Trig Expressions

Tom G · #1 $Old$ October 3rd, 2007, 01:00 PM

Hi, I've got to re-write 3sin x cos x + 4cos^2 x in the form a sin2x+b sin2x +c and I'm struggling. I've already asked my maths teacher for help and she told me how to do it but I can only remember half of the process and am now confused. Any help would be appreciated.

Soroban · #2 $Old$ October 3rd, 2007, 01:52 PM

Hello, Tom G!

Quote:

I've got to re-write $3\sin x\cos x + 4\cos^2\!x$
in the form $a\sin2x + b\sin2x +c$ . . . . This can't be right!
. . Do you mean: . ${\color{blue}a\sin^2\!x + b\sin2x + c}$

You're expected to know these two identities: . $\begin{array}{ccc}2\sin\theta\cos\theta & = & \sin2\theta\\ \cos^2\!\theta & = & 1 - \sin^2\!\theta\end{array}$

Then: . $3\sin x\cos x + 4\cos^2\!x \;=\;\frac{3}{2}(2\sin x\cos x) + 4\left(1 - \sin^2\!x\right)$

. . $= \;\frac{3}{2}\sin2x + 4 - 4\sin^2\!x \;=\; -4\sin^2\!x + \frac{3}{2}\sin2x + 4$

Tom G · #3 $Old$ October 3rd, 2007, 02:25 PM

Hi, I've checked the wording of the answer and have copied it down exactly:
"Express 3sin x cos x + 4sin^2 x in the form a sin2x +b cos2x +c where a,b and c are constants to be found." - sorry i copied it down wrong.

Soroban · #4 $Old$ October 3rd, 2007, 02:39 PM

Hello, Tom G!

Okay, that makes it clearer . . .

Quote:

Express . $3\sin x\cos x + 4\sin^2\!x$ .in the form . $a\sin2x + b\cos2x + c$

We will use: . $2\sin x\cos x \:=\: \sin2x\qquad \sin^2\!x \:=\: \frac{1 - \cos2x}{2}$

We have: . $3\sin x\cos x + 4\sin^2x \;=\;\frac{3}{2}(2\sin x\cos x) + 4\left(\frac{1 - \cos2x}{2}\right)$

. . $= \;\frac{3}{2}\sin2x + 2(1 - \cos2x) \;=\;\frac{3}{2}\sin2x - 2\cos2x + 2$