Question

Need help with Trig. Question is in attached file I need help with Part C

Answer 1

I need help in Part C Cos x/2

Answer 2

\[\cos(2x)=\cos^2(x)-\sin^2(x) \\ \cos(2x)=\cos^2(x)-(1-\cos^2(x)) \\ \cos(2x)=2\cos^2(x)-1 \]

let u=2x
then u/2=x
so we have
\[\cos(u)=2\cos^2(\frac{u}{2})-1 \\ \cos(u)+1=2\cos^2(\frac{u}{2}) \\ \frac{\cos(u)+1}{2}=\cos^2(\frac{u}{2}) \\ \text{ this going \to change the u back \to x} \\ \frac{\cos(x)+1}{2}=\cos^2(\frac{x}{2})\]

Answer 3

so you first need to know the value for cos(x)

Answer 4

which you drew a triangle
but this triangle is in the second quadrant
so the value for cos(x) should be negative ?

Answer 5

oh ok, yea my bad i drew the triangle wrong, root 5 is suppose to be negative since its in the second quadrant

Answer 6

Thanks for helping :)

Answer 7

ok we also those cos(x/2) will be positive since
we had \[\frac{\pi}{2} \le x \le \pi \\ \text{ so that means that } \\ \frac{\pi}{4} \le \frac{x}{2} \le \frac{\pi}{2}\]
and that means x/2 is in the first quadrant
and we know the cos value there is positive

Answer 8

so when you take the square root of both sides you know you take the + square root and not the negative square root

Answer 9

Alright got the right answer, thanks for helping

Answer 10

np

Answer 11

oh and for part D, would I expand the sin3x to sin(2x+x) and then use the addition formula for sin?

Answer 12

that is what i would do

Answer 13

and you already found sin(2x) and cos(2x)
so it shouldn't be too bad