# Math Help - [SOLVED] (7cosx+5sinx)²=Acos2x+Bsin2x+C

1. ## [SOLVED] (7cosx+5sinx)²=Acos2x+Bsin2x+C

Hello there

I need help with the following trig question...

(7cosx+5sinx)²=Acos2x+Bsin2x+C

(7cosx+5sinx)²=24cos²x+35sin2x+25

I need to do something about the 24cos²x but im unsure how to proceed... I was thinking i could use cos²(x/2)=(1+cosx)/2

but im not sure...

Janne the Swede

2. Originally Posted by JannetheSwede
Hello there

I need help with the following trig question...

(7cosx+5sinx)²=Acos2x+Bsin2x+C

(7cosx+5sinx)²=24cos²x+35sin2x+25

I need to do something about the 24cos²x but im unsure how to proceed... I was thinking i could use cos²(x/2)=(1+cosx)/2

but im not sure...

Janne the Swede
$(7\cos{x} + 5\sin{x})^2 = 49\cos^2{x} + 70\cos{x}\sin{x} + 25\sin^2{x}$

$= 24\cos^2{x} + 25\cos^2{x} + 35\cdot 2\cos{x}\sin{x} + 25\sin^2{x}$

$= 24\cos^2{x} + 35\sin{(2x)} + 25(\cos^2{x} + \sin^2{x})$

$= 24\cos^2{x} + 35\sin{(2x)} + 25\cdot 1$

$= 24\cos^2{x} + 35\sin{(2x)} + 25$.

3. mmh, thats about as far as I got, but my awnser have to be A*cos (2x)+B*sin (2x)+C where A, B and C are integers...

somehow i have to get rid of 24*cos²x but i don't know how

4. Originally Posted by JannetheSwede
mmh, thats about as far as I got, but my awnser have to be A*cos (2x)+B*sin (2x)+C where A, B and C are integers...

somehow i have to get rid of 24*cos²x but i don't know how
Oh yeah, you need to remember

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

And since $\sin^2{x} = 1 - \cos^2{x}$ we get

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

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

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

So substitute this back into the equation we found:

$24\cos^2{x} + 35\sin{(2x)} + 25 = 24\left(\frac{1}{2} + \frac{1}{2}\cos{(2x)}\right) + 35\sin{(2x)} + 25$

$= 12 + 12\cos{(2x)} + 35\sin{(2x)} + 25$

$= 12\cos{(2x)} + 35\sin{(2x)} + 37$.

So $A = 12, B = 35, C = 37$.