SENSEI AI
About App

We need to prove two trigonometric identities. 1. $\frac{2 + \sin 2x - 2 \cos 2x}{1 + 3 \sin^2 x - \cos 2x} = \frac{2}{5} (2 + \frac{1}{\tan x})$ for $x \neq k \pi, k \in Z$.

TrigonometryTrigonometric IdentitiesDouble Angle FormulasHalf Angle FormulasTrigonometric Simplification
2025/4/14

1. Problem Description

We need to prove two trigonometric identities.

1. $\frac{2 + \sin 2x - 2 \cos 2x}{1 + 3 \sin^2 x - \cos 2x} = \frac{2}{5} (2 + \frac{1}{\tan x})$ for $x \neq k \pi, k \in Z$.

2. $\frac{1 + \cos x - \sin x}{1 - \cos x - \sin x} = - \frac{\cos \frac{x}{2}}{\sin \frac{x}{2}}$ for $x \neq \frac{\pi}{2} + 2 k \pi, k \in Z$.

2. Solution Steps

1. For the first identity, we start with the left-hand side (LHS):

2+sin⁡2x−2cos⁡2x1+3sin⁡2x−cos⁡2x\frac{2 + \sin 2x - 2 \cos 2x}{1 + 3 \sin^2 x - \cos 2x}
We use the double angle formulas:
sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2 \sin x \cos x
cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = \cos^2 x - \sin^2 x
Substituting these into the LHS, we get:
2+2sin⁡xcos⁡x−2(cos⁡2x−sin⁡2x)1+3sin⁡2x−(cos⁡2x−sin⁡2x)\frac{2 + 2 \sin x \cos x - 2 (\cos^2 x - \sin^2 x)}{1 + 3 \sin^2 x - (\cos^2 x - \sin^2 x)}
2+2sin⁡xcos⁡x−2cos⁡2x+2sin⁡2x1+3sin⁡2x−cos⁡2x+sin⁡2x\frac{2 + 2 \sin x \cos x - 2 \cos^2 x + 2 \sin^2 x}{1 + 3 \sin^2 x - \cos^2 x + \sin^2 x}
2(1+sin⁡xcos⁡x−cos⁡2x+sin⁡2x)1−cos⁡2x+4sin⁡2x\frac{2 (1 + \sin x \cos x - \cos^2 x + \sin^2 x)}{1 - \cos^2 x + 4 \sin^2 x}
2(1−cos⁡2x+sin⁡2x+sin⁡xcos⁡x)sin⁡2x+4sin⁡2x\frac{2 (1 - \cos^2 x + \sin^2 x + \sin x \cos x)}{\sin^2 x + 4 \sin^2 x}
2(sin⁡2x+sin⁡2x+sin⁡xcos⁡x)5sin⁡2x\frac{2 (\sin^2 x + \sin^2 x + \sin x \cos x)}{5 \sin^2 x}
2(2sin⁡2x+sin⁡xcos⁡x)5sin⁡2x\frac{2 (2 \sin^2 x + \sin x \cos x)}{5 \sin^2 x}
2sin⁡x(2sin⁡x+cos⁡x)5sin⁡2x\frac{2 \sin x (2 \sin x + \cos x)}{5 \sin^2 x}
2(2sin⁡x+cos⁡x)5sin⁡x\frac{2 (2 \sin x + \cos x)}{5 \sin x}
252sin⁡x+cos⁡xsin⁡x\frac{2}{5} \frac{2 \sin x + \cos x}{\sin x}
25(2+cos⁡xsin⁡x)\frac{2}{5} (2 + \frac{\cos x}{\sin x})
25(2+1tan⁡x)\frac{2}{5} (2 + \frac{1}{\tan x})
This is the right-hand side (RHS). Therefore, the first identity is proven.

