Question

I need help with the following proofs.
a.) cos(nθ)=cos((n-2)θ)-2sinθsin((n-1)θ)
b.) (sin(x/2)+sin(x))/(cos(x/2)-cos(x))=cot(x/4)
c.) (sin∝-cos∝+1)/(sin∝+cos∝-1)=(sin∝+1)/cos∝
d.) tan(x/2)=(1-cos(x)+sin(x))/(1+cos(x)+sin(x))

Answer 1

Question#38948, Math, Trigonometry

I need help with the following proofs.

a.) $\cos(n\theta) = \cos((n-2)\theta) - 2\sin\theta \sin((n-1)\theta)$

b.) $(\sin(x/2) + \sin(x)) / (\cos(x/2) - \cos(x)) = \cot(x/4)$

c.) $(\sin \alpha - \cos \alpha + 1) / (\sin \alpha + \cos \alpha - 1) = (\sin \alpha + 1) / \cos \alpha$

d.) $\tan(x/2) = (1 - \cos(x) + \sin(x)) / (1 + \cos(x) + \sin(x))$

Solution

a.) $\cos(n\theta) = \cos((n-2)\theta) - 2\sin\theta \sin((n-1)\theta)$

First we rewrite the identity in the form

$$\cos(n\theta) - \cos((n-2)\theta) = -2\sin\theta \sin((n-1)\theta).$$

Then using

$$\cos A - \cos B = -2\sin\left(\frac{A - B}{2}\right)\sin\left(\frac{A + B}{2}\right),$$

we obtain

$$\begin{aligned} \cos(n\theta) - \cos((n-2)\theta) &= -2\sin\left(\frac{n\theta - n\theta + 2\theta}{2}\right)\sin\left(\frac{2n\theta - 2\theta}{2}\right) \\ &= -2\sin\theta \sin((n-1)\theta). \end{aligned}$$

Thus, we started with the left side and deduced the right side, so the proof is complete.

b.) $(\sin(x/2) + \sin(x)) / (\cos(x/2) - \cos(x)) = \cot(x/4)$

Using

$$\sin A + \sin B = 2\sin\left(\frac{A + B}{2}\right)\cos\left(\frac{A - B}{2}\right),$$

the identity (1) and cancelling we obtain

$$\frac{\sin \frac{x}{2} + \sin x}{\cos \frac{x}{2} - \cos x} = \frac{2\sin \frac{3x}{4} \cos\left(-\frac{x}{4}\right)}{-2\sin\left(-\frac{x}{4}\right)\sin \frac{3x}{4}} = \frac{\cos \frac{x}{4}}{\sin \frac{x}{4}} = \cot \frac{x}{4}.$$

We started with the left side and deduced the right side, so the proof is complete.

c.) $(\sin \alpha - \cos \alpha + 1) / (\sin \alpha + \cos \alpha - 1) = (\sin \alpha + 1) / \cos \alpha$

Because $\sin \alpha + \cos \alpha - 1 \neq 0$ and $\cos \alpha \neq 0$, multiplying the both sides of the identity by their product, we rewrite the identity in the form

$$(\sin \alpha - \cos \alpha + 1) \cos \alpha = (\sin \alpha + \cos \alpha - 1) \cdot (\sin \alpha + 1).$$

Then we start with the left side.

$$(\sin \alpha - \cos \alpha + 1) \cos \alpha = \sin \alpha \cos \alpha + \cos \alpha - \cos^ {2} \alpha .$$

Now we stop and work with the right side.

$$\begin{array}{l} (\sin \alpha + \cos \alpha - 1) \cdot (\sin \alpha + 1) = \sin^ {2} \alpha + \sin \alpha \cos \alpha + \cos \alpha - 1 \\ = \sin \alpha \cos \alpha + \cos \alpha - (1 - \sin^ {2} \alpha) \\ = \sin \alpha \cos \alpha + \cos \alpha - \cos^ {2} \alpha . \end{array}$$

We have used the Pythagorean identity.

We have deduced the same expression from each side, so the proof is complete.

d.) $\tan (x / 2) = (1 - \cos (x) + \sin (x)) / (1 + \cos (x) + \sin (x))$

We start with the right side. Because $x = 2 \cdot \frac{x}{2}$ from double-angle identities

$$\sin 2 x = 2 \sin x \cos x, \cos 2 x = 2 \cos^ {2} x - 1 \text{ and } \cos 2 x = 1 - 2 \sin^ {2} x$$

we easily get

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

Using these identities and cancelling we obtain

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

We started with the right side and deduced the left one, so the proof is complete.