What is the general solution for sin(φ)-cos(φ)=sin(φ)tan(φ) ?

The general solution for sin(φ)-cos(φ)=sin(φ)tan(φ) is No Solution for φ\in\mathbb{R}

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Popular >

sin(φ)-cos(φ)=sin(φ)tan(φ)

Solution

sin (φ) - cos (φ) = sin (φ) tan (φ)

Solution

No Solution for φ \in R

Solution steps

sin (φ) - cos (φ) = sin (φ) tan (φ)

Subtract

sin (φ) tan (φ)

from both sides

sin (φ) - cos (φ) - sin (φ) tan (φ) = 0

Express with sin, cos

- cos (φ) + sin (φ) - sin (φ) tan (φ)

Use the basic trigonometric identity:

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

= - cos (φ) + sin (φ) - sin (φ) \frac{sin ( φ )}{cos ( φ )}

Simplify

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

- cos (φ) + sin (φ) - sin (φ) \frac{sin ( φ )}{cos ( φ )}

sin (φ) \frac{sin ( φ )}{cos ( φ )} = \frac{sin ^{2} ( φ )}{cos ( φ )}

sin (φ) \frac{sin ( φ )}{cos ( φ )}

Multiply fractions:

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

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

sin (φ) sin (φ) = sin^{2} (φ)

sin (φ) sin (φ)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

sin (φ) sin (φ) = sin^{1 + 1} (φ)

= sin^{1 + 1} (φ)

Add the numbers:

1 + 1 = 2

= sin^{2} (φ)

= \frac{sin ^{2} ( φ )}{cos ( φ )}

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

Convert element to fraction:

cos (φ) = \frac{cos ( φ ) cos ( φ )}{cos ( φ )}, sin (φ) = \frac{sin ( φ ) cos ( φ )}{cos ( φ )}

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

Since the denominators are equal, combine the fractions:

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

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

- cos (φ) cos (φ) + sin (φ) cos (φ) - sin^{2} (φ) = - cos^{2} (φ) + sin (φ) cos (φ) - sin^{2} (φ)

- cos (φ) cos (φ) + sin (φ) cos (φ) - sin^{2} (φ)

cos (φ) cos (φ) = cos^{2} (φ)

cos (φ) cos (φ)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos (φ) cos (φ) = cos^{1 + 1} (φ)

= cos^{1 + 1} (φ)

Add the numbers:

1 + 1 = 2

= cos^{2} (φ)

= - cos^{2} (φ) + sin (φ) cos (φ) - sin^{2} (φ)

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

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

\frac{- cos ^{2} ( φ ) - sin ^{2} ( φ ) + cos ( φ ) sin ( φ )}{cos ( φ )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

- cos^{2} (φ) - sin^{2} (φ) + cos (φ) sin (φ) = 0

Rewrite using trig identities

cos (φ) sin (φ) - 1

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

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

= - 1 + \frac{sin ( 2 φ )}{2}

- 1 + \frac{sin ( 2 φ )}{2} = 0

Move

1

to the right side

- 1 + \frac{sin ( 2 φ )}{2} = 0

Add

1

to both sides

- 1 + \frac{sin ( 2 φ )}{2} + 1 = 0 + 1

Simplify

\frac{sin ( 2 φ )}{2} = 1

\frac{sin ( 2 φ )}{2} = 1

Multiply both sides by

2

\frac{sin ( 2 φ )}{2} = 1

Multiply both sides by

2

\frac{2 sin ( 2 φ )}{2} = 1 \cdot 2

Simplify

sin (2 φ) = 2

sin (2 φ) = 2

- 1 \leq sin (x) \leq 1

No Solution for φ \in R

Popular Examples

cos(x)-cos(x/2)=0

cos (x) - cos (\frac{x}{2}) = 0

sin(x)=0,-pi/4 <x<(9pi)/4

sin (x) = 0, - \frac{π}{4} < x < \frac{9 π}{4}

sqrt(2)tan(x)cos(x)=tan(x)

2 tan (x) cos (x) = tan (x)

sin(x)= 5/4

sin (x) = \frac{5}{4}

sin(2θ)=3cos(θ)

sin (2 θ) = 3 cos (θ)

Frequently Asked Questions (FAQ)

What is the general solution for sin(φ)-cos(φ)=sin(φ)tan(φ) ?
The general solution for sin(φ)-cos(φ)=sin(φ)tan(φ) is No Solution for φ\in\mathbb{R}