you making arccos(x) = A?

I though that cos(2arccos(x)) = 2x, why wouldn't this work?

6. Hello, Paymemoney!

The first one requires an imaginative step . . .

$\displaystyle \text{1) Express the function: }\:y \:=\:\cos\theta + \sin\theta\,\text{ in the form: }\:y \:=\: R\cos(\theta - \alpha)$

We are given: .$\displaystyle y \;=\;\cos\theta + \sin\theta$

Divide by $\displaystyle \sqrt{2}\!:\;\;\frac{1}{\sqrt{2}}\,y \;=\;\frac{1}{\sqrt{2}}\,\cos\theta + \frac{1}{\sqrt{2}}\,\sin\theta$ .[1]

. . We know that: .$\displaystyle \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}},\;\; \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}$

Substitute into [1]: .$\displaystyle \frac{1}{\sqrt{2}}\,y \;=\;\cos\frac{\pi}{4}\cos\theta + \sin\frac{\pi}{4}\sin\theta$

And this becomes: .$\displaystyle \frac{1}{\sqrt{2}}\,y \;=\; \cos\left(\theta - \frac{\pi}{4}\right)$

Multiply by $\displaystyle \sqrt{2}:\;\;y \;=\;\sqrt{2}\cos\left(\theta - \frac{\pi}{4}\right)$

7. Hello Paymemoney
Originally Posted by Paymemoney
so from the formula: cos2A are you making arccos(x) = A?
OK. Let's see if we can make it easier. I'll set it out slightly differently.

Let
$\displaystyle \arccos(x)=A$
Now $\displaystyle \arccos...$ means 'the angle whose cosine is ...'. So
$\displaystyle \arccos(x) = A$
means
the angle whose cosine is $\displaystyle x$ is $\displaystyle A$
In other words
$\displaystyle \cos A = x$, which I'll call equation (1)
Now the question asks us for
$\displaystyle \cos(2\arccos(x))$
In other words, for
$\displaystyle \cos2A$
So we use the well-known double angle formula
$\displaystyle \cos2A=2\cos^2A-1$
and then use equation (1) to replace $\displaystyle \cos A$ by $\displaystyle x$. So we get:
$\displaystyle \cos(2\arccos(x)) = \cos2A$
$\displaystyle =2\cos^2A -1$

$\displaystyle =2x^2-1$

OK now?
Originally Posted by Paymemoney
I though that cos(2arccos(x)) = 2x, why wouldn't this work?
I'm not sure why you should think this. This would mean that if you double an angle you double its cosine. So $\displaystyle \cos2A$ would equal $\displaystyle 2\cos A$. Which it doesn't, does it?