Is (sin(2θ))/(sin(θ))-(cos(2θ))/(cos(θ))=sec(θ) ?

The answer to whether (sin(2θ))/(sin(θ))-(cos(2θ))/(cos(θ))=sec(θ) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove (sin(2θ))/(sin(θ))-(cos(2θ))/(cos(θ))=sec(θ)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Simplify

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

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

Least Common Multiplier of

cos (θ), sin (θ) : cos (θ) sin (θ)

cos (θ), sin (θ)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

cos (θ)

sin (θ)

= cos (θ) sin (θ)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

cos (θ) sin (θ)

For

\frac{cos ( 2 θ )}{cos ( θ )} :

multiply the denominator and numerator by

sin (θ)

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

For

\frac{sin ( 2 θ )}{sin ( θ )} :

multiply the denominator and numerator by

cos (θ)

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

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

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

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

Rewrite using trig identities

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

Use the Angle Difference identity:

sin (s) cos (t) - cos (s) sin (t) = sin (s - t)

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

Simplify

\frac{sin ( 2 θ - θ )}{cos ( θ ) sin ( θ )} : \frac{1}{cos ( θ )}

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

Add similar elements:

2 θ - θ = θ

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

Cancel the common factor:

sin (θ)

= \frac{1}{cos ( θ )}

= \frac{1}{cos ( θ )}

= \frac{1}{cos ( θ )}

Rewrite using trig identities

Use the basic trigonometric identity:

cos (x) = \frac{1}{sec ( x )}

\frac{1}{\frac{1}{s e c ( θ )}}

Simplify

\frac{1}{\frac{1}{s e c ( θ )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{sec ( θ )}{1}

Apply rule

\frac{a}{1} = a

= sec (θ)

sec (θ)

sec (θ)

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

prove (sec(x))/(csc(x))+(sin(x))/(cos(x))=2tan(x)prove

\frac{sec ( x )}{csc ( x )} + \frac{sin ( x )}{cos ( x )} = 2 tan (x)

prove (sin(θ)+tan(θ))/(1+sec(θ))=sin(θ)prove

\frac{sin ( θ ) + tan ( θ )}{1 + sec ( θ )} = sin (θ)

prove sec(2θ)=(sec^2(θ))/(1-tan^2(θ))prove

sec (2 θ) = \frac{sec ^{2} ( θ )}{1 - tan ^{2} ( θ )}

prove cos(2a)=cos^2(a)-sin^2(a)prove

cos (2 a) = cos^{2} (a) - sin^{2} (a)

prove tan(θ)+cot(θ)=sec(θ)*csc(θ)prove

tan (θ) + cot (θ) = sec (θ) \cdot csc (θ)

Frequently Asked Questions (FAQ)

Is (sin(2θ))/(sin(θ))-(cos(2θ))/(cos(θ))=sec(θ) ?
The answer to whether (sin(2θ))/(sin(θ))-(cos(2θ))/(cos(θ))=sec(θ) is True