so on my pretest for our finals in my precalc/trig class i came across a concept with half angle formulas and simplifying expressions with them. for example:"use the half angle formulas to simplify the expression." -(cos3xsin6x)/(1+cos6x). i need help on how to do this type of problem!
half angle formulas!
- Thread starter chayce
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mmm4444bot
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Do you recognize that 3x is half of 6x ?
In other words, the factor of cos(3x) in the numerator of the given ratio can be rewritten as cos(6x/2).
Use the half-angle formula for cosine.
Replace the factor cos(3x) with the negative radical.
Now, rationalize the numerator in the resulting ratio, and simplify algebraically.
I get the following.
\(\displaystyle \frac{sin(6x)}{2 \cdot cos(3x)}\)
Please show whatever work you can, or explain why you're stuck, if you want more help.
Cheers ~ Mark
In other words, the factor of cos(3x) in the numerator of the given ratio can be rewritten as cos(6x/2).
Use the half-angle formula for cosine.
Replace the factor cos(3x) with the negative radical.
Now, rationalize the numerator in the resulting ratio, and simplify algebraically.
I get the following.
\(\displaystyle \frac{sin(6x)}{2 \cdot cos(3x)}\)
Please show whatever work you can, or explain why you're stuck, if you want more help.
Cheers ~ Mark
Hello, chayce!
You're expected to know these formulas:
. . \(\displaystyle \sin2\theta \;=\; 2\sin\theta\cos\theta\)
. . \(\displaystyle \cos2\theta \;=\;\cos^2\!\theta - \sin^2\!\theta \;=\;2\cos^2\!\theta -1 \;=\;1-2\!\sin^2\theta\)
\(\displaystyle \text{Use the formulas to replace }\sin6x\text{ and }\cos6x.\)
\(\displaystyle \text{We have: }\;\frac{\cos3x\overbrace{(2\sin3x\cos3x)}^{\sin6x}}{1 + \underbrace{(2\cos^2\!3x - 1)}_{\cos6x}} \;=\;\frac{2\sin3x\cos^2\!3x}{2\cos^2\!3x} \;=\;\sin3x\)
You're expected to know these formulas:
. . \(\displaystyle \sin2\theta \;=\; 2\sin\theta\cos\theta\)
. . \(\displaystyle \cos2\theta \;=\;\cos^2\!\theta - \sin^2\!\theta \;=\;2\cos^2\!\theta -1 \;=\;1-2\!\sin^2\theta\)
\(\displaystyle \text{Use the formulas to replace }\sin6x\text{ and }\cos6x.\)
\(\displaystyle \text{We have: }\;\frac{\cos3x\overbrace{(2\sin3x\cos3x)}^{\sin6x}}{1 + \underbrace{(2\cos^2\!3x - 1)}_{\cos6x}} \;=\;\frac{2\sin3x\cos^2\!3x}{2\cos^2\!3x} \;=\;\sin3x\)
mmm4444bot
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chayce said:
Oops. Maybe the "example" above is only an example of an instruction (heh, heh, heh).