Is (sin(4θ)+sin(2θ))/(cos(4θ)+cos(2θ))=tan(3θ) ?

The answer to whether (sin(4θ)+sin(2θ))/(cos(4θ)+cos(2θ))=tan(3θ) is True

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Popular >

prove (sin(4θ)+sin(2θ))/(cos(4θ)+cos(2θ))=tan(3θ)

Solution

prove

\frac{sin ( 4 θ ) + sin ( 2 θ )}{cos ( 4 θ ) + cos ( 2 θ )} = tan (3 θ)

Solution

True

Solution steps

\frac{sin ( 4 θ ) + sin ( 2 θ )}{cos ( 4 θ ) + cos ( 2 θ )} = tan (3 θ)

Manipulating left side

\frac{sin ( 4 θ ) + sin ( 2 θ )}{cos ( 4 θ ) + cos ( 2 θ )}

Rewrite using trig identities

\frac{sin ( 4 θ ) + sin ( 2 θ )}{cos ( 4 θ ) + cos ( 2 θ )}

Use the Sum to Product identity:

cos (s) + cos (t) = 2 cos (\frac{s + t}{2}) cos (\frac{s - t}{2})

= \frac{sin ( 4 θ ) + sin ( 2 θ )}{2 cos ( \frac{2 θ + 4 θ}{2} ) cos ( \frac{2 θ - 4 θ}{2} )}

Simplify

\frac{sin ( 4 θ ) + sin ( 2 θ )}{2 cos ( \frac{2 θ + 4 θ}{2} ) cos ( \frac{2 θ - 4 θ}{2} )} : \frac{sin ( 2 θ ) + sin ( 4 θ )}{2 cos ( θ ) cos ( 3 θ )}

\frac{sin ( 4 θ ) + sin ( 2 θ )}{2 cos ( \frac{2 θ + 4 θ}{2} ) cos ( \frac{2 θ - 4 θ}{2} )}

2 cos (\frac{2 θ + 4 θ}{2}) cos (\frac{2 θ - 4 θ}{2}) = 2 cos (\frac{6 θ}{2}) cos (- \frac{2 θ}{2})

2 cos (\frac{2 θ + 4 θ}{2}) cos (\frac{2 θ - 4 θ}{2})

Add similar elements:

2 θ + 4 θ = 6 θ

= 2 cos (\frac{6 θ}{2}) cos (\frac{2 θ - 4 θ}{2})

\frac{2 θ - 4 θ}{2} = - \frac{2 θ}{2}

\frac{2 θ - 4 θ}{2}

Add similar elements:

2 θ - 4 θ = - 2 θ

= \frac{- 2 θ}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2 θ}{2}

= 2 cos (\frac{6 θ}{2}) cos (- \frac{2 θ}{2})

= \frac{sin ( 4 θ ) + sin ( 2 θ )}{2 cos ( \frac{6 θ}{2} ) cos ( - \frac{2 θ}{2} )}

Use the negative angle identity:

cos (- x) = cos (x)

= \frac{sin ( 2 θ ) + sin ( 4 θ )}{2 cos ( \frac{2 θ}{2} ) cos ( \frac{6 θ}{2} )}

Divide the numbers:

\frac{2}{2} = 1

= \frac{sin ( 2 θ ) + sin ( 4 θ )}{2 cos ( θ ) cos ( \frac{6 θ}{2} )}

Divide the numbers:

\frac{6}{2} = 3

= \frac{sin ( 2 θ ) + sin ( 4 θ )}{2 cos ( θ ) cos ( 3 θ )}

= \frac{sin ( 2 θ ) + sin ( 4 θ )}{2 cos ( θ ) cos ( 3 θ )}

Use the Sum to Product identity:

sin (s) + sin (t) = 2 sin (\frac{s + t}{2}) cos (\frac{s - t}{2})

= \frac{2 sin ( \frac{2 θ + 4 θ}{2} ) cos ( \frac{2 θ - 4 θ}{2} )}{2 cos ( θ ) cos ( 3 θ )}

Simplify

\frac{2 sin ( \frac{2 θ + 4 θ}{2} ) cos ( \frac{2 θ - 4 θ}{2} )}{2 cos ( θ ) cos ( 3 θ )} : \frac{sin ( 3 θ )}{cos ( 3 θ )}

\frac{2 sin ( \frac{2 θ + 4 θ}{2} ) cos ( \frac{2 θ - 4 θ}{2} )}{2 cos ( θ ) cos ( 3 θ )}

2 sin (\frac{2 θ + 4 θ}{2}) cos (\frac{2 θ - 4 θ}{2}) = 2 sin (\frac{6 θ}{2}) cos (- \frac{2 θ}{2})

2 sin (\frac{2 θ + 4 θ}{2}) cos (\frac{2 θ - 4 θ}{2})

Add similar elements:

2 θ + 4 θ = 6 θ

= 2 sin (\frac{6 θ}{2}) cos (\frac{2 θ - 4 θ}{2})

\frac{2 θ - 4 θ}{2} = - \frac{2 θ}{2}

\frac{2 θ - 4 θ}{2}

Add similar elements:

2 θ - 4 θ = - 2 θ

= \frac{- 2 θ}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2 θ}{2}

= 2 sin (\frac{6 θ}{2}) cos (- \frac{2 θ}{2})

= \frac{2 sin ( \frac{6 θ}{2} ) cos ( - \frac{2 θ}{2} )}{2 cos ( θ ) cos ( 3 θ )}

Divide the numbers:

\frac{2}{2} = 1

= \frac{sin ( \frac{6 θ}{2} ) cos ( - \frac{2 θ}{2} )}{cos ( θ ) cos ( 3 θ )}

Use the negative angle identity:

cos (- x) = cos (x)

= \frac{cos ( \frac{2 θ}{2} ) sin ( \frac{6 θ}{2} )}{cos ( 3 θ ) cos ( θ )}

Divide the numbers:

\frac{2}{2} = 1

= \frac{cos ( θ ) sin ( \frac{6 θ}{2} )}{cos ( 3 θ ) cos ( θ )}

Divide the numbers:

\frac{6}{2} = 3

= \frac{cos ( θ ) sin ( 3 θ )}{cos ( 3 θ ) cos ( θ )}

Cancel the common factor:

cos (θ)

= \frac{sin ( 3 θ )}{cos ( 3 θ )}

= \frac{sin ( 3 θ )}{cos ( 3 θ )}

= \frac{sin ( 3 θ )}{cos ( 3 θ )}

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

= tan (3 θ)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is (sin(4θ)+sin(2θ))/(cos(4θ)+cos(2θ))=tan(3θ) ?
The answer to whether (sin(4θ)+sin(2θ))/(cos(4θ)+cos(2θ))=tan(3θ) is True