Solve the equation by first using a Sum-to-Product Formula. \cos 5 \theta - \cos 7 \theta = 0 \cos 5 \theta - \cos 7 \theta = 0

Aneeka Hunt 2021-08-04 Answered
Using Sum-to-Product Formulas Solve the equation by first using a Sum-to-Product Formula.
\(\displaystyle{\cos{{5}}}\theta-{\cos{{7}}}\theta={0}{\cos{{5}}}\theta-{\cos{{7}}}\theta={0}\)

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Plainmath recommends

  • Ask your own question for free.
  • Get a detailed answer even on the hardest topics.
  • Ask an expert for a step-by-step guidance to learn to do it yourself.
Ask Question

Expert Answer

cyhuddwyr9
Answered 2021-08-05 Author has 11518 answers

Approach:
The range of the trigonometric functions of \(\displaystyle{\sin{\theta}}\) is lie between \(\displaystyle{\left[-{1},{1}\right]}{\left[-{1},{1}\right]}\). No solution exists beyond this range.
Simplify the equation.
Obtain the factors of the equation.
The sum-to-product formulas for cosine is,
\(\displaystyle{\cos{{u}}}+{\cos{{v}}}=-{2}{\sin{{\frac{{{u}+{v}}}{{{2}}}}}}{\sin{{\frac{{{u}+{v}}}{{{2}}}}}}{\cos{{u}}}-{\cos{{v}}}=-{2}{\sin{{u}}}+{v}{2}{\sin{{u}}}-{v}{2}\)
Cosine function has period \(\displaystyle{2}\pi{2}\pi\), thus find the solution in any interval of length \(\displaystyle{2}\pi{2}\pi\). Sine function is positive in first and second quadrant.
Calculation:
Consider the equation.
\(\displaystyle{\cos{{5}}}\theta-{\cos{{7}}}\theta={0}{\cos{{5}}}\theta-{\cos{{7}}}\theta={0}\)
Use Sum-to-Product formulas in the above equation,
\(\displaystyle{\cos{{5}}}\theta-{\cos{{7}}}\theta={0}\)
\(\displaystyle-{2}{\sin{{\frac{{{5}\theta+{7}\theta}}{{{2}}}}}}{\sin{{\frac{{{5}\theta+{7}\theta}}{{{2}}}}}}={0}{\cos{{5}}}\theta-{\cos{{7}}}\theta={0}-{2}{\sin{{5}}}\theta+{7}\theta{2}{\sin{{5}}}\theta-{7}\theta{2}={0}-{2}{\sin{{6}}}\theta{\sin{{\left(-\theta\right)}}}={0}\)
\(\displaystyle-{2}{\sin{{6}}}\theta{\sin{{\left(-\theta\right)}}}={0}\)
Use the zero product property,
\(\displaystyle{\sin{{6}}}\theta={0}{\sin{{6}}}\theta={0}\ldots{\left({1}\right)}\ldots{\left({1}\right)}\)
\(\displaystyle{\sin{\theta}}={0}{\sin{\theta}}={0}\ldots{\left({2}\right)}\ldots{\left({2}\right)}\)
Consider equation (1).
\(\displaystyle{\sin{{6}}}\theta={0}{\sin{{6}}}\theta={0}\)
Taking sine inverse both sides,
\(\sin^{-1} \sin 6 \theta = \sin^{-1} (0)\)
\(\displaystyle{6}\theta={{\sin}^{{-{1}}}{\left({0}\right)}}{\sin{-}}{1}{\sin{{6}}}\theta={\sin{-}}{1}{\left({0}\right)}{6}\theta={\sin{-}}{1}{\left({0}\right)}={0},\pi\)
\(\displaystyle={0},\pi\)
The solution of the equation is obtained by adding in the integer multiples of \(\displaystyle\pi\),
\(\displaystyle{6}\theta={k}\pi\)
\(\displaystyle\theta={\frac{{{k}\pi}}{{{6}}}}{6}\theta={k}\pi\theta={k}\pi{6}\)
Consider equation (2).
\(\displaystyle{\sin{\theta}}={0}{\sin{\theta}}={0}\)
Taking \(\displaystyle{\sin{}}\) inverse both sides,
\(\displaystyle{{\sin}^{{-{1}}}{\sin{\theta}}}={{\sin}^{{-{1}}}{\left({0}\right)}}\)
\(\displaystyle\theta={{\sin}^{{-{1}}}{\left({0}\right)}}{\sin{-}}{1}{\sin{\theta}}={\sin{-}}{1}{\left({0}\right)}\theta={\sin{-}}{1}{\left({0}\right)}\theta=\pi\)
\(\displaystyle\theta=\pi\)
The solution of the equation is obtained by adding in the integer multiples of \(\displaystyle\pi\pi\),
\(\displaystyle\theta={k}\pi\theta={k}\pi\)
Therefore, the solutions of the trigonometry equation \(\displaystyle{\cos{{5}}}\theta-{\cos{{7}}}\theta={0}{\cos{{5}}}\theta-{\cos{{7}}}\theta={0}\) is \(\displaystyle\theta=\pi{k},{\frac{{\pi{k}}}{{{6}}}}\theta=\pi{k},\pi{k}{6}\).

Have a similar question?
Ask An Expert
43
 

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

asked 2021-08-11
Using Sum-to-Product Formulas Solve the equation by first using a Sum-to-Product Formula.
\(\displaystyle{\sin{\theta}}+{\sin{{3}}}\theta={0}\)
asked 2021-07-31
To solve:
The equation \(\displaystyle{4}{\cos{\theta}}{\sin{\theta}}+{3}{\cos{\theta}}={0}\)
asked 2021-08-11
To solve:
The equation \(\displaystyle{\cos{\theta}}{\left({2}{\sin{\theta}}+{1}\right)}={0}\)
asked 2021-08-10
Write the exprissuon as a sum and/or difference of logarithms.
\(\displaystyle{\log{{\left({\frac{{{7}\sqrt{{{x}+{3}}}}}{{{x}^{{{4}}}{\left({x}-{5}\right)}^{{{2}}}}}}\right)}}}{x}{>}{5}\)
asked 2021-08-06
Given \(\displaystyle{P}{\left({x}\right)}={3}{x}^{{{5}}}-{4}{x}^{{{4}}}+{9}{x}^{{{3}}}-{12}{x}^{{{2}}}-{12}{x}+{16}\), and that 2i is a zero, write P in factored form (as a product of linear factors). Be sure to write the full equation, including \(\displaystyle{P}{\left({x}\right)}=\).
asked 2021-08-02
Given \(\displaystyle{P}{\left({x}\right)}={3}{x}^{{{5}}}-{x}^{{{4}}}+{81}{x}^{{{3}}}-{27}{x}^{{{2}}}-{972}{x}+{324}\), and that 6i is a zero, write P in factored form (as a product of linear factors). Be sure to write the full equation, including \(\displaystyle{P}{\left({x}\right)}=\).
asked 2021-08-11
Given \(\displaystyle{P}{\left({x}\right)}={3}{x}^{{{5}}}−{5}{x}^{{{4}}}+{37}{x}^{{{3}}}−{83}{x}^{{{2}}}−{176}{x}−{48}\), and that 4i is a zero, write P in factored form (as a product of linear factors). Be sure to write the full equation, including \(\displaystyle{P}{\left({x}\right)}=\).
...