2. For the second identity, we start with the LHS:

1+cos⁡x−sin⁡x1−cos⁡x−sin⁡x\frac{1 + \cos x - \sin x}{1 - \cos x - \sin x}
Multiply both numerator and denominator by 1+cos⁡x+sin⁡x1 + \cos x + \sin x:
(1+cos⁡x−sin⁡x)(1+cos⁡x+sin⁡x)(1−cos⁡x−sin⁡x)(1+cos⁡x+sin⁡x)\frac{(1 + \cos x - \sin x) (1 + \cos x + \sin x)}{(1 - \cos x - \sin x) (1 + \cos x + \sin x)}
(1+cos⁡x)2−sin⁡2x(1+sin⁡x)2−cos⁡2x\frac{(1 + \cos x)^2 - \sin^2 x}{(1 + \sin x)^2 - \cos^2 x}
1+2cos⁡x+cos⁡2x−sin⁡2x1+2sin⁡x+sin⁡2x−cos⁡2x\frac{1 + 2 \cos x + \cos^2 x - \sin^2 x}{1 + 2 \sin x + \sin^2 x - \cos^2 x}
1+2cos⁡x+cos⁡2x−(1−cos⁡2x)1+2sin⁡x+sin⁡2x−(1−sin⁡2x)\frac{1 + 2 \cos x + \cos^2 x - (1 - \cos^2 x)}{1 + 2 \sin x + \sin^2 x - (1 - \sin^2 x)}
2cos⁡x+2cos⁡2x2sin⁡x+2sin⁡2x\frac{2 \cos x + 2 \cos^2 x}{2 \sin x + 2 \sin^2 x}
2cos⁡x(1+cos⁡x)2sin⁡x(1+sin⁡x)\frac{2 \cos x (1 + \cos x)}{2 \sin x (1 + \sin x)}
cos⁡x(1+cos⁡x)sin⁡x(1+sin⁡x)\frac{\cos x (1 + \cos x)}{\sin x (1 + \sin x)}
Multiply both numerator and denominator by 1−sin⁡x1 - \sin x:
cos⁡x(1+cos⁡x)(1−sin⁡x)sin⁡x(1+sin⁡x)(1−sin⁡x)\frac{\cos x (1 + \cos x) (1 - \sin x)}{\sin x (1 + \sin x) (1 - \sin x)}
However, this approach appears to be more complicated. Instead, use the half-angle formulas:
cos⁡x=2cos⁡2x2−1=1−2sin⁡2x2\cos x = 2 \cos^2 \frac{x}{2} - 1 = 1 - 2 \sin^2 \frac{x}{2}
sin⁡x=2sin⁡x2cos⁡x2\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}
Substituting these into LHS:
1+1−2sin⁡2x2−2sin⁡x2cos⁡x21−(1−2sin⁡2x2)−2sin⁡x2cos⁡x2\frac{1 + 1 - 2 \sin^2 \frac{x}{2} - 2 \sin \frac{x}{2} \cos \frac{x}{2}}{1 - (1 - 2 \sin^2 \frac{x}{2}) - 2 \sin \frac{x}{2} \cos \frac{x}{2}}
2−2sin⁡2x2−2sin⁡x2cos⁡x22sin⁡2x2−2sin⁡x2cos⁡x2\frac{2 - 2 \sin^2 \frac{x}{2} - 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin^2 \frac{x}{2} - 2 \sin \frac{x}{2} \cos \frac{x}{2}}
2(1−sin⁡2x2−sin⁡x2cos⁡x2)2(sin⁡2x2−sin⁡x2cos⁡x2)\frac{2 (1 - \sin^2 \frac{x}{2} - \sin \frac{x}{2} \cos \frac{x}{2})}{2 (\sin^2 \frac{x}{2} - \sin \frac{x}{2} \cos \frac{x}{2})}
cos⁡2x2−sin⁡x2cos⁡x2sin⁡2x2−sin⁡x2cos⁡x2\frac{\cos^2 \frac{x}{2} - \sin \frac{x}{2} \cos \frac{x}{2}}{\sin^2 \frac{x}{2} - \sin \frac{x}{2} \cos \frac{x}{2}}
cos⁡x2(cos⁡x2−sin⁡x2)sin⁡x2(sin⁡x2−cos⁡x2)\frac{\cos \frac{x}{2} (\cos \frac{x}{2} - \sin \frac{x}{2})}{\sin \frac{x}{2} (\sin \frac{x}{2} - \cos \frac{x}{2})}
cos⁡x2(cos⁡x2−sin⁡x2)sin⁡x2(−(cos⁡x2−sin⁡x2))\frac{\cos \frac{x}{2} (\cos \frac{x}{2} - \sin \frac{x}{2})}{\sin \frac{x}{2} (- (\cos \frac{x}{2} - \sin \frac{x}{2}))}
−cos⁡x2sin⁡x2- \frac{\cos \frac{x}{2}}{\sin \frac{x}{2}}
This is the RHS. Therefore, the second identity is proven.

3. Final Answer

1. $\frac{2 + \sin 2x - 2 \cos 2x}{1 + 3 \sin^2 x - \cos 2x} = \frac{2}{5} (2 + \frac{1}{\tan x})$

2. $\frac{1 + \cos x - \sin x}{1 - \cos x - \sin x} = - \frac{\cos \frac{x}{2}}{\sin \frac{x}{2}}$

Related problems in "Trigonometry"

We are asked to simplify the expression $\frac{2 \tan 60^{\circ} + \cos 30^{\circ}}{\sin 60^{\circ}}...

TrigonometryTrigonometric IdentitiesSimplification
2025/4/21

The problem asks us to evaluate several trigonometric expressions: a. $sin(390^\circ)$ b. $cos(\frac...

TrigonometryTrigonometric FunctionsSineCosineTangentAngle ReductionUnit Circle
2025/4/19

We are asked to prove the trigonometric identity $\sin^4 A - \cos^4 A = 2\sin^2 A - 1$.

Trigonometric IdentitiesPythagorean IdentityDifference of SquaresAlgebraic Manipulation
2025/4/15

We are given the equation $\cos(4x) = \sin(x)$ and we need to solve it in the interval $[-\frac{\pi}...

Trigonometric EquationsTrigonometric IdentitiesSolving EquationsIntervalSineCosineRoots
2025/4/14

The problem asks us to solve several trigonometric equations in $\mathbb{R}$ and represent the solut...

Trigonometric EquationsUnit CircleTrigonometric IdentitiesSolving Equations
2025/4/14

We are asked to solve the following trigonometric equations in $R$ and represent the solutions on th...

Trigonometric EquationsUnit CircleTrigonometric IdentitiesSolutions
2025/4/14

The problem is divided into three parts: 1. Show that $cos(5x) = cos(x)(16cos^4(x) - 20cos^2(x) + 5)...

Trigonometric IdentitiesMultiple Angle FormulasTrigonometric EquationsSolving Equations
2025/4/14

The problem asks to prove that $16 \sin(\frac{\pi}{24}) \sin(\frac{5\pi}{24}) \sin(\frac{7\pi}{24}) ...

Trigonometric IdentitiesTrigonometric FunctionsProduct-to-Sum FormulasAngle Sum and Difference Identities
2025/4/14

Exercise 22 is asking us to solve a series of problems. First, we need to show that $\sin(5x) = 16\s...

Trigonometric IdentitiesTrigonometric EquationsSine FunctionRoots of Equations
2025/4/14

The problem asks us to compute the value of $A = \cos(\frac{\pi}{12})\cos(\frac{5\pi}{12}) + \sin(\f...

Trigonometric IdentitiesAngle Sum and Difference FormulasTrigonometric ValuesCosineSine
2025/4/14