证明这些三角恒等式。

  1. (a)

    1−cos⁡xsin⁡x+sin⁡x1−cos⁡x≡2sin⁡x\frac{1-\cos x}{\sin x} + \frac{\sin x}{1-\cos x} \equiv \frac{2}{\sin x}

  2. (b)

    sin⁡x+cos⁡x≡tan⁡x−cot⁡xsec⁡x−csc⁡x\sin x + \cos x \equiv \frac{\tan x - \cot x}{\sec x - \csc x}

  3. (c)

    12(1+sec⁡2x)≡cot⁡2xcot⁡2x−1\tfrac{1}{2}\left(1+\sec 2x\right) \equiv \dfrac{\cot^{2}x}{\cot^{2}x-1}

  4. (d)

    sec⁡2x≡tan⁡x+cot⁡xcot⁡x−tan⁡x\sec 2x \equiv \dfrac{\tan x + \cot x}{\cot x - \tan x}

  5. (e)

    sin⁡x−sin⁡2x+sin⁡3xcos⁡x−cos⁡2x+cos⁡3x≡tan⁡2x\dfrac{\sin x - \sin 2x + \sin 3x}{\cos x - \cos 2x + \cos 3x} \equiv \tan 2x

函数概念English
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完整解法

1(a)

1. Combine the terms on the Left Hand Side into a single fraction with a common denominator.

LHS=1−cos⁡xsin⁡x+sin⁡x1−cos⁡x\text{LHS} = \frac{1-\cos x}{\sin x} + \frac{\sin x}{1-\cos x}
=(1−cos⁡x)2+sin⁡2xsin⁡x(1−cos⁡x)= \frac{(1-\cos x)^2 + \sin^2 x}{\sin x(1-\cos x)}

2. Expand the squared binomial in the numerator.

=1−2cos⁡x+cos⁡2x+sin⁡2xsin⁡x(1−cos⁡x)= \frac{1 - 2\cos x + \cos^2 x + \sin^2 x}{\sin x(1-\cos x)}

3. Simplify the numerator using the Pythagorean identity.

Using the identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1:
=1−2cos⁡x+1sin⁡x(1−cos⁡x)= \frac{1 - 2\cos x + 1}{\sin x(1-\cos x)}
=2−2cos⁡xsin⁡x(1−cos⁡x)= \frac{2 - 2\cos x}{\sin x(1-\cos x)}

4. Factorise the numerator and cancel the common factor with the denominator to reach the RHS.

=2(1−cos⁡x)sin⁡x(1−cos⁡x)= \frac{2(1-\cos x)}{\sin x(1-\cos x)}
=2sin⁡x= \frac{2}{\sin x}
=RHS (proven)= \text{RHS \quad (proven)}

答案

\text{Proven}

1(b)

1. Express the right-hand side entirely in terms of $\sin x$ and $\cos x$.

RHS=tan⁡x−cot⁡xsec⁡x−csc⁡x=sin⁡xcos⁡x−cos⁡xsin⁡x1cos⁡x−1sin⁡x\text{RHS} = \frac{\tan x - \cot x}{\sec x - \csc x} = \frac{\frac{\sin x}{\cos x} - \frac{\cos x}{\sin x}}{\frac{1}{\cos x} - \frac{1}{\sin x}}

2. Combine the fractions in both the numerator and the denominator over a common denominator.

=sin⁡2x−cos⁡2xsin⁡xcos⁡xsin⁡x−cos⁡xsin⁡xcos⁡x= \frac{ \frac{\sin^2 x - \cos^2 x}{\sin x \cos x} }{ \frac{\sin x - \cos x}{\sin x \cos x} }

3. Simplify the complex fraction, factorise using the difference of two squares, and cancel the common factor.

=sin⁡2x−cos⁡2xsin⁡x−cos⁡x=(sin⁡x−cos⁡x)(sin⁡x+cos⁡x)sin⁡x−cos⁡x=sin⁡x+cos⁡x=LHS= \frac{\sin^2 x - \cos^2 x}{\sin x - \cos x} \\ = \frac{(\sin x - \cos x)(\sin x + \cos x)}{\sin x - \cos x} \\ = \sin x + \cos x \\ = \text{LHS}

答案

Proof complete.\text{Proof complete.}

1(c)

1. Express the left-hand side in terms of $\cos 2x$ as a single fraction.

LHS=12(1+sec⁡2x)=12(1+1cos⁡2x)=cos⁡2x+12cos⁡2x\text{LHS} = \frac{1}{2}(1 + \sec 2x) = \frac{1}{2} \left( 1 + \frac{1}{\cos 2x} \right) = \frac{\cos 2x + 1}{2\cos 2x}

2. Apply the double angle identity to the numerator to cancel the factor of 2.

=(2cos⁡2x−1)+12cos⁡2x=2cos⁡2x2cos⁡2x=cos⁡2xcos⁡2x= \frac{(2\cos^2 x - 1) + 1}{2\cos 2x} = \frac{2\cos^2 x}{2\cos 2x} = \frac{\cos^2 x}{\cos 2x}

3. Express the denominator using the double angle identity in terms of sine and cosine.

=cos⁡2xcos⁡2x−sin⁡2x= \frac{\cos^2 x}{\cos^2 x - \sin^2 x}

4. Divide the numerator and denominator by $\sin^2 x$ to obtain the right-hand side.

=cos⁡2xsin⁡2xcos⁡2xsin⁡2x−sin⁡2xsin⁡2x=cot⁡2xcot⁡2x−1=RHS\begin{aligned} &= \frac{\frac{\cos^2 x}{\sin^2 x}}{\frac{\cos^2 x}{\sin^2 x} - \frac{\sin^2 x}{\sin^2 x}} \\ &= \frac{\cot^2 x}{\cot^2 x - 1} \\ &= \text{RHS} \end{aligned}

答案

Proof completed.\text{Proof completed.}

1(d)

1. Express the right-hand side in terms of $\sin x$ and $\cos x$.

RHS=tan⁡x+cot⁡xcot⁡x−tan⁡x=sin⁡xcos⁡x+cos⁡xsin⁡xcos⁡xsin⁡x−sin⁡xcos⁡x\begin{align*} \text{RHS} &= \frac{\tan x + \cot x}{\cot x - \tan x} \\ &= \frac{\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}}{\frac{\cos x}{\sin x} - \frac{\sin x}{\cos x}} \end{align*}

2. Combine the fractions in the numerator and the denominator.

RHS=sin⁡2x+cos⁡2xsin⁡xcos⁡xcos⁡2x−sin⁡2xsin⁡xcos⁡x\begin{align*} \text{RHS} &= \frac{\frac{\sin^2 x + \cos^2 x}{\sin x \cos x}}{\frac{\cos^2 x - \sin^2 x}{\sin x \cos x}} \end{align*}

3. Simplify the complex fraction and apply the Pythagorean identity.

RHS=sin⁡2x+cos⁡2xcos⁡2x−sin⁡2x=1cos⁡2x−sin⁡2x\begin{align*} \text{RHS} &= \frac{\sin^2 x + \cos^2 x}{\cos^2 x - \sin^2 x} \\ &= \frac{1}{\cos^2 x - \sin^2 x} \end{align*}

4. Apply the double angle formula to reach the left-hand side.

RHS=1cos⁡2x=sec⁡2x=LHS\begin{align*} \text{RHS} &= \frac{1}{\cos 2x} \\ &= \sec 2x \\ &= \text{LHS} \end{align*}
答案

Proved\text{Proved}

1(e)

1. Group the terms in the numerator and the denominator to prepare for the sum-to-product formulae.

LHS=(sin⁡3x+sin⁡x)−sin⁡2x(cos⁡3x+cos⁡x)−cos⁡2x\text{LHS} = \frac{(\sin 3x + \sin x) - \sin 2x}{(\cos 3x + \cos x) - \cos 2x}

2. Apply the sum-to-product (factor) formulae to the grouped terms.

=2sin⁡(3x+x2)cos⁡(3x−x2)−sin⁡2x2cos⁡(3x+x2)cos⁡(3x−x2)−cos⁡2x= \frac{2\sin\left(\frac{3x+x}{2}\right)\cos\left(\frac{3x-x}{2}\right) - \sin 2x}{2\cos\left(\frac{3x+x}{2}\right)\cos\left(\frac{3x-x}{2}\right) - \cos 2x}
=2sin⁡2xcos⁡x−sin⁡2x2cos⁡2xcos⁡x−cos⁡2x= \frac{2\sin 2x \cos x - \sin 2x}{2\cos 2x \cos x - \cos 2x}

3. Factorise the common terms in the numerator and denominator, then simplify to obtain the RHS.

=sin⁡2x(2cos⁡x−1)cos⁡2x(2cos⁡x−1)= \frac{\sin 2x (2\cos x - 1)}{\cos 2x (2\cos x - 1)}
=sin⁡2xcos⁡2x= \frac{\sin 2x}{\cos 2x}
=tan⁡2x≡RHS= \tan 2x \equiv \text{RHS}

答案

\text{Proof completed}

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