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)

Grandad