What is the general solution for cos(θ)-sin(θ)=cot(θ)cos(θ) ?

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

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Popular >

cos(θ)-sin(θ)=cot(θ)cos(θ)

Solution

cos (θ) - sin (θ) = cot (θ) cos (θ)

Solution

No Solution for θ \in R

Solution steps

cos (θ) - sin (θ) = cot (θ) cos (θ)

Subtract

cot (θ) cos (θ)

from both sides

cos (θ) - sin (θ) - cot (θ) cos (θ) = 0

Express with sin, cos

cos (θ) - sin (θ) - cos (θ) cot (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

cos (θ) \frac{cos ( θ )}{sin ( θ )}

Multiply fractions:

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

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

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

Apply exponent rule:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= cos^{2} (θ)

= \frac{cos ^{2} ( θ )}{sin ( θ )}

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

cos (θ) sin (θ) - sin (θ) sin (θ) - cos^{2} (θ)

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sin^{2} (θ)

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

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

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

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

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

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

Rewrite using trig identities

cos (θ) sin (θ) - 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 θ )}{2}

- 1 + \frac{sin ( 2 θ )}{2} = 0

Move

1

to the right side

- 1 + \frac{sin ( 2 θ )}{2} = 0

Add

1

to both sides

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

Simplify

\frac{sin ( 2 θ )}{2} = 1

\frac{sin ( 2 θ )}{2} = 1

Multiply both sides by

2

\frac{sin ( 2 θ )}{2} = 1

Multiply both sides by

2

\frac{2 sin ( 2 θ )}{2} = 1 \cdot 2

Simplify

sin (2 θ) = 2

sin (2 θ) = 2

- 1 \leq sin (x) \leq 1

No Solution for θ \in R

Popular Examples

-2tan(x)+1+2tan^2(x)=tan^2(x)

- 2 tan (x) + 1 + 2 tan^{2} (x) = tan^{2} (x)

(sin(a))/(50)=(sin(25))/(35)

\frac{sin ( a )}{50} = \frac{sin ( 2 5 ^{\circ} )}{35}

sin(x+pi/4)=\pm 1

sin (x + \frac{π}{4}) = \pm 1

sin(2x+64)=(sqrt(3))/2

sin (2 x + 64) = \frac{3}{2}

sin(θ)=0,6

sin (θ) = 0, 6

Frequently Asked Questions (FAQ)

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