1. ## Help!

Use the formula for $\displaystyle cos(A+B)$ to show that $\displaystyle cos^2x=\frac{1}{2}(1+cos2x)$, and write down a similar expression for $\displaystyle sin^2x$.

Given that $\displaystyle f(x) = 10cos^2x+2sin^2x=6sinxcosx$, express f(x) in the form $\displaystyle Q + Rsin(2x+a)$, where Q, R are constants $\displaystyle tana=\frac{4}{3}$.

Hence find the greatest and least values of$\displaystyle f(x)$.

I could solve the first part using other formulas but not cos(A+B) formula...

Having problems with expressing sin^2x in a similar way...

Also asking if if the first paragraph is related to the second paragraph of questions...

2. Originally Posted by Punch
Use the formula for $\displaystyle cos(A+B)$ to show that $\displaystyle cos^2x=\frac{1}{2}(1+cos2x)$, and write down a similar expression for $\displaystyle sin^2x$.

Given that $\displaystyle f(x) = 10cos^2x+2sin^2x=6sinxcosx$, express f(x) in the form $\displaystyle Q + Rsin(2x+a)$, where Q, R are constants $\displaystyle tana=\frac{4}{3}$.

Hence find the greatest and least values of$\displaystyle f(x)$.

I could solve the first part using other formulas but not cos(A+B) formula...

Having problems with expressing sin^2x in a similar way...

Also asking if if the first paragraph is related to the second paragraph of questions...
1. $\displaystyle \cos{(\alpha + \beta)} = \cos{\alpha}\cos{\beta} - \sin{\alpha}\sin{\beta}$.

So if $\displaystyle \alpha = \beta = x$, then

$\displaystyle \cos{(x + x)} = \cos{x}\cos{x} - \sin{x}\sin{x}$

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

Solving for $\displaystyle \cos^2{x}$...

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

$\displaystyle \cos{2x} = \cos^2{x} - (1 - \cos^2{x})$

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

$\displaystyle 1 + \cos{2x} = 2\cos^2{x}$

$\displaystyle \cos^2{x} = \frac{1}{2}(1 + \cos{2x})$.

Solving for $\displaystyle \sin^2{x}$...

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

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

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

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

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