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Math  /  Trigonometry

Questioncos⁡(x−y)cos⁡(x+y)=1+tan⁡xtan⁡y1−tan⁡xtan⁡y\frac{\cos (x-y)}{\cos (x+y)}=\frac{1+\tan x \tan y}{1-\tan x \tan y}

Studdy Solution

STEP 1

What is this asking? We need to prove that an equation involving cosine, tangent, and the difference and sum of angles is true. Watch out! Remember your angle sum and difference formulas for cosine!
Also, don't forget how tangent is related to sine and cosine.

STEP 2

1. Expand the cosines
2. Rewrite using tangent
3. Simplify to the final form

STEP 3

Let's **expand** the numerator cos⁡(x−y)\cos(x-y) using the cosine angle subtraction formula: cos⁡(a−b)=cos⁡(a)cos⁡(b)+sin⁡(a)sin⁡(b)\cos(a-b) = \cos(a)\cos(b) + \sin(a)\sin(b).
So, cos⁡(x−y)=cos⁡(x)cos⁡(y)+sin⁡(x)sin⁡(y)\cos(x-y) = \cos(x)\cos(y) + \sin(x)\sin(y).
This is a great **first step** because it gets rid of the angle subtraction inside the cosine!

STEP 4

Now, let's **expand** the denominator cos⁡(x+y)\cos(x+y) using the cosine angle addition formula: cos⁡(a+b)=cos⁡(a)cos⁡(b)−sin⁡(a)sin⁡(b)\cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b).
Therefore, cos⁡(x+y)=cos⁡(x)cos⁡(y)−sin⁡(x)sin⁡(y)\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y).
We're making **great progress**!

STEP 5

With both the numerator and denominator expanded, we can rewrite the left side of the original equation as: cos⁡(x−y)cos⁡(x+y)=cos⁡(x)cos⁡(y)+sin⁡(x)sin⁡(y)cos⁡(x)cos⁡(y)−sin⁡(x)sin⁡(y). \frac{\cos(x-y)}{\cos(x+y)} = \frac{\cos(x)\cos(y) + \sin(x)\sin(y)}{\cos(x)\cos(y) - \sin(x)\sin(y)}. This looks much more manageable!

STEP 6

To introduce tangent into our expression, remember that tan⁡(θ)=sin⁡(θ)cos⁡(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}.
Let's **divide** both the numerator and the denominator of our fraction by cos⁡(x)cos⁡(y)\cos(x)\cos(y).
Remember, dividing both the top and bottom by the same value is the same as multiplying by one, so it doesn't change the value of the fraction!

STEP 7

After dividing, we get: cos⁡(x)cos⁡(y)cos⁡(x)cos⁡(y)+sin⁡(x)sin⁡(y)cos⁡(x)cos⁡(y)cos⁡(x)cos⁡(y)cos⁡(x)cos⁡(y)−sin⁡(x)sin⁡(y)cos⁡(x)cos⁡(y). \frac{\frac{\cos(x)\cos(y)}{\cos(x)\cos(y)} + \frac{\sin(x)\sin(y)}{\cos(x)\cos(y)}}{\frac{\cos(x)\cos(y)}{\cos(x)\cos(y)} - \frac{\sin(x)\sin(y)}{\cos(x)\cos(y)}}. Now, we can simplify by adding to zero and dividing to one: 1+sin⁡(x)cos⁡(x)⋅sin⁡(y)cos⁡(y)1−sin⁡(x)cos⁡(x)⋅sin⁡(y)cos⁡(y). \frac{1 + \frac{\sin(x)}{\cos(x)} \cdot \frac{\sin(y)}{\cos(y)}}{1 - \frac{\sin(x)}{\cos(x)} \cdot \frac{\sin(y)}{\cos(y)}}.

STEP 8

Finally, using the definition of tangent, tan⁡(θ)=sin⁡(θ)cos⁡(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}, we can **rewrite** the expression as: 1+tan⁡(x)tan⁡(y)1−tan⁡(x)tan⁡(y). \frac{1 + \tan(x)\tan(y)}{1 - \tan(x)\tan(y)}. This is exactly what we wanted to show!

STEP 9

We have successfully proven the identity: cos⁡(x−y)cos⁡(x+y)=1+tan⁡(x)tan⁡(y)1−tan⁡(x)tan⁡(y). \frac{\cos(x-y)}{\cos(x+y)} = \frac{1 + \tan(x)\tan(y)}{1 - \tan(x)\tan(y)}.

